Chapter 16, Problem 81QAP

Interpretation Introduction

Interpretation:

The first and second ionization constants for sulfuric acid, H₂ SO₄ need to be calculated.

Concept introduction:

1. Ionization constant or dissociation constant is a value which is used to measure the strength of acids (K_a ) and bases (K_b ). These are essentially equilibrium constants which are expressed as the ratio between the dissociated and undissociated forms.
2. The ionization constant is related to the standard Gibbs free energy ΔG° by the following equation:

$\begin{array}{l}{\text{ΔG}}^{\text{0}}{\text{ = -RTlnK}}_{\text{eq}}\text{ -----(1)}\\ {\text{where K}}_{\text{eq}}{\text{ = K}}_{\text{a}}{\text{ or K}}_{\text{b}}\end{array}$

3. The standard Gibbs free energy ${\text{ΔG}}^{\text{0}}$ for a given chemical reaction can be expressed as a function of temperature, T via the Gibbs-Helmholtz equation:

${\text{ΔG}}^{\text{0}}{\text{ = ΔH}}^{\text{0}}{\text{ - TΔS}}^{\text{0}}\text{ -------(2)}$

where, ${\text{ΔH}}^{\text{0}}$ is the standard enthalpy change, and ${\text{ΔS}}^{\text{0}}$ is the standard entropy change

4. The standard enthalpy change for a reaction ${\text{ΔH}}^{\text{0}}$ is given in terms of the difference in the standard enthalpy of formation of the products and that of reactants.

$\Delta {\text{H}}^0\text{ = }\sum {\text{n}}_{\text{p}}{\text{ΔH}}_{\text{f}}^{\text{0}}\text{(products) - }\sum {\text{n}}_{\text{r}}{\text{ΔH}}_{\text{f}}^{\text{0}}\text{(reactants) ------(3)}$

Where n_p and n_r are the number of moles of the products and reactants

5. The standard entropy change for a reaction ${\text{ΔS}}^{\text{0}}$ is given in terms of the difference in the standard entropy of formation of the products and that of reactants.

$\Delta {\text{S}}^0\text{ = }\sum {\text{n}}_{\text{p}}{\text{S}}_{\text{f}}^{\text{0}}\text{(products) - }\sum {\text{n}}_{\text{r}}{\text{S}}_{\text{f}}^{\text{0}}\text{(reactants) -------(4)}$

Where n_p and n_r are the number of moles of the products and reactants.

Answer to Problem 81QAP

$\begin{array}{l}{\text{K}}_{\text{a1}}\text{ = 1}\text{.0 }\times {\text{10}}^3\\ {\text{K}}_{\text{a2}}\text{ = 1}\text{.0}\times {\text{10}}^{-2}\end{array}$

Explanation of Solution

Given:

$\begin{array}{l}{\text{ΔH}}_{\text{f}}^{\text{0}}{\text{(H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{) = -902}\text{.9 kJ}\\ {\text{S}}^{\text{0}}{\text{(H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{) = 17}\text{.14 J/K}\end{array}$

The dissociation of H₂ SO₄ can be expressed as

${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{(aq)}\rightleftharpoons 2{\text{H}}^{\text{+}}{\text{(aq) + SO}}_{\text{4}}^{\text{2-}}\text{(aq)}$

This is a two-step process each associated with a K_a value as shown below:

$\begin{array}{l}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{(aq)}\rightleftharpoons {\text{H}}^{\text{+}}{\text{(aq) + HSO}}_{\text{4}}^{\text{-}}{\text{(aq) K}}_{\text{a1}}\text{ -------(1)}\\ {\text{HSO}}_{\text{4}}^{\text{-}}\text{(aq) }\rightleftharpoons {\text{H}}^{\text{+}}{\text{(aq) + SO}}_{\text{4}}^{\text{2-}}{\text{(aq) K}}_{\text{a2}}\text{ --------(2)}\end{array}$

The value of K_a1 and K_a2 can be calculated by deducing the ΔG° value for reactions (1) and (2) and substituting the appropriate values in equation (1):

Step 1: Calculate ΔG° for the overall dissociation of H₂ SO₄.

The dissociation of H₂ SO₄ is represented as ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{(aq)}\rightleftharpoons 2{\text{H}}^{\text{+}}{\text{(aq) + SO}}_{\text{4}}^{\text{2-}}\text{(aq)}$

From equation (3), ΔH° for the above reaction is

$\begin{array}{l}\Delta {\text{H}}^0{\text{ = [2ΔH}}_{\text{f}}^{\text{0}}{\text{(H}}^{\text{+}}{\text{(aq)) + 1ΔH}}_{\text{f}}^{\text{0}}{\text{(SO}}_{\text{4}}^{\text{2-}}{\text{(aq))] - [1ΔH}}_{\text{f}}^{\text{0}}{\text{(H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{(aq)) ]}\\ {\text{Based on the ΔH}}_{\text{f}}^{\text{0}}\text{ values:}\\ \Delta {\text{H}}^0\text{ = [2 mole }\times (0\text{ kJ/mol) + 1 mole }\times \text{(-907}\text{.51 kJ/mol)]-[1 mole }\times \text{(}-902.9\text{ kJ/mol)] }\\ \text{ = 0 - 907}\text{.51+ 902}\text{.9 = -4}\text{.61 kJ }\end{array}$

From equation (4), S° for the above reaction is

$\begin{array}{l}\Delta {\text{S}}^0{\text{ = [2S}}_{\text{f}}^{\text{0}}{\text{(H}}^{\text{+}}{\text{(aq)) + 1S}}_{\text{f}}^{\text{0}}{\text{(SO}}_{\text{4}}^{\text{2-}}{\text{(aq))] - [1S}}_{\text{f}}^{\text{0}}{\text{(H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{(aq)) ]}\\ {\text{Based on the ΔH}}_{\text{f}}^{\text{0}}\text{ values:}\\ \Delta {\text{S}}^0\text{ = [2 mole }\times (0\text{ J/mol-K) + 1 mole }\times \text{(20}\text{.1 J/mol-K)]-[1 mole }\times \text{(}17.14\text{ J/mol-K)] }\\ \text{ = 0 + 20}\text{.1 - 17}\text{.14 = 2}\text{.96 J/mol-K }\end{array}$

At T = 298 K,

${\text{ΔG}}^{\text{0}}{\text{ = ΔH}}^{\text{0}}{\text{ - TΔS}}^{\text{0}}=-4.61-(298)(0.00296)=-5.49\text{ kJ}$

Step 2: Calculate ΔG° for the reaction (1) depicted in the two step process.

The reaction (1) is represented as

${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{(aq)}\rightleftharpoons {\text{H}}^{\text{+}}{\text{(aq) + HSO}}_{\text{4}}^{\text{-}}\text{(aq)}$

From equation (3), ΔH° for the above reaction is

$\begin{array}{l}\Delta {\text{H}}^0{\text{ = [2ΔH}}_{\text{f}}^{\text{0}}{\text{(H}}^{\text{+}}{\text{(aq)) + 1ΔH}}_{\text{f}}^{\text{0}}{\text{(HSO}}_{\text{4}}^{\text{-}}{\text{(aq))] - [1ΔH}}_{\text{f}}^{\text{0}}{\text{(H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{(aq)) ]}\\ {\text{Based on the ΔH}}_{\text{f}}^{\text{0}}\text{ values,}\\ \Delta {\text{H}}^0\text{ = [2 mole }\times (0\text{ kJ/mol) + 1 mole }\times \text{(-885}\text{.75 kJ/mol)]-[1 mole }\times \text{(}-902.9\text{ kJ/mol)] }\\ \text{ = 0 - 885}\text{.75 + 902}\text{.9 = 17}\text{.15 kJ }\end{array}$

