Chapter 10, Problem 109AE

(a)

Interpretation Introduction

Interpretation:

The value of change in Gibbs free energy of the reaction needs to be determined at given condition.

[HF] = [H⁺] = [F^-] = 1.0 M

Concept Introduction :

For a reaction, the value of change in Gibbs free energy can be calculated as follows:

  ${\text{ΔG=ΔG}}^{\text{0}}\text{+ RTlnQ}$

Here, R is Universal gas constant, T is temperature and Q is reaction quotient.

Answer to Problem 109AE

  $\Delta \text{G=17}\text{.929 kJ or 18 kJ}$

Equilibrium shifts towards left because of +ve value of $\Delta \text{G}$ .

Explanation of Solution

The change in Gibbs free energy can be calculated as follows:

  ${\text{ΔG=ΔG}}^{\text{0}}\text{+ RTlnQ}$

For,

  $\text{HF(aq) }𑨀{\text{H}}^{\text{+}}{\text{(aq) + F}}^{\text{-}}\text{(aq)}$

K_a = $\text{7}\text{.2 }\times {\text{ 10}}^{-4}{\text{ at 25}}^{\text{0}}\text{C}$ or 298 K

Here,

  $\begin{array}{l}{\text{ΔG}}^{\text{0}}{\text{ = -RTlnK}}_{\text{a}}\\ \text{= - (8}\text{.3145)(298)ln(7}{\text{.2×10}}^{\text{-4}}\text{)}\\ \text{= 17929J}\\ \text{= 17}\text{.9kJ}\end{array}$

For the below condition:

  $\begin{array}{l}\left[\text{HF}\right]\text{ = }\left[{\text{H}}^{\text{+}}\right]\text{ = }\left[{\text{F}}^{\text{-}}\right]\text{ = 1}\text{.0 M}\\ \begin{array}{l}\text{= 17929 + }\left(\text{8}\text{.3145}\right)\text{(298)ln(1)}\hfill \\ \begin{array}{l}\text{= 17929 + 0}\\ \Delta \text{G=17}\text{.929 kJ or 18 kJ}\end{array}\hfill \end{array}\end{array}$

Equilibrium shifts towards left because of +ve value of $\Delta \text{G}$ .

(b)

Interpretation Introduction

Interpretation:

The value of change in Gibbs free energy of the reaction needs to be determined at given condition.

[HF] = 0.98 M, [H⁺] = [F^-] = $\text{2}\text{.7 }\times {\text{ 10}}^{\text{-2}}\text{m}$

Concept Introduction :

For a reaction, the value of change in Gibbs free energy can be calculated as follows:

  ${\text{ΔG=ΔG}}^{\text{0}}\text{+ RTlnQ}$

Here, R is Universal gas constant, T is temperature and Q is reaction quotient.

Answer to Problem 109AE

  $\Delta G$ = 0

There is no effect of equilibrium because of zero $\Delta G$ value.

Explanation of Solution

[HF] = 0.98 M, [H⁺] = [F^-] = $\text{2}\text{.7 }\times {\text{ 10}}^{\text{-2}}\text{m}$

  $\begin{array}{l}\Delta G=17929+(8.314)(298)\ln \frac{{\left(2.7\times {10}^{-2}\right)}^2}{0.98}\\ \Delta G=17929-17930.3\\ =0\end{array}$

There is no effect of equilibrium because of zero $\Delta G$ value.

(c)

Interpretation Introduction

Interpretation:

The value of change in Gibbs free energy of the reaction needs to be determined at given condition.

  $\left[\text{HF}\right]\text{ = }\left[{\text{H}}^{\text{+}}\right]\left[{\text{F}}^{\text{-}}\right]\text{ = 1}\text{.0 }\times {\text{ 10}}^{\text{-5}}\text{m}$

Concept Introduction :

For a reaction, the value of change in Gibbs free energy can be calculated as follows:

  ${\text{ΔG=ΔG}}^{\text{0}}\text{+ RTlnQ}$

Here, R is Universal gas constant, T is temperature and Q is reaction quotient.

Answer to Problem 109AE

  $\Delta G$ = -10596.8 J

The equilibrium will shift towards right side because of -ve $\Delta G$ value.

Explanation of Solution

  $\left[\text{HF}\right]\text{ = }\left[{\text{H}}^{\text{+}}\right]\left[{\text{F}}^{\text{-}}\right]\text{ = 1}\text{.0 }\times {\text{ 10}}^{\text{-5}}\text{m}$

  ${\text{ΔG=ΔG}}^{\text{0}}\text{+ RTlnQ}$

  $\begin{array}{l}\text{= 17929+(8}\text{.314)(298)ln}\frac{\left[{\text{H}}^{\text{+}}\right]\left[{\text{F}}^{\text{-}}\right]}{\left[\text{HF}\right]}\\ \text{= 17929+(8}\text{.314)(298)ln}\frac{{\left(\text{1}{\text{.0×10}}^{\text{-5}}\right)}^{\text{2}}}{\left(\text{1}{\text{.0×10}}^{\text{-5}}\right)}\\ \text{= -10596}\text{.8J}\end{array}$

The equilibrium will shift towards right side because of −ve $\Delta G$ value.

(d)

Interpretation Introduction

Interpretation:

The value of change in Gibbs free energy of the reaction needs to be determined at given condition.

[HF] = [F^-] = 0.27 m, [H⁺] = $\text{7}\text{.2 }\times {\text{ 10}}^{\text{-4}}\text{m}$

Concept Introduction :

For a reaction, the value of change in Gibbs free energy can be calculated as follows:

  ${\text{ΔG=ΔG}}^{\text{0}}\text{+ RTