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Chapter 5 Thermodynamics
Thermodynamics is the branch of science that deals with the study of energy transformations. It specifically focuses on energy changes in macroscopic systems (those involving a large number of particles) rather than microscopic ones. Thermodynamics is concerned with the initial and final states of a system undergoing a change, not the pathway or rate of the change. Its laws apply when a system is in equilibrium or moves between equilibrium states. Macroscopic properties like pressure and temperature are constant for a system in equilibrium.
Key questions addressed by thermodynamics in chemistry include: how to determine energy changes in reactions, whether a reaction will occur spontaneously, what drives a spontaneous process, and to what extent a reaction will proceed.
Forms Of Energy And Their Interrelation
Molecules store chemical energy, which can be released during reactions (e.g., combustion). Energy can be transformed between various forms, such as chemical, thermal (heat), mechanical (work), or electrical energy.
Thermodynamic Terms
To study energy changes in chemical systems, it's essential to define some fundamental thermodynamic terms.
The System And The Surroundings
The system is the specific part of the universe being studied, where observations are made. The surroundings are everything else in the universe outside the system. Together, the system and surroundings constitute the universe (Universe = System + Surroundings).
For practical purposes, the surroundings usually refer to the region near the system that can interact with it. The system is separated from the surroundings by a boundary, which can be real (like the walls of a container) or imaginary, allowing us to track matter and energy exchange.
Types Of The System
Systems are classified based on the exchange of matter and energy with their surroundings:
- Open System: Exchanges both matter and energy with the surroundings. Example: Reactants in an open beaker.
- Closed System: Exchanges energy but not matter with the surroundings. Example: Reactants in a sealed container made of a conducting material (like metal).
- Isolated System: Exchanges neither matter nor energy with the surroundings. Example: Reactants in a sealed, insulated container (like a thermos flask).
The State Of The System
The state of a thermodynamic system is defined by its measurable or macroscopic properties, such as pressure ($p$), volume ($V$), temperature ($T$), and composition (amount of substance, $n$). These properties are called state variables or state functions because their values depend only on the current state of the system, not on the path taken to reach that state. Once a minimum number of independent state variables are fixed, the values of other properties are also fixed.
The Internal Energy As A State Function
The total energy of a system, encompassing all forms of energy (chemical, kinetic, potential, etc.), is called its Internal Energy (U). The internal energy is a state function, meaning its value depends only on the system's state and not on the process that brought it to that state. The internal energy of a system can change through the transfer of heat, work done on or by the system, or the transfer of matter.
Work
One way to change the internal energy of a system is by doing work ($w$) on or by the system. An adiabatic process is one where no heat exchange occurs with the surroundings. For an adiabatic process, the change in internal energy ($\Delta U$) is equal to the adiabatic work ($w_{ad}$) done on the system: $\Delta U = U_{final} - U_{initial} = w_{ad}$. J.P. Joule's experiments showed that the same amount of work, regardless of the method, produced the same change in the state of the system (measured by temperature change), confirming internal energy as a state function. By IUPAC convention, $w$ is positive when work is done on the system (internal energy increases) and negative when work is done by the system (internal energy decreases).
Heat
Another way to change internal energy is by transferring heat ($q$) across the system boundary due to a temperature difference between the system and surroundings. This occurs through thermally conducting walls. For a system at constant volume where no work is done, the change in internal energy ($\Delta U$) is equal to the heat ($q_V$) transferred: $\Delta U = q_V$. By IUPAC convention, $q$ is positive when heat is transferred to the system (internal energy increases) and negative when heat is transferred from the system (internal energy decreases).
The General Case
When the state of a system changes due to both heat transfer and work, the total change in internal energy is given by the sum of heat and work:
$$ \Delta U = q + w $$This equation is the mathematical statement of the First Law of Thermodynamics, which states that the energy of an isolated system is constant. It is a restatement of the law of conservation of energy: energy cannot be created or destroyed, only transferred or transformed.
For a given change in state, the values of $q$ and $w$ can vary depending on the process (path), but their sum ($\Delta U$) is constant and path-independent because $U$ is a state function.
Example 1. Express the change in internal energy of a system when
(i) No heat is absorbed by the system from the surroundings, but work (w) is done on the system. What type of wall does the system have ?
(ii) No work is done on the system, but q amount of heat is taken out from the system and given to the surroundings. What type of wall does the system have?
(iii) w amount of work is done by the system and q amount of heat is supplied to the system. What type of system would it be?
Answer:
The First Law of Thermodynamics is $\Delta U = q + w$.
(i) No heat absorbed means $q=0$. Work $w$ is done on the system, so $w$ is positive. $\Delta U = 0 + w = w$. Since $q=0$, the process is adiabatic. The wall is adiabatic.
(ii) No work done means $w=0$. Heat $q$ is taken out from the system, so $q$ is negative, let's denote it as $-q_{out}$. $\Delta U = -q_{out} + 0 = -q_{out}$. Since heat transfer occurs and no work is done, the process happens at constant volume in a container with thermally conducting walls. The wall is thermally conducting.
(iii) Work $w$ is done by the system, so $w$ is negative, let's denote it as $-w_{by}$. Heat $q$ is supplied to the system, so $q$ is positive. $\Delta U = q + (-w_{by}) = q - w_{by}$. The system exchanges both heat and work but doesn't mention mass exchange. Assuming no mass exchange, it is a closed system.
Applications
Chemical reactions often involve energy changes in the form of heat and work (especially if gases are produced or consumed). Thermodynamics helps quantify these changes.
Work
In chemical thermodynamics, the most common type of work is pressure-volume (PV) work, which occurs when a system expands or contracts against an external pressure. For a gas expanding or being compressed in a cylinder with a piston:
Work done on the system ($w$) = $-p_{ex} \Delta V = -p_{ex} (V_f - V_i)$, where $p_{ex}$ is the external pressure, $V_i$ is the initial volume, and $V_f$ is the final volume.
According to the sign convention, compression ($V_f < V_i$, $\Delta V$ is negative) results in positive work (work done on the system). Expansion ($V_f > V_i$, $\Delta V$ is positive) results in negative work (work done by the system).
If the process involves multiple steps or occurs reversibly (where the external pressure is always infinitesimally close to the internal pressure, $p_{ex} \approx p_{in}$), the work must be calculated by integration:
$$ w = -\int_{V_i}^{V_f} p_{ex} dV $$For a reversible process of an ideal gas at constant temperature (isothermal): $p_{ex} = p_{in} = \frac{nRT}{V}$.
$$ w_{rev} = -\int_{V_i}^{V_f} \frac{nRT}{V} dV = -nRT \int_{V_i}^{V_f} \frac{1}{V} dV = -nRT [\ln V]_{V_i}^{V_f} = -nRT \ln \left(\frac{V_f}{V_i}\right) $$ $$ w_{rev} = -2.303 nRT \log \left(\frac{V_f}{V_i}\right) $$Free expansion is expansion against zero external pressure ($p_{ex}=0$), which occurs in a vacuum. In free expansion, $w=0$. For an ideal gas, free expansion is also isothermal ($\Delta U=0$).
Reversible Process: A process that can be reversed at any point by an infinitesimal change in conditions, proceeding infinitely slowly through a series of equilibrium states.
Irreversible Process: Any process that is not reversible, typically occurring rapidly or in a single step against a constant external pressure.
Example 2. Two litres of an ideal gas at a pressure of 10 atm expands isothermally at 25 °C into a vacuum until its total volume is 10 litres. How much heat is absorbed and how much work is done in the expansion ?
Answer:
Initial volume ($V_i$) = 2 L, Initial pressure = 10 atm, Temperature = 25°C (isothermal expansion).
Final volume ($V_f$) = 10 L.
Expansion into a vacuum means the external pressure ($p_{ex}$) is 0.
Work done ($w$) = $-p_{ex} \Delta V = -p_{ex} (V_f - V_i) = -0 \times (10 \text{ L} - 2 \text{ L}) = 0$ J.
Since the gas is ideal and the expansion is isothermal, the change in internal energy ($\Delta U$) is 0 for an ideal gas under isothermal conditions.
According to the First Law, $\Delta U = q + w$.
$0 = q + 0$, so $q = 0$ J.
No work is done, and no heat is absorbed (or released).
Example 3. Consider the same expansion, but this time against a constant external pressure of 1 atm.
Answer:
Initial volume ($V_i$) = 2 L, Final volume ($V_f$) = 10 L. Constant external pressure ($p_{ex}$) = 1 atm. Isothermal expansion at 25°C.
Work done ($w$) = $-p_{ex} \Delta V = -p_{ex} (V_f - V_i) = -1 \text{ atm} \times (10 \text{ L} - 2 \text{ L}) = -1 \text{ atm} \times 8 \text{ L} = -8$ L atm.
To convert L atm to Joules (1 L atm $\approx 101.3$ J): $w = -8 \text{ L atm} \times 101.3 \text{ J/L atm} \approx -810.4$ J.
Since the process is isothermal expansion of an ideal gas, $\Delta U = 0$.
According to the First Law, $\Delta U = q + w$.
$0 = q + (-8 \text{ L atm})$, so $q = +8$ L atm.
Heat absorbed ($q$) is 8 L atm (or approximately 810.4 J).
Example 4. Consider the expansion given in problem 5.2, for 1 mol of an ideal gas conducted reversibly.
Answer:
Initial volume ($V_i$) = 2 L, Final volume ($V_f$) = 10 L. 1 mol of ideal gas ($n=1$). Temperature = 25°C = 298 K. Reversible, isothermal expansion.
Work done ($w_{rev}$) = $-nRT \ln \left(\frac{V_f}{V_i}\right) = -2.303 nRT \log \left(\frac{V_f}{V_i}\right)$
Using $R = 0.08206$ L atm K⁻¹ mol⁻¹ or $R = 8.314$ J K⁻¹ mol⁻¹ (Let's use L atm first as in the example, though J is SI unit):
$w_{rev} = -2.303 \times 1 \text{ mol} \times 0.08206 \text{ L atm K}^{-1} \text{ mol}^{-1} \times 298 \text{ K} \times \log \left(\frac{10 \text{ L}}{2 \text{ L}}\right)$
$w_{rev} = -2.303 \times 0.08206 \times 298 \times \log(5)$ L atm
$w_{rev} = -2.303 \times 0.08206 \times 298 \times 0.6990$ L atm
$w_{rev} \approx -39.366$ L atm.
Converting to Joules: $w_{rev} \approx -39.366 \text{ L atm} \times 101.3 \text{ J/L atm} \approx -3988$ J.
Since the process is isothermal expansion of an ideal gas, $\Delta U = 0$.
According to the First Law, $\Delta U = q + w$.
$0 = q + (-39.366 \text{ L atm})$, so $q = +39.366$ L atm.
Heat absorbed ($q$) is 39.366 L atm (or approximately 3988 J).
Note: The work done in reversible expansion is the maximum possible work done by the system for a given volume change, and the heat absorbed is also the maximum possible heat absorbed for an isothermal process.
Enthalpy, H
Most chemical reactions are carried out at constant atmospheric pressure, not constant volume. To quantify heat changes under constant pressure conditions, a new state function called Enthalpy (H) is defined.
A Useful New State Function
Enthalpy (H) is defined as the sum of internal energy (U) and the product of pressure (p) and volume (V):
$$ H = U + pV $$Since U, p, and V are state functions, H is also a state function. For a process occurring at constant pressure ($\Delta p = 0$), the change in enthalpy ($\Delta H$) is given by:
$$ \Delta H = \Delta U + p\Delta V $$Combining this with the First Law ($\Delta U = q + w$) and the definition of PV work ($w = -p\Delta V$) when only PV work is involved:
$\Delta H = (q + w) + p\Delta V = (q - p\Delta V) + p\Delta V = q$
So, for a process at constant pressure, the heat absorbed ($q_p$) is equal to the change in enthalpy:
$$ \Delta H = q_p $$This is a very useful relationship in chemistry. If heat is absorbed by the system at constant pressure (endothermic reaction), $\Delta H$ is positive. If heat is released by the system (exothermic reaction), $\Delta H$ is negative. For reactions involving only solids or liquids, volume changes are usually small, so $p\Delta V$ is negligible, and $\Delta H \approx \Delta U$. The difference is significant when gases are involved.
For reactions involving gases at constant temperature and pressure, the volume change ($\Delta V$) is related to the change in the number of moles of gas ($\Delta n_g$) by the ideal gas law: $p\Delta V = \Delta n_g RT$. Here, $\Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})$.
Substituting this into the enthalpy equation:
$$ \Delta H = \Delta U + \Delta n_g RT $$This equation allows interconversion between $\Delta H$ and $\Delta U$ for reactions involving gases.
Example 5. If water vapour is assumed to be a perfect gas, molar enthalpy change for vapourisation of 1 mol of water at 1bar and 100°C is 41kJ mol⁻¹. Calculate the internal energy change, when 1 mol of water is vapourised at 1 bar pressure and 100°C.
Answer:
Process: H₂O (l) $\rightarrow$ H₂O (g)
Given: $\Delta H = 41$ kJ mol⁻¹ for 1 mol at constant pressure (1 bar) and temperature (100°C).
Temperature ($T$) = 100°C = 273.15 + 100 = 373.15 K. Pressure ($p$) = 1 bar.
We need to calculate $\Delta U$. The relationship is $\Delta H = \Delta U + \Delta n_g RT$. So, $\Delta U = \Delta H - \Delta n_g RT$.
For the reaction H₂O (l) $\rightarrow$ H₂O (g):
Number of moles of gaseous products ($n_{products, gas}$) = 1 (from 1 mol H₂O(g))
Number of moles of gaseous reactants ($n_{reactants, gas}$) = 0 (from 1 mol H₂O(l))
Change in number of moles of gas ($\Delta n_g$) = $1 - 0 = 1$ mol.
Gas constant ($R$) = 8.314 J K⁻¹ mol⁻¹ = 0.008314 kJ K⁻¹ mol⁻¹.
$\Delta U = 41 \text{ kJ mol}^{-1} - (1 \text{ mol}) \times (0.008314 \text{ kJ K}^{-1} \text{ mol}^{-1}) \times (373.15 \text{ K})$
$\Delta U = 41 \text{ kJ mol}^{-1} - 3.102 \text{ kJ mol}^{-1}$
$\Delta U = 37.898 \text{ kJ mol}^{-1}$.
The internal energy change for the vaporisation is approximately 37.9 kJ mol⁻¹.
Extensive And Intensive Properties
Thermodynamic properties can be classified based on their dependence on the amount of substance:
- Extensive Properties: Properties that depend on the quantity or size of matter in the system. Examples: mass, volume, internal energy (U), enthalpy (H), heat capacity (C). If you double the amount of substance, the value of the extensive property typically doubles.
- Intensive Properties: Properties that do not depend on the quantity or size of matter. Examples: temperature (T), pressure (p), density, concentration, molar volume ($V_m$, volume per mole), molar heat capacity ($C_m$, heat capacity per mole). These properties are characteristic of the substance itself, regardless of how much is present.
Heat Capacity
When heat is transferred to a system, its temperature usually changes. The increase in temperature ($\Delta T$) is proportional to the heat transferred ($q$). The proportionality constant is the heat capacity (C):
$$ q = C \Delta T $$Heat capacity is an extensive property; it depends on the amount of substance and its composition. A larger heat capacity means more heat is needed to cause a given temperature rise.
- Molar heat capacity ($C_m$): Heat capacity per mole of substance ($C_m = C/n$). This is an intensive property, representing the heat needed to raise the temperature of one mole by one degree Celsius (or Kelvin).
- Specific heat (or specific heat capacity, $c$): Heat capacity per unit mass of substance ($c = C/m$). This is also an intensive property, representing the heat needed to raise the temperature of one gram by one degree Celsius (or Kelvin). $q = c \times m \times \Delta T$.
The Relationship Between Cp And CV For An Ideal Gas
Heat capacity can be measured under constant volume ($C_V$) or constant pressure ($C_p$) conditions.
- At constant volume: $\Delta U = q_V = C_V \Delta T$.
- At constant pressure: $\Delta H = q_p = C_p \Delta T$.
For an ideal gas, the relationship between $C_p$ and $C_V$ is:
$$ C_p - C_V = R $$where R is the ideal gas constant. This difference arises because, at constant pressure, some of the heat added is used to do expansion work, whereas at constant volume, all the heat added increases the internal energy (and temperature).
Measurement Of ΔU And ΔH: Calorimetry
Calorimetry is the experimental technique used to measure the heat changes ($q$) associated with chemical or physical processes. A calorimeter is a device where the process is carried out, typically immersed in a liquid of known heat capacity. By measuring the temperature change of the liquid and the calorimeter, the heat absorbed or evolved can be determined using the heat capacity formula ($q = C \Delta T$). Measurements are usually done under constant volume or constant pressure conditions.
ΔU Measurements
Heat changes at constant volume ($q_V = \Delta U$) are measured using a bomb calorimeter. This consists of a sealed steel vessel (the "bomb") where the reaction (often combustion) takes place, immersed in a water bath within an insulated container (the calorimeter). Since the bomb is sealed, the volume is constant ($\Delta V = 0$), so no PV work is done ($w=0$). The heat released or absorbed by the reaction changes the temperature of the bomb and the surrounding water. By knowing the total heat capacity of the calorimeter (bomb + water), the heat of reaction at constant volume ($q_V = \Delta U$) is calculated from the measured temperature change.
ΔH Measurements
Heat changes at constant pressure ($q_p = \Delta H$) are measured using a different type of calorimeter, often a simple insulated container open to the atmosphere (or maintained at constant pressure). This is suitable for reactions in solution or those not involving significant gas volume changes at constant pressure. The heat absorbed or evolved ($q_p = \Delta H$) is calculated from the temperature change of the solution and the calorimeter, using their heat capacities.
For an exothermic reaction, the system releases heat, $q_p$ is negative, and $\Delta H$ is negative. For an endothermic reaction, the system absorbs heat, $q_p$ is positive, and $\Delta H$ is positive.
Example 6. 1g of graphite is burnt in a bomb calorimeter in excess of oxygen at 298 K and 1 atmospheric pressure according to the equation C (graphite) + O₂(g) → CO₂(g). During the reaction, temperature rises from 298 K to 299 K. If the heat capacity of the bomb calorimeter is 20.7kJ/K, what is the enthalpy change for the above reaction at 298 K and 1 atm?
Answer:
Reaction: C (graphite) + O₂(g) → CO₂(g)
This reaction is carried out in a bomb calorimeter, which measures heat at constant volume ($q_V = \Delta U$).
Mass of graphite burnt = 1 g.
Initial temperature = 298 K, Final temperature = 299 K. $\Delta T = 299 \text{ K} - 298 \text{ K} = 1 \text{ K}$.
Heat capacity of the bomb calorimeter ($C_V$) = 20.7 kJ/K.
The heat absorbed by the calorimeter is $q_{calorimeter} = C_V \times \Delta T = 20.7 \text{ kJ/K} \times 1 \text{ K} = 20.7$ kJ.
The heat released by the reaction ($q_V$) is equal in magnitude but opposite in sign to the heat absorbed by the calorimeter: $q_V = -20.7$ kJ.
At constant volume, $\Delta U = q_V$. So, $\Delta U$ for burning 1 g of graphite is $-20.7$ kJ.
We need to find the enthalpy change ($\Delta H$) for the combustion of 1 mole of graphite. The molar mass of graphite (Carbon) is 12.01 g/mol (approximately 12 g/mol in many contexts, consistent with the problem). Let's use 12 g/mol for calculation as implied by the result's order of magnitude.
Moles of graphite burnt = $\frac{1 \text{ g}}{12 \text{ g/mol}} = \frac{1}{12}$ mol.
$\Delta U$ per mole = $\frac{\Delta U \text{ for } 1 \text{ g}}{\text{Moles of graphite}} = \frac{-20.7 \text{ kJ}}{1/12 \text{ mol}} = -20.7 \times 12 \text{ kJ/mol} = -248.4$ kJ mol⁻¹.
So, $\Delta U$ for the combustion of 1 mole of graphite is $-248.4$ kJ mol⁻¹.
Now, we relate $\Delta H$ and $\Delta U$ using $\Delta H = \Delta U + \Delta n_g RT$.
For the reaction C (graphite) + O₂(g) → CO₂(g):
- Gaseous reactants: O₂ (1 mol). $n_{reactants, gas} = 1$.
- Gaseous products: CO₂ (1 mol). $n_{products, gas} = 1$.
- $\Delta n_g = n_{products, gas} - n_{reactants, gas} = 1 - 1 = 0$.
$\Delta H = \Delta U + 0 \times RT = \Delta U$.
Therefore, the enthalpy change for the combustion of graphite is equal to the internal energy change.
$\Delta H = -248.4$ kJ mol⁻¹.
The enthalpy change for the reaction C (graphite) + O₂(g) → CO₂(g) at 298 K and 1 atm is –248.4 kJ mol⁻¹.
(Note: The example in the prompt uses 248 x 10² kJ/mol, which seems like a typo and should likely be 2.48 x 10² kJ/mol or 248 kJ/mol. Our calculated value matches this corrected magnitude).
Enthalpy Change, ΔrH Of A Reaction – Reaction Enthalpy
The enthalpy change of a chemical reaction, denoted as $\Delta_r H$, is the difference between the sum of the enthalpies of the products and the sum of the enthalpies of the reactants, considering their stoichiometry.
For a general reaction: $aA + bB \rightarrow cC + dD$
$$ \Delta_r H = \sum (\text{enthalpies of products}) - \sum (\text{enthalpies of reactants}) $$ $$ \Delta_r H = [c H_m(C) + d H_m(D)] - [a H_m(A) + b H_m(B)] $$where $H_m$ is the molar enthalpy of the substance, and $a, b, c, d$ are the stoichiometric coefficients.
$\Delta_r H$ is negative for exothermic reactions (heat released) and positive for endothermic reactions (heat absorbed).
Standard Enthalpy Of Reactions
The standard enthalpy of reaction ($\Delta_r H^\ominus$) is the enthalpy change for a reaction carried out under standard conditions. The standard state of a substance at a specified temperature (usually 298 K) is its pure form at 1 bar pressure. Standard state is indicated by the superscript $\ominus$ (e.g., $\Delta H^\ominus$).
Enthalpy Changes During Phase Transformations
Changes in the physical state of a substance (phase transformations) also involve enthalpy changes. These typically occur at constant temperature and pressure.
- Standard enthalpy of fusion ($\Delta_{fus} H^\ominus$): Enthalpy change when one mole of a solid melts at its melting point under standard pressure. Melting is endothermic, so $\Delta_{fus} H^\ominus$ is always positive. (e.g., H₂O(s) $\rightarrow$ H₂O(l); $\Delta_{fus} H^\ominus = +6.00$ kJ mol⁻¹ at 273 K). Freezing is the reverse process with $\Delta_{freeze} H^\ominus = -\Delta_{fus} H^\ominus$.
- Standard enthalpy of vaporization ($\Delta_{vap} H^\ominus$): Enthalpy change when one mole of a liquid vaporizes at its boiling point under standard pressure. Vaporization is endothermic, so $\Delta_{vap} H^\ominus$ is always positive. (e.g., H₂O(l) $\rightarrow$ H₂O(g); $\Delta_{vap} H^\ominus = +40.79$ kJ mol⁻¹ at 373 K). Condensation is the reverse process with $\Delta_{cond} H^\ominus = -\Delta_{vap} H^\ominus$.
