Reaction Enthalpy and Hess's Law

Enthalpy Change, $\Delta H$, Of A Reaction – Reaction Enthalpy

Reaction Enthalpy ($\Delta H_{rxn}$): The enthalpy change for a chemical reaction is the amount of heat absorbed or released by the reaction when it occurs at constant pressure. It is also known as the heat of reaction.

Relationship to Internal Energy: As derived earlier, $\Delta H = \Delta U + \Delta(PV)$. For reactions involving gases where there is a change in the number of moles of gas, $\Delta H$ will differ from $\Delta U$. If there is no change in the number of moles of gas, or if the reaction occurs in solution where volume changes are negligible, then $\Delta H \approx \Delta U$.

Exothermic vs. Endothermic Reactions:

Exothermic Reaction: Releases heat to the surroundings. The enthalpy change is negative ($\Delta H < 0$). The products have lower enthalpy than the reactants.
Endothermic Reaction: Absorbs heat from the surroundings. The enthalpy change is positive ($\Delta H > 0$). The products have higher enthalpy than the reactants.

Enthalpy Diagram: A visual representation showing the relative enthalpies of reactants and products. For an exothermic reaction, products are lower on the diagram than reactants; for an endothermic reaction, products are higher.

Standard Enthalpy Of Reactions ($\Delta H^\circ_{rxn}$)

Definition: The standard enthalpy of reaction is the enthalpy change that occurs when reactants in their standard states are converted to products in their standard states.

Standard State Conditions:

Pressure: 1 bar (or 1 atm, though 1 bar is the current IUPAC standard).
Temperature: Usually specified, most commonly 298.15 K (25°C).
Concentration: For solutions, 1 molar (1 M).
State: The most stable form of the substance under standard conditions (e.g., graphite for carbon, $O_2(g)$ for oxygen).

Significance: Standard enthalpies of reaction allow for the comparison of enthalpy changes for different reactions under a common set of conditions.

Enthalpy Changes During Phase Transformations

Phase Transitions: These are physical processes where a substance changes from one state of matter to another (e.g., melting, boiling, sublimation). These processes involve changes in enthalpy.

Enthalpy of Fusion ($\Delta H_{fus}$): The enthalpy change when one mole of a solid melts into a liquid at constant pressure and temperature (melting point). $Solid \rightarrow Liquid$. This is always an endothermic process ($\Delta H_{fus} > 0$).
Enthalpy of Vaporization ($\Delta H_{vap}$): The enthalpy change when one mole of a liquid vaporizes into a gas at constant pressure and temperature (boiling point). $Liquid \rightarrow Gas$. This is always an endothermic process ($\Delta H_{vap} > 0$).
Enthalpy of Sublimation ($\Delta H_{sub}$): The enthalpy change when one mole of a solid transforms directly into a gas at constant pressure and temperature. $Solid \rightarrow Gas$.

Relationship: According to Hess's Law, the enthalpy of sublimation is the sum of the enthalpy of fusion and the enthalpy of vaporization:

$$\Delta H_{sub} = \Delta H_{fus} + \Delta H_{vap}$$

Similarly, the reverse processes are exothermic:

Enthalpy of Freezing ($\Delta H_{freez}$) = $-\Delta H_{fus}$
Enthalpy of Condensation ($\Delta H_{cond}$) = $-\Delta H_{vap}$
Enthalpy of Deposition ($\Delta H_{dep}$) = $-\Delta H_{sub}$

Standard Enthalpy Of Formation ($\Delta H^\circ_f$)

Definition: The standard enthalpy of formation of a compound is the enthalpy change when one mole of the compound is formed from its constituent elements in their most stable standard states.

Example:

Formation of water: $H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l)$ $\Delta H^\circ_f [H_2O(l)] = -285.8 \text{ kJ/mol}$
Formation of carbon dioxide: $C(s, \text{graphite}) + O_2(g) \rightarrow CO_2(g)$ $\Delta H^\circ_f [CO_2(g)] = -393.5 \text{ kJ/mol}$

Key Points:

The standard enthalpy of formation of any element in its most stable standard state is defined as zero. For example, $\Delta H^\circ_f [O_2(g)] = 0$, $\Delta H^\circ_f [C(s, \text{graphite})] = 0$, $\Delta H^\circ_f [Na(s)] = 0$.
Standard enthalpies of formation are usually tabulated at 298.15 K.