From equation (4), S° for the above reaction is

$\begin{array}{l}\Delta {\text{S}}^0{\text{ = [2S}}_{\text{f}}^{\text{0}}{\text{(H}}^{\text{+}}{\text{(aq)) + 1S}}_{\text{f}}^{\text{0}}{\text{(HSO}}_{\text{4}}^{\text{-}}{\text{(aq))] - [1S}}_{\text{f}}^{\text{0}}{\text{(H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{(aq)) ]}\\ {\text{Based on the S}}_{\text{f}}^{\text{0}}\text{ values,}\\ \Delta {\text{S}}^0\text{ = [2 mole }\times (0\text{ J/mol-K) + 1 mole }\times \text{(131}\text{.8 J/mol-K)]-[1 mole }\times \text{(}17.14\text{ J/mol-K)] }\\ \text{ = 0 + 131}\text{.8 - 17}\text{.14 = 114}\text{.66 J/mol-K }\end{array}$

At T = 298 K,

$\begin{array}{l}{\text{ΔG}}^{\text{0}}{\text{ = ΔH}}^{\text{0}}{\text{ - TΔS}}^{\text{0}}=17.15-(298)(0.1147)=0.12\text{ kJ}\\ \text{Therefore, }\Delta {\text{G}}_1^0=\text{ -17}\text{.02 kJ}\end{array}$

Step 3: Calculate ΔG° for the reaction (2) depicted in the two step process.

The overall reaction is the sum of reactions (1) and (2). Therefore,

$\begin{array}{l}{\text{ΔG = ΔG}}_{\text{1}}^{\text{0}}{\text{ + ΔG}}_{\text{2}}^{\text{0}}\\ {\text{ΔG}}_{\text{2}}^{\text{0}}=\Delta \text{G}-{\text{ΔG}}_{\text{1}}^{\text{0}}\\ {\text{ΔG}}_{\text{2}}^{\text{0}}=(-5.49)-(-\text{17}\text{.02)}=+11.53\text{ kJ}\end{array}$

Step 4: Calculate K_a1 and K_a2.

$\begin{array}{l}{\text{ΔG}}_{\text{1}}^{\text{0}}{\text{ = -RTlnK}}_{\text{a1}}\\ {\text{K}}_{\text{a1}}{\text{ = e}}^{\frac{{\text{-ΔG}}_{\text{1}}^{\text{0}}}{\text{RT}}}\\ {\text{ΔG}}_{\text{1}}^{\text{0}}\text{ =-11}\text{.53 kJ; R = 0}\text{.008314 kJ/K-mol; T = 298 K}\end{array}$

$\begin{array}{l}{\text{K}}_{\text{a1}}{\text{ = e}}^{\frac{11.53}{\text{0}\text{.008314}\times 298}}=\text{962}\simeq \text{1}\times {\text{10}}^3\\ \\ {\text{ΔG}}_2^{\text{0}}{\text{ = -RTlnK}}_{\text{a2}}\\ {\text{K}}_{\text{a2}}{\text{ = e}}^{\frac{{\text{-ΔG}}_2^{\text{0}}}{\text{RT}}}\\ {\text{ΔG}}_2^{\text{0}}\text{ =11}\text{.53 kJ; R = 0}\text{.008314 kJ/K-mol; T = 298 K}\\ {\text{K}}_{\text{a2}}{\text{ = e}}^{\frac{-11.53}{\text{0}\text{.008314}\times 298}}=\text{0}\text{.00953}\simeq \text{1}\text{.0}\times {\text{10}}^{-2}\end{array}$

Conclusion

${\text{The ionization constants for H}}_{\text{2}}{\text{SO}}_{\text{4}}{\text{ are: K}}_{\text{a1}}\text{ = 1}\text{.0 }\times {\text{10}}^3{\text{ and K}}_{\text{a2}}\text{ = 1}\text{.0}\times {\text{10}}^{-2}$.