- Standard enthalpy of sublimation ($\Delta_{sub} H^\ominus$): Enthalpy change when one mole of a solid directly converts to its vapor at a constant temperature and standard pressure. Sublimation is endothermic, so $\Delta_{sub} H^\ominus$ is always positive. (e.g., CO₂(s) $\rightarrow$ CO₂(g); $\Delta_{sub} H^\ominus = +25.2$ kJ mol⁻¹ at 195 K).
These enthalpy changes reflect the strength of intermolecular forces; substances with stronger forces have higher enthalpies of fusion and vaporization.
Example 7. A swimmer coming out from a pool is covered with a film of water weighing about 18g. How much heat must be supplied to evaporate this water at 298 K ? Calculate the internal energy of vaporisation at 298K. $\Delta_{vap} H^\ominus$ for water at 298K= 44.01kJ mol⁻¹
Answer:
Mass of water film = 18 g. Molar mass of water (H₂O) = $2 \times 1.008 + 16.00 = 18.016$ g/mol (approximately 18 g/mol as used in the problem).
Number of moles of water = $\frac{18 \text{ g}}{18 \text{ g/mol}} = 1$ mol.
The process is evaporation of 1 mol of water at constant pressure (atmospheric pressure is close to 1 bar) and constant temperature (298 K).
The heat supplied to evaporate the water is equal to the enthalpy of vaporization at 298 K.
Heat supplied = $\Delta_{vap} H^\ominus = 44.01$ kJ mol⁻¹.
Since we are evaporating 1 mole, the total heat supplied is 44.01 kJ.
Now, calculate the internal energy of vaporization ($\Delta_{vap} U$). We use the relationship $\Delta H = \Delta U + \Delta n_g RT$, or $\Delta U = \Delta H - \Delta n_g RT$.
Process: H₂O (l) $\rightarrow$ H₂O (g)
Assume water vapor is a perfect gas. Liquid water is not a gas.
$\Delta n_g = (\text{moles of gaseous product}) - (\text{moles of gaseous reactant}) = 1 \text{ mol (H₂O g)} - 0 \text{ mol (H₂O l)} = 1$ mol.
$R = 8.314$ J K⁻¹ mol⁻¹ = $0.008314$ kJ K⁻¹ mol⁻¹.
$T = 298$ K.
$\Delta_{vap} U = 44.01 \text{ kJ mol}^{-1} - (1 \text{ mol}) \times (0.008314 \text{ kJ K}^{-1} \text{ mol}^{-1}) \times (298 \text{ K})$
$\Delta_{vap} U = 44.01 \text{ kJ mol}^{-1} - 2.477 \text{ kJ mol}^{-1}$
$\Delta_{vap} U = 41.533 \text{ kJ mol}^{-1}$.
The heat supplied is 44.01 kJ, and the internal energy of vaporization is approximately 41.53 kJ mol⁻¹.
Example 8. Assuming the water vapour to be a perfect gas, calculate the internal energy change when 1 mol of water at 100°C and 1 bar pressure is converted to ice at 0°C. Given the enthalpy of fusion of ice is 6.00 kJ mol⁻¹ heat capacity of water is 4.2 J/g°C.
Answer:
The overall process is the conversion of 1 mol of water from gas at 100°C to solid (ice) at 0°C.
H₂O (g, 100°C, 1 bar) $\rightarrow$ H₂O (s, 0°C)
This process can be broken down into steps using enthalpy as a state function (Hess's Law):
Step 1: Condensation of water vapor at 100°C.
H₂O (g, 100°C) $\rightarrow$ H₂O (l, 100°C)
The enthalpy change for vaporization at 100°C (373 K) is 41 kJ mol⁻¹ (from a previous example). So, the enthalpy change for condensation is $\Delta H_1 = -\Delta_{vap} H^\ominus = -41.00$ kJ mol⁻¹.
Step 2: Cooling of liquid water from 100°C to 0°C.
H₂O (l, 100°C) $\rightarrow$ H₂O (l, 0°C)
This involves a temperature change at constant pressure. The enthalpy change is $q_p = m \times c \times \Delta T$ or $n \times C_p \times \Delta T$. We are given the heat capacity of water ($c$) per gram per °C. Molar mass of water is 18 g/mol. Molar heat capacity of liquid water ($C_p$) = $18 \text{ g/mol} \times 4.2 \text{ J/g°C} = 75.6 \text{ J mol}^{-1} \text{°C}^{-1}$. Note that a change of 1°C is the same as a change of 1 K, so $75.6$ J mol⁻¹ K⁻¹.
Number of moles ($n$) = 1 mol. $\Delta T = T_{final} - T_{initial} = 0°C - 100°C = -100°C$ or $-100$ K.
$\Delta H_2 = n \times C_p(\text{liquid}) \times \Delta T = 1 \text{ mol} \times 75.6 \text{ J mol}^{-1} \text{K}^{-1} \times (-100 \text{ K})$
$\Delta H_2 = -7560$ J mol⁻¹ = $-7.56$ kJ mol⁻¹.
Step 3: Freezing of liquid water to ice at 0°C.
H₂O (l, 0°C) $\rightarrow$ H₂O (s, 0°C)
The enthalpy of fusion of ice is given as 6.00 kJ mol⁻¹. Freezing is the reverse, so $\Delta H_3 = -\Delta_{fus} H^\ominus = -6.00$ kJ mol⁻¹.
Total enthalpy change ($\Delta H_{total}$) for the overall process = $\Delta H_1 + \Delta H_2 + \Delta H_3$.
$\Delta H_{total} = -41.00 \text{ kJ mol}^{-1} + (-7.56 \text{ kJ mol}^{-1}) + (-6.00 \text{ kJ mol}^{-1})$
$\Delta H_{total} = -41.00 - 7.56 - 6.00 \text{ kJ mol}^{-1} = -54.56$ kJ mol⁻¹.
We need the internal energy change ($\Delta U$). The relationship is $\Delta U = \Delta H - \Delta n_g RT$.
Let's look at the phase changes involved in the overall process: H₂O (g, 100°C) $\rightarrow$ H₂O (s, 0°C).
Gaseous reactants: 1 mol H₂O(g). $n_{reactants, gas} = 1$.
Gaseous products: 0 mol. $n_{products, gas} = 0$.
$\Delta n_g = 0 - 1 = -1$ mol.
However, the temperatures are different (100°C vs 0°C). The equation $\Delta H = \Delta U + \Delta n_g RT$ strictly applies at constant temperature.
To find $\Delta U_{total}$, we can calculate $\Delta U$ for each step.
- Step 1 (Condensation): H₂O (g) $\rightarrow$ H₂O (l) at 373 K. $\Delta H_1 = -41$ kJ mol⁻¹. $\Delta n_g = 0 - 1 = -1$ mol. $\Delta U_1 = \Delta H_1 - \Delta n_g RT = -41 \text{ kJ} - (-1 \text{ mol})(0.008314 \text{ kJ/mol K})(373 \text{ K}) = -41 + 3.102 = -37.898$ kJ.
- Step 2 (Cooling liquid): H₂O (l) $\rightarrow$ H₂O (l). No gas involved, $\Delta V$ is negligible. $\Delta n_g = 0$. $\Delta U_2 \approx \Delta H_2 = -7.56$ kJ.
- Step 3 (Freezing): H₂O (l) $\rightarrow$ H₂O (s). No gas involved, $\Delta V$ is negligible. $\Delta n_g = 0$. $\Delta U_3 \approx \Delta H_3 = -6.00$ kJ.
Total internal energy change $\Delta U_{total} = \Delta U_1 + \Delta U_2 + \Delta U_3 = -37.898 - 7.56 - 6.00 = -51.458$ kJ mol⁻¹.
The internal energy change when 1 mol of water vapor at 100°C is converted to ice at 0°C is approximately -51.46 kJ mol⁻¹.
(The example's intermediate calculation for ΔH1 seems to use specific heat for a temperature change, which is incorrect for a phase change. It also combines values differently. Our calculation using the standard enthalpy of vaporization at 100°C and correct steps should be accurate).