Calculating Standard Enthalpy of Reaction ($\Delta H^\circ_{rxn}$) from Standard Enthalpies of Formation:

For a reaction:

$$aA + bB \rightarrow cC + dD$$

The standard enthalpy of reaction can be calculated as:

$$\Delta H^\circ_{rxn} = \sum (\text{stoichiometric coefficient} \times \Delta H^\circ_f \text{ of products}) - \sum (\text{stoichiometric coefficient} \times \Delta H^\circ_f \text{ of reactants})$$ $$\Delta H^\circ_{rxn} = [c \Delta H^\circ_f(C) + d \Delta H^\circ_f(D)] - [a \Delta H^\circ_f(A) + b \Delta H^\circ_f(B)]$$

Example: Calculate the standard enthalpy of combustion of methane ($CH_4$) given the standard enthalpies of formation:

$CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$

$\Delta H^\circ_f [CH_4(g)] = -74.8 \text{ kJ/mol}$

$\Delta H^\circ_f [O_2(g)] = 0 \text{ kJ/mol}$

$\Delta H^\circ_f [CO_2(g)] = -393.5 \text{ kJ/mol}$

$\Delta H^\circ_f [H_2O(l)] = -285.8 \text{ kJ/mol}$

Example 1. Calculate the standard enthalpy of combustion of methane ($CH_4$) given the standard enthalpies of formation.

Answer:

Using the formula $\Delta H^\circ_{rxn} = \sum (\text{coeff} \times \Delta H^\circ_f \text{ products}) - \sum (\text{coeff} \times \Delta H^\circ_f \text{ reactants})$:

$\Delta H^\circ_{rxn} = [1 \times \Delta H^\circ_f(CO_2(g)) + 2 \times \Delta H^\circ_f(H_2O(l))] - [1 \times \Delta H^\circ_f(CH_4(g)) + 2 \times \Delta H^\circ_f(O_2(g))]$

$\Delta H^\circ_{rxn} = [1 \times (-393.5 \text{ kJ/mol}) + 2 \times (-285.8 \text{ kJ/mol})] - [1 \times (-74.8 \text{ kJ/mol}) + 2 \times (0 \text{ kJ/mol})]$

$\Delta H^\circ_{rxn} = [-393.5 - 571.6] - [-74.8]$

$\Delta H^\circ_{rxn} = -965.1 + 74.8$

$\Delta H^\circ_{rxn} = -890.3 \text{ kJ/mol}$

The standard enthalpy of combustion of methane is -890.3 kJ/mol.

Thermochemical Equations

Definition: A thermochemical equation is a balanced chemical equation that includes the enthalpy change ($\Delta H$) for the reaction. It shows the physical states of reactants and products, as the enthalpy change can vary depending on the states.

Conventions:

The value of $\Delta H$ is written on the right side of the equation.
The $\Delta H$ value corresponds to the molar quantities shown in the balanced equation.
The sign of $\Delta H$ is included.
Physical states (solid (s), liquid (l), gas (g), aqueous (aq)) are specified.

Examples:

Combustion of methane: $CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$ $\Delta H = -890.3 \text{ kJ/mol}$
Formation of water: $H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l)$ $\Delta H = -285.8 \text{ kJ/mol}$
Melting of ice: $H_2O(s) \rightarrow H_2O(l)$ $\Delta H = +6.01 \text{ kJ/mol}$

Hess’s Law Of Constant Heat Summation

Statement: Hess's Law states that the total enthalpy change for a chemical reaction is independent of the pathway or the number of steps taken to achieve the overall change. It is a direct consequence of enthalpy being a state function.

Application: This law is extremely useful for calculating enthalpy changes for reactions that are difficult or impossible to measure directly. We can combine known enthalpy changes of other reactions (which sum up to the desired reaction) to find the unknown enthalpy change.

Rules for Manipulating Thermochemical Equations:

Reversing a reaction: If a reaction is reversed, the sign of its $\Delta H$ is reversed.
Multiplying a reaction: If a reaction is multiplied by a factor, its $\Delta H$ is multiplied by the same factor.
Adding reactions: If two or more reactions are added together, their corresponding $\Delta H$ values are also added.

Example: Calculate the enthalpy of formation of $CO(g)$ from the following data:

$2C(s, \text{graphite}) + O_2(g) \rightarrow 2CO(g)$ $\Delta H_1 = -221.0 \text{ kJ}$
$2CO(g) + O_2(g) \rightarrow 2CO_2(g)$ $\Delta H_2 = -566.0 \text{ kJ}$
$C(s, \text{graphite}) + O_2(g) \rightarrow CO_2(g)$ $\Delta H_3 = -393.5 \text{ kJ}$

We want to find the $\Delta H$ for: $C(s, \text{graphite}) + \frac{1}{2}O_2(g) \rightarrow CO(g)$

Example 2. Calculate the enthalpy of formation of $CO(g)$ from the given thermochemical equations.

Answer:

Target Reaction: $C(s, \text{graphite}) + \frac{1}{2}O_2(g) \rightarrow CO(g)$

Step 1: Manipulate the given equations to match the target reaction.