Standard Enthalpy Of Formation (Symbol : ΔfH)
The standard molar enthalpy of formation ($\Delta_f H^\ominus$) of a compound is the enthalpy change that occurs when one mole of the compound is formed from its constituent elements in their most stable states of aggregation (standard states) at a specified temperature (usually 298 K) and 1 bar pressure.
The most stable state of aggregation for an element at 298 K and 1 bar is its standard state. For example:
- Hydrogen: H₂(g)
- Oxygen: O₂(g)
- Carbon: C(graphite, s)
- Sulfur: S(rhombic, s)
- Bromine: Br₂(l)
- Iodine: I₂(s)
- Metals: Solid state (e.g., Na(s), Fe(s))
By definition, the standard enthalpy of formation of an element in its most stable standard state is taken as zero ($\Delta_f H^\ominus \text{ (element in standard state)} = 0$).
Example formation reactions:
- H₂(g) + ½O₂(g) $\rightarrow$ H₂O(l); $\Delta_f H^\ominus(\text{H}_2\text{O, l})$
- C(graphite, s) + 2H₂(g) $\rightarrow$ CH₄(g); $\Delta_f H^\ominus(\text{CH}_4\text{, g})$
$\Delta_f H^\ominus$ is a specific type of reaction enthalpy where the reactants are defined as elements in their standard states, and the product is one mole of the compound in its standard state.
Standard enthalpies of formation are used to calculate the standard enthalpy change of any reaction using the formula derived from Hess's Law:
$$ \Delta_r H^\ominus = \sum (\text{stoich. coeff. of products}) \times \Delta_f H^\ominus (\text{products}) - \sum (\text{stoich. coeff. of reactants}) \times \Delta_f H^\ominus (\text{reactants}) $$Thermochemical Equations
A thermochemical equation is a balanced chemical equation that includes the physical states of all reactants and products and the enthalpy change ($\Delta_r H$) for the reaction as written.
Conventions for thermochemical equations:
- Coefficients refer to the number of moles of substances.
- Physical states (s, l, g, aq) and allotropic forms (e.g., graphite, diamond for carbon) are specified.
- The sign of $\Delta_r H$ indicates whether the reaction is exothermic (-) or endothermic (+).
- If the equation is reversed, the sign of $\Delta_r H$ is reversed.
- If the stoichiometric coefficients are multiplied by a factor, $\Delta_r H$ is also multiplied by the same factor (enthalpy is an extensive property).
Example: C₂H₅OH(l) + 3O₂(g) $\rightarrow$ 2CO₂(g) + 3H₂O(l); $\Delta_r H^\ominus = -1367$ kJ mol⁻¹
This indicates that when 1 mole of liquid ethanol reacts completely with 3 moles of oxygen gas to produce 2 moles of carbon dioxide gas and 3 moles of liquid water at standard conditions, 1367 kJ of heat is released.
Hess’s Law Of Constant Heat Summation
Hess's Law states that the total enthalpy change for a reaction is the same whether the reaction occurs in one step or in a series of steps, provided the initial and final states are the same. This is a direct consequence of enthalpy being a state function.
If a reaction A $\rightarrow$ B has an enthalpy change $\Delta_r H$, and this reaction can also occur through a series of steps A $\rightarrow$ C $\rightarrow$ D $\rightarrow$ B with enthalpy changes $\Delta H_1, \Delta H_2, \Delta H_3$, respectively, then:
$$ \Delta_r H = \Delta H_1 + \Delta H_2 + \Delta H_3 $$Hess's law is extremely useful for calculating the enthalpy changes of reactions that are difficult or impossible to measure directly by calorimetry (e.g., formation of carbon monoxide from graphite).
Enthalpies For Different Types Of Reactions
Various types of reactions have specific enthalpy changes associated with them, and these are given specific names.
Standard Enthalpy Of Combustion (Symbol : ΔcH)
The standard enthalpy of combustion ($\Delta_c H^\ominus$) is the enthalpy change when one mole of a substance undergoes complete combustion (typically reacting with excess oxygen) in its standard state to form products in their standard states under standard conditions. Combustion reactions are usually highly exothermic ($\Delta_c H^\ominus$ is negative).
Example: C₄H₁₀(g) + ¹³/₂ O₂(g) $\rightarrow$ 4CO₂(g) + 5H₂O(l); $\Delta_c H^\ominus = -2658.0$ kJ mol⁻¹ (Combustion of butane)
Enthalpy Of Atomization (Symbol: ΔaH)
The enthalpy of atomization ($\Delta_a H^\ominus$) is the enthalpy change when one mole of a substance is completely converted into gaseous atoms in their standard states. This involves breaking all bonds in the substance.
Example: H₂(g) $\rightarrow$ 2H(g); $\Delta_a H^\ominus = +435.0$ kJ mol⁻¹
Example: CH₄(g) $\rightarrow$ C(g) + 4H(g); $\Delta_a H^\ominus = +1665$ kJ mol⁻¹
For diatomic molecules, $\Delta_a H^\ominus$ is equal to the bond dissociation enthalpy.
For solids like metals, atomization is the conversion of the solid into gaseous atoms:
Na(s) $\rightarrow$ Na(g); $\Delta_a H^\ominus = +108.4$ kJ mol⁻¹ (This is also enthalpy of sublimation for metals)
Bond Enthalpy (Symbol: ΔbondH)
Chemical reactions involve breaking bonds in reactants and forming new bonds in products. Energy is absorbed to break bonds and released when bonds are formed. Bond enthalpy relates to the strength of chemical bonds.
- Bond dissociation enthalpy: The enthalpy change required to break one specific bond in a gaseous molecule. For diatomic molecules, this is the same as enthalpy of atomization (e.g., H₂ $\rightarrow$ 2H, $\Delta H = \Delta_{H-H} H^\ominus$). For polyatomic molecules, the energy to break the same type of bond can vary depending on the environment (e.g., breaking the first C-H bond in CH₄ is different from breaking the fourth).
- Mean bond enthalpy: In polyatomic molecules with multiple bonds of the same type (e.g., C-H bonds in CH₄), the mean bond enthalpy is the average energy required to break that type of bond. It is calculated as the total atomization enthalpy of the molecule divided by the number of bonds of that type. For CH₄, mean C-H bond enthalpy = $\Delta_a H^\ominus(\text{CH}_4) / 4 = 1665 / 4 = 416$ kJ mol⁻¹.
Mean bond enthalpies are approximately constant for a given bond type across different molecules and can be used to estimate the enthalpy of reaction for gas-phase reactions:
$$ \Delta_r H^\ominus \approx \sum (\text{mean bond enthalpies of reactants}) - \sum (\text{mean bond enthalpies of products}) $$This is an approximation because mean bond enthalpies are averages and may differ slightly from actual bond dissociation energies in specific molecules.
Lattice Enthalpy
The Lattice Enthalpy ($\Delta_{lattice} H^\ominus$) of an ionic compound is the enthalpy change when one mole of the solid ionic compound separates into its constituent gaseous ions. This process is highly endothermic ($\Delta_{lattice} H^\ominus$ is positive). It's the reverse of the formation of the ionic lattice from gaseous ions.
Example: NaCl(s) $\rightarrow$ Na⁺(g) + Cl⁻(g); $\Delta_{lattice} H^\ominus = +788$ kJ mol⁻¹.
Lattice enthalpy cannot be measured directly. It is calculated indirectly using Hess's Law in a series of steps called the Born-Haber Cycle. A Born-Haber cycle links the standard enthalpy of formation of an ionic compound to the enthalpy changes for sublimation (or vaporization), ionization, dissociation, electron gain, and lattice formation of the constituent elements and ions.
Applying Hess's Law to the cycle (sum of enthalpy changes around the cycle is zero) allows calculation of the unknown lattice enthalpy.
Enthalpy Of Solution (Symbol : ΔsolH )
The enthalpy of solution ($\Delta_{sol} H^\ominus$) is the enthalpy change when one mole of a substance dissolves in a specified amount of solvent. For ionic compounds, dissolution involves breaking the crystal lattice (requiring energy, related to lattice enthalpy) and hydrating/solvating the ions (releasing energy, enthalpy of hydration/solvation, $\Delta_{hyd} H^\ominus$).
For an ionic compound AB(s) dissolving in water:
AB(s) $\rightarrow$ A⁺(g) + B⁻(g); $\Delta H = \Delta_{lattice} H^\ominus$ (Lattice dissociation)
A⁺(g) + B⁻(g) + aq $\rightarrow$ A⁺(aq) + B⁻(aq); $\Delta H = \Delta_{hyd} H^\ominus$ (Hydration of ions)
Overall: AB(s) + aq $\rightarrow$ A⁺(aq) + B⁻(aq); $\Delta_{sol} H^\ominus = \Delta_{lattice} H^\ominus + \Delta_{hyd} H^\ominus$
If $\Delta_{sol} H^\ominus$ is positive, dissolution is endothermic (solubility often increases with temperature). If it's negative, dissolution is exothermic. High lattice enthalpy generally leads to lower solubility.
Enthalpy Of Dilution
The enthalpy of dilution is the enthalpy change that occurs when an additional amount of solvent is added to a solution of a given concentration. It represents the heat absorbed or released when the solution is diluted. The enthalpy of dilution depends on the initial concentration and the amount of solvent added.
Spontaneity
The First Law of Thermodynamics deals with energy conservation but doesn't predict whether a process will occur spontaneously or the direction of a spontaneous change. Many natural processes occur spontaneously in one direction only (e.g., gas expansion, heat flow from hot to cold, a stone falling downhill).
A spontaneous process is one that has the potential to proceed without continuous external assistance. It does not imply the process is fast. Spontaneous processes are generally irreversible; they can only be reversed by intervention from the surroundings.
Is Decrease In Enthalpy A Criterion For Spontaneity ?
Many spontaneous processes are exothermic (release heat, $\Delta H < 0$), leading to a decrease in enthalpy (e.g., combustion, neutralization). This suggests that a decrease in energy might be a driving force for spontaneity.
However, some spontaneous processes are endothermic (absorb heat, $\Delta H > 0$), like the dissolution of some salts (e.g., NH₄NO₃ in water makes the water colder) or the reaction N₂(g) + O₂(g) $\rightarrow$ 2NO(g) ($\Delta H$ is positive). This shows that while a decrease in enthalpy is often associated with spontaneity, it is not the sole criterion.
Entropy And Spontaneity
If not just enthalpy, what else drives spontaneity? Consider processes with $\Delta H = 0$, like the mixing of two gases by diffusion in an isolated container. This process is spontaneous but involves no enthalpy change.
Mixing increases the disorder or randomness of the system. Before mixing, you know which gas molecule is in which compartment. After mixing, the molecules are distributed throughout, and the system is less ordered and more unpredictable. This suggests that a driving force for spontaneity is the tendency towards increased disorder or randomness.
A thermodynamic function called Entropy (S) is introduced as a measure of the degree of randomness or disorder in a system. Entropy is a state function; its change ($\Delta S$) depends only on the initial and final states. Solids have the lowest entropy (most ordered), and gases have the highest entropy (most disordered). An increase in the number of particles during a reaction also generally increases entropy.
Entropy change ($\Delta S$) is related to heat transfer ($q$) and temperature ($T$). For a reversible process, the entropy change of the system is given by:
$$ \Delta S_{sys} = \frac{q_{rev}}{T} $$For a spontaneous (irreversible) process, the total entropy change of the universe (system + surroundings) must be positive:
$$ \Delta S_{total} = \Delta S_{system} + \Delta S_{surroundings} > 0 $$For a system at equilibrium, $\Delta S_{total} = 0$.
Example 10. Predict in which of the following, entropy increases/decreases :
(i) A liquid crystallizes into a solid.
(ii) Temperature of a crystalline solid is raised from 0 K to 115 K.
iii NaHCO₃(s) $\rightarrow$ Na₂CO₃(s) + CO₂(g) + H₂O(g)
(iv) H₂(g) $\rightarrow$ 2H(g)
Answer:
Entropy is a measure of disorder. More ordered states have lower entropy, more disordered states have higher entropy. Gases are more disordered than liquids, which are more disordered than solids. Increasing temperature generally increases disorder.
(i) Liquid (more disordered) crystallizes into a solid (more ordered). Entropy decreases.
(ii) Raising the temperature of a solid from 0 K (perfect order in a pure crystal, minimum entropy) increases the motion of particles and thus the disorder. Entropy increases.
(iii) Reactant is a solid (low entropy). Products include a solid and two gases. The formation of gases significantly increases the disorder compared to the solid reactant. Entropy increases.
(iv) One molecule of gas (H₂) breaks into two separate atoms of gas (2H). The number of gaseous particles increases, leading to greater disorder and more ways for the atoms to be arranged. Entropy increases.
Example 11. For oxidation of iron, 4 Fe(s) + 3 O₂(g) $\rightarrow$ 2 Fe₂O₃(s), entropy change is – 549.4 JK⁻¹ mol⁻¹ at 298 K. Inspite of negative entropy change of this reaction, why is the reaction spontaneous? ($\Delta_r H^\ominus$ for this reaction is –1648 × 10³ J mol⁻¹)
Answer:
The spontaneity of a process is determined by the total entropy change of the universe ($\Delta S_{total} = \Delta S_{system} + \Delta S_{surroundings}$). For a spontaneous process, $\Delta S_{total} > 0$.
The given entropy change is for the system: $\Delta S_{system} = -549.4$ J K⁻¹ mol⁻¹.
The enthalpy change for the reaction is $\Delta_r H^\ominus = -1648 \times 10^3$ J mol⁻¹ = $-1648$ kJ mol⁻¹.
This is an exothermic reaction, meaning heat is released by the system and absorbed by the surroundings. The heat absorbed by the surroundings is $q_{surroundings} = -\Delta_r H^\ominus = +1648 \times 10^3$ J mol⁻¹.
The entropy change of the surroundings is $\Delta S_{surroundings} = \frac{q_{surroundings}}{T}$. Assuming the surroundings are large enough that their temperature remains constant at 298 K.
$\Delta S_{surroundings} = \frac{+1648 \times 10^3 \text{ J mol}^{-1}}{298 \text{ K}} \approx +5530.2$ J K⁻¹ mol⁻¹.
Now, calculate the total entropy change:
$\Delta S_{total} = \Delta S_{system} + \Delta S_{surroundings} = (-549.4 \text{ J K}^{-1} \text{ mol}^{-1}) + (+5530.2 \text{ J K}^{-1} \text{ mol}^{-1})$
$\Delta S_{total} = +4980.8$ J K⁻¹ mol⁻¹.
Since $\Delta S_{total} > 0$, the reaction is spontaneous.
Even though the entropy of the system decreases (products are more ordered solids formed from a gas and a solid), the large amount of heat released significantly increases the entropy of the surroundings, making the overall entropy change positive and driving the spontaneity.
Gibbs Energy And Spontaneity
For processes occurring at constant temperature and pressure (common in chemistry labs), a more convenient criterion for spontaneity is the Gibbs energy (G), also known as Gibbs function or free energy. It combines enthalpy and entropy and is defined as:
$$ G = H - TS $$Gibbs energy is a state function and an extensive property. For a process at constant temperature, the change in Gibbs energy ($\Delta G$) is related to the changes in enthalpy ($\Delta H$) and entropy ($\Delta S$) of the system by the Gibbs equation:
$$ \Delta G = \Delta H - T\Delta S $$This equation is derived from the second law of thermodynamics applied to constant temperature and pressure conditions. The sign of $\Delta G$ provides a criterion for spontaneity:
- If $\Delta G < 0$ (negative), the process is spontaneous (proceeds in the forward direction).
- If $\Delta G > 0$ (positive), the process is non-spontaneous (spontaneous in the reverse direction).
- If $\Delta G = 0$, the system is at equilibrium (no net change).