Take equation 1 and divide by 2:

$C(s, \text{graphite}) + O_2(g) \rightarrow CO(g)$ $\Delta H_1/2 = -221.0 \text{ kJ} / 2 = -110.5 \text{ kJ}$

We need $CO(g)$ as a product, and equation 1 gives it as a product. However, we need $CO(g)$ with a coefficient of 1, which we achieved by dividing by 2.
We need $O_2(g)$ as a reactant with a coefficient of 1/2. Equation 2 has $O_2(g)$ as a reactant but with a coefficient of 1 and $CO(g)$ as a reactant. We need to eliminate $CO(g)$ from the final sum and $O_2(g)$ as a product.
Take equation 2 and reverse it, then divide by 2:

$CO_2(g) \rightarrow CO(g) + \frac{1}{2}O_2(g)$ $-\Delta H_2 / 2 = -(-566.0 \text{ kJ}) / 2 = +283.0 \text{ kJ}$

Equation 3 has $C(s, \text{graphite})$ and $O_2(g)$ as reactants and $CO_2(g)$ as a product. We need to cancel out $CO_2(g)$. We can use equation 3 as is, but we need to cancel the $CO_2(g)$ produced in the target formation of $CO(g)$. However, the target reaction does not involve $CO_2(g)$ as a reactant or product in its final form. Let's re-evaluate.

Let's restart with the target: $C(s, \text{graphite}) + \frac{1}{2}O_2(g) \rightarrow CO(g)$

From reaction 1: $2C(s, \text{graphite}) + O_2(g) \rightarrow 2CO(g)$ $\Delta H_1 = -221.0 \text{ kJ}$

Divide reaction 1 by 2: $C(s, \text{graphite}) + \frac{1}{2}O_2(g) \rightarrow CO(g)$ $\Delta H = -110.5 \text{ kJ}$

This directly gives us the target reaction if we assume the $\Delta H$ from reaction 1 is for the formation of 2 moles of CO. If $\Delta H_1$ is indeed the enthalpy change for the reaction as written, then the enthalpy of formation of CO is $\Delta H_1 / 2$. However, often these values are given with respect to the formation of 1 mole of the product.

Let's assume the intent is to use all data. We need to get $CO(g)$ on the product side with a coefficient of 1. Reaction 1 is the source.

Desired Reaction: $C(s) + \frac{1}{2}O_2(g) \rightarrow CO(g)$

Given Reactions:

$2C(s) + O_2(g) \rightarrow 2CO(g)$ $\Delta H_1 = -221.0 \text{ kJ}$
$2CO(g) + O_2(g) \rightarrow 2CO_2(g)$ $\Delta H_2 = -566.0 \text{ kJ}$
$C(s) + O_2(g) \rightarrow CO_2(g)$ $\Delta H_3 = -393.5 \text{ kJ}$

Step 1: Take reaction (i) and divide it by 2:

$C(s) + \frac{1}{2}O_2(g) \rightarrow CO(g)$ $\Delta H = \frac{-221.0 \text{ kJ}}{2} = -110.5 \text{ kJ}$

This is precisely the target reaction! It seems reaction (ii) and (iii) might be provided for context or to ensure one uses Hess's Law. Let's verify if they are consistent.

If we want to confirm Hess's Law, we can try to construct the target from all reactions:

From (i) divide by 2: $C(s) + \frac{1}{2}O_2(g) \rightarrow CO(g)$ ($\Delta H = -110.5 \text{ kJ}$)
From (ii) reverse and divide by 2: $CO_2(g) \rightarrow CO(g) + \frac{1}{2}O_2(g)$ ($\Delta H = +283.0 \text{ kJ}$)
From (iii): $C(s) + O_2(g) \rightarrow CO_2(g)$ ($\Delta H = -393.5 \text{ kJ}$)

Let's try to combine (ii) reversed/2 and (iii) to get the target:

(ii) reversed/2: $CO_2(g) \rightarrow CO(g) + \frac{1}{2}O_2(g)$ ($\Delta H = +283.0 \text{ kJ}$)

(iii): $C(s) + O_2(g) \rightarrow CO_2(g)$ ($\Delta H = -393.5 \text{ kJ}$)

Summing these two:

$C(s) + O_2(g) + CO_2(g) \rightarrow CO(g) + \frac{1}{2}O_2(g) + CO_2(g)$

Cancel $CO_2(g)$ and $\frac{1}{2}O_2(g)$ from both sides:

$C(s) + \frac{1}{2}O_2(g) \rightarrow CO(g)$

The enthalpy change for this combination is: $+283.0 \text{ kJ} + (-393.5 \text{ kJ}) = -110.5 \text{ kJ}$.

This matches the result from simply dividing reaction (i) by 2. Therefore, the enthalpy of formation of $CO(g)$ is -110.5 kJ/mol.