$\Delta G$ represents the maximum amount of non-expansion work that can be extracted from a process at constant temperature and pressure. It is often referred to as "free energy" because it's the energy available to do useful work.
The sign of $\Delta G$ depends on the signs of $\Delta H$ and $\Delta S$ and the temperature (T > 0 K). The effect of $\Delta H$, $\Delta S$, and T on spontaneity is summarized in the table below:
| $\Delta_r H$ | $\Delta_r S$ | $\Delta_r G = \Delta_r H - T\Delta_r S$ | Spontaneity |
|---|---|---|---|
| Negative (-) | Positive (+) | Negative (-) at all T | Spontaneous at all temperatures |
| Negative (-) | Negative (-) | Negative (-) at low T Positive (+) at high T | Spontaneous at low temperatures Non-spontaneous at high temperatures |
| Positive (+) | Positive (+) | Positive (+) at low T Negative (-) at high T | Non-spontaneous at low temperatures Spontaneous at high temperatures |
| Positive (+) | Negative (-) | Positive (+) at all T | Non-spontaneous at all temperatures |
Entropy And Second Law Of Thermodynamics
The Second Law of Thermodynamics can be stated in several ways, but a key implication is that for any spontaneous process occurring in an isolated system, the entropy of the system increases ($\Delta S_{system} > 0$). Since an isolated system has no energy or matter exchange, $\Delta S_{total} = \Delta S_{system}$, so $\Delta S_{total} > 0$ for spontaneous processes in isolated systems. This law explains why natural processes tend towards states of greater disorder.
Absolute Entropy And Third Law Of Thermodynamics
The Third Law of Thermodynamics states that the entropy of any pure crystalline substance approaches zero as the temperature approaches absolute zero (0 K). At absolute zero, a perfect crystal has perfect order, and the motion of its constituent particles (translational, rotational, vibrational) is minimal or ceases, resulting in minimum entropy.
This law allows for the determination of absolute values of entropy for pure substances based on heat capacity measurements from 0 K up to a desired temperature. Standard molar entropies ($S^\ominus$) are the absolute entropies of substances in their standard states at a specified temperature (usually 298 K). Standard entropy changes for reactions ($\Delta_r S^\ominus$) can be calculated using a similar formula to Hess's law, summing standard entropies of products and reactants.
Gibbs Energy Change And Equilibrium
The Gibbs energy change ($\Delta G$) is also related to the equilibrium state of a reversible reaction. At equilibrium, the net change in Gibbs energy is zero ($\Delta G = 0$), meaning there is no longer a driving force for the reaction to proceed in either direction.
For a reaction A + B $\rightleftharpoons$ C + D, at equilibrium, $\Delta_r G = 0$.
The standard Gibbs energy change ($\Delta_r G^\ominus$), which is the Gibbs energy change when reactants and products are in their standard states, is related to the equilibrium constant (K) of the reaction by the following equation:
$$ \Delta_r G^\ominus = -RT \ln K $$ $$ \Delta_r G^\ominus = -2.303 RT \log K $$where R is the ideal gas constant and T is the absolute temperature.
This fundamental equation connects thermodynamics (energy changes) to chemical equilibrium (the extent of reaction). It shows:
- If $\Delta_r G^\ominus < 0$, $\ln K > 0$, so $K > 1$. Products are favored at equilibrium.
- If $\Delta_r G^\ominus > 0$, $\ln K < 0$, so $K < 1$. Reactants are favored at equilibrium.
- If $\Delta_r G^\ominus = 0$, $\ln K = 0$, so $K = 1$. Equilibrium lies roughly in the middle.
Since $\Delta_r G^\ominus = \Delta_r H^\ominus - T \Delta_r S^\ominus$, the equilibrium constant K is affected by both the standard enthalpy and entropy changes of the reaction, as well as the temperature.
This relationship allows us to calculate K from thermodynamic data ($\Delta_r H^\ominus$, $\Delta_r S^\ominus$) or calculate $\Delta_r G^\ominus$ from experimental equilibrium data (K).
Example 12. Calculate ΔrG for conversion of oxygen to ozone, 3/2 O₂(g) → O₃(g) at 298 K. if Kp for this conversion is $2.47 \times 10^{-29}$.
Answer:
Reaction: ³/₂ O₂(g) $\rightarrow$ O₃(g)
Temperature ($T$) = 298 K.
Equilibrium constant ($K_p$) = $2.47 \times 10^{-29}$.
Gas constant ($R$) = 8.314 J K⁻¹ mol⁻¹.
Use the equation: $\Delta_r G^\ominus = -2.303 RT \log K_p$
$\Delta_r G^\ominus = -2.303 \times (8.314 \text{ J K}^{-1} \text{ mol}^{-1}) \times (298 \text{ K}) \times \log (2.47 \times 10^{-29})$
$\log (2.47 \times 10^{-29}) = \log(2.47) + \log(10^{-29}) \approx 0.393 - 29 = -28.607$.
$\Delta_r G^\ominus = -2.303 \times 8.314 \times 298 \times (-28.607)$ J mol⁻¹
$\Delta_r G^\ominus \approx 5705.8 \times (-28.607)$ J mol⁻¹
$\Delta_r G^\ominus \approx 163200$ J mol⁻¹ = 163.2 kJ mol⁻¹.
The standard Gibbs energy change for the conversion of oxygen to ozone at 298 K is approximately 163.2 kJ mol⁻¹.
Example 13. Find out the value of equilibrium constant for the following reaction at 298 K. Standard Gibbs energy change, ΔrG at the given temperature is –13.6 kJ mol⁻¹.
Answer:
Standard Gibbs energy change ($\Delta_r G^\ominus$) = -13.6 kJ mol⁻¹ = -13600 J mol⁻¹.
Temperature ($T$) = 298 K.
Gas constant ($R$) = 8.314 J K⁻¹ mol⁻¹.
Use the equation: $\Delta_r G^\ominus = -2.303 RT \log K$
Rearrange to solve for $\log K$:
$\log K = \frac{-\Delta_r G^\ominus}{2.303 RT}$
$\log K = \frac{-(-13600 \text{ J mol}^{-1})}{2.303 \times (8.314 \text{ J K}^{-1} \text{ mol}^{-1}) \times (298 \text{ K})}$
$\log K = \frac{13600}{2.303 \times 8.314 \times 298} = \frac{13600}{5705.84} \approx 2.3836$
Now, find K by taking the antilog (10 raised to the power of $\log K$):
$K = 10^{2.3836} = 10^{0.3836} \times 10^2 \approx 2.418 \times 10^2$.
The value of the equilibrium constant at 298 K is approximately $2.42 \times 10^2$.
Example 14. At 60°C, dinitrogen tetroxide is 50 per cent dissociated. Calculate the standard free energy change at this temperature and at one atmosphere.
Answer:
Reaction: N₂O₄(g) $\rightleftharpoons$ 2NO₂(g)
Temperature ($T$) = 60°C = 273.15 + 60 = 333.15 K. Total pressure = 1 atm.
Dinitrogen tetroxide (N₂O₄) is 50% dissociated. Let the initial moles of N₂O₄ be 1.
Initial: 1 mol N₂O₄, 0 mol NO₂.
At equilibrium: (1 - 0.50) mol N₂O₄, $2 \times 0.50$ mol NO₂.
At equilibrium: 0.50 mol N₂O₄, 1.00 mol NO₂.
Total moles of gas at equilibrium = 0.50 + 1.00 = 1.50 mol.
Partial pressure of N₂O₄ ($p_{N₂O₄}$) = (mole fraction of N₂O₄) $\times$ (total pressure) = $\frac{0.50 \text{ mol}}{1.50 \text{ mol}} \times 1 \text{ atm} = \frac{1}{3}$ atm.
Partial pressure of NO₂ ($p_{NO₂}$) = (mole fraction of NO₂) $\times$ (total pressure) = $\frac{1.00 \text{ mol}}{1.50 \text{ mol}} \times 1 \text{ atm} = \frac{2}{3}$ atm.
The equilibrium constant $K_p$ is expressed in terms of partial pressures:
$K_p = \frac{(p_{NO₂})^2}{p_{N₂O₄}} = \frac{(2/3 \text{ atm})^2}{(1/3 \text{ atm})} = \frac{4/9}{1/3} \text{ atm} = \frac{4}{9} \times 3 \text{ atm} = \frac{4}{3}$ atm $\approx 1.333$ atm.
Now, calculate the standard Gibbs energy change ($\Delta_r G^\ominus$) using $\Delta_r G^\ominus = -RT \ln K_p$.
$R = 8.314$ J K⁻¹ mol⁻¹.
$\Delta_r G^\ominus = -(8.314 \text{ J K}^{-1} \text{ mol}^{-1}) \times (333.15 \text{ K}) \times \ln (4/3)$
$\ln(4/3) = \ln(1.333) \approx 0.2877$.
$\Delta_r G^\ominus = -(8.314) \times (333.15) \times (0.2877)$ J mol⁻¹
$\Delta_r G^\ominus \approx -2388.7 \times 0.2877$ J mol⁻¹ $\approx -687.2$ J mol⁻¹.
The standard free energy change at 60°C and 1 atm is approximately -687.2 J mol⁻¹.
Note: The prompt's example uses 333K instead of 333.15K and gets -763.8 kJ mol⁻¹. This is likely a significant calculation error in the source material. Our calculation using standard values and correct temperature gives a value in Joules, not Kilojoules, and a smaller magnitude.
Exercises
Question 5.1. Choose the correct answer. A thermodynamic state function is a quantity
(i) used to determine heat changes
(ii) whose value is independent of path
(iii) used to determine pressure volume work
(iv) whose value depends on temperature only.
Answer:
Question 5.2. For the process to occur under adiabatic conditions, the correct condition is:
(i) $\Delta T = 0$
(ii) $\Delta p = 0$
(iii) q = 0
(iv) w = 0
Answer:
Question 5.3. The enthalpies of all elements in their standard states are:
(i) unity
(ii) zero
(iii) < 0
(iv) different for each element
Answer:
Question 5.4. $\Delta U^\ominus$ of combustion of methane is $– X \text{ kJ mol}^{-1}$. The value of $\Delta H^\ominus$ is
(i) $= \Delta U^\ominus$
(ii) $> \Delta U^\ominus$
(iii) $< \Delta U^\ominus$
(iv) $= 0$
Answer:
Question 5.5. The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, –890.3 kJ mol$^{-1}$ –393.5 kJ mol$^{-1}$, and –285.8 kJ mol$^{-1}$ respectively. Enthalpy of formation of $CH_4(g)$ will be
(i) –74.8 kJ mol$^{-1}$
(ii) –52.27 kJ mol$^{-1}$
(iii) +74.8 kJ mol$^{-1}$
(iv) +52.26 kJ mol$^{-1}$.
Answer:
Question 5.6. A reaction, A + B $\rightarrow$ C + D + q is found to have a positive entropy change. The reaction will be
(i) possible at high temperature
(ii) possible only at low temperature
(iii) not possible at any temperature
(v) possible at any temperature
Answer:
Question 5.7. In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process?
Answer:
Question 5.8. The reaction of cyanamide, $NH_2CN (s)$, with dioxygen was carried out in a bomb calorimeter, and $\Delta U$ was found to be –742.7 kJ mol$^{-1}$ at 298 K. Calculate enthalpy change for the reaction at 298 K.
$NH_2CN(g) + \frac{3}{2}O_2(g) \rightarrow N_2(g) + CO_2(g) + H_2O(l)$
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Question 5.9. Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from 35°C to 55°C. Molar heat capacity of Al is 24 J mol$^{-1}$ K$^{-1}$.
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Question 5.10. Calculate the enthalpy change on freezing of 1.0 mol of water at10.0°C to ice at –10.0°C. $\Delta_{fus}H$ = 6.03 kJ mol$^{-1}$ at 0°C.
$C_p [H_2O(l)]$ = 75.3 J mol$^{-1}$ K$^{-1}$
$C_p [H_2O(s)]$ = 36.8 J mol$^{-1}$ K$^{-1}$
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Question 5.11. Enthalpy of combustion of carbon to $CO_2$ is –393.5 kJ mol$^{-1}$. Calculate the heat released upon formation of 35.2 g of $CO_2$ from carbon and dioxygen gas.
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Question 5.12. Enthalpies of formation of $CO(g)$, $CO_2(g)$, $N_2O(g)$ and $N_2O_4(g)$ are –110, – 393, 81 and 9.7 kJ mol$^{-1}$ respectively. Find the value of $\Delta_rH$ for the reaction:
$N_2O_4(g) + 3CO(g) \rightarrow N_2O(g) + 3CO_2(g)$
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Question 5.13. Given
$N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$; $\Delta_r H^\ominus = –92.4 \text{ kJ mol}^{–1}$
What is the standard enthalpy of formation of $NH_3$ gas?
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Question 5.14. Calculate the standard enthalpy of formation of $CH_3OH(l)$ from the following data:
$CH_3OH (l) + \frac{3}{2} O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$ ; $\Delta_r H^\ominus = –726 \text{ kJ mol}^{–1}$
$C(\text{graphite}) + O_2(g) \rightarrow CO_2(g)$ ; $\Delta_c H^\ominus = –393 \text{ kJ mol}^{–1}$
$H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(l)$; $\Delta_f H^\ominus = –286 \text{ kJ mol}^{–1}$.
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Question 5.15. Calculate the enthalpy change for the process
$CCl_4(g) \rightarrow C(g) + 4 Cl(g)$
and calculate bond enthalpy of C – Cl in $CCl_4(g)$.
$\Delta_{vap}H^\ominus(CCl_4) = 30.5 \text{ kJ mol}^{–1}$.
$\Delta_f H^\ominus (CCl_4) = –135.5 \text{ kJ mol}^{–1}$.
$\Delta_a H^\ominus (C) = 715.0 \text{ kJ mol}^{–1}$, where $\Delta_a H^\ominus$ is enthalpy of atomisation
$\Delta_a H^\ominus (Cl_2) = 242 \text{ kJ mol}^{–1}$
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Question 5.16. For an isolated system, $\Delta U = 0$, what will be $\Delta S$ ?
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Question 5.17. For the reaction at 298 K,
2A + B $\rightarrow$ C
$\Delta H = 400$ kJ mol$^{-1}$ and $\Delta S = 0.2$ kJ K$^{-1}$ mol$^{-1}$
At what temperature will the reaction become spontaneous considering $\Delta H$ and $\Delta S$ to be constant over the temperature range.
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Question 5.18. For the reaction,
$2 Cl(g) \rightarrow Cl_2(g)$, what are the signs of $\Delta H$ and $\Delta S$ ?
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Question 5.19. For the reaction
$2 A(g) + B(g) \rightarrow 2D(g)$
$\Delta U^\ominus = –10.5 \text{ kJ}$ and $\Delta S^\ominus = –44.1 \text{ JK}^{–1}$.
Calculate $\Delta G^\ominus$ for the reaction, and predict whether the reaction may occur spontaneously.
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Question 5.20. The equilibrium constant for a reaction is 10. What will be the value of $\Delta G^\ominus$ ?
$R = 8.314 \text{ JK}^{–1} \text{ mol}^{–1}$, $T = 300 \text{ K}$.
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Question 5.21. Comment on the thermodynamic stability of NO(g), given
$\frac{1}{2} N_2(g) + \frac{1}{2} O_2(g) \rightarrow NO(g)$; $\Delta_r H^\ominus = 90 \text{ kJ mol}^{–1}$
$NO(g) + \frac{1}{2} O_2(g) \rightarrow NO_2(g): \Delta_r H^\ominus = –74 \text{ kJ mol}^{–1}$
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Question 5.22. Calculate the entropy change in surroundings when 1.00 mol of $H_2O(l)$ is formed under standard conditions. $\Delta_f H^\ominus = –286 \text{ kJ mol}^{–1}$.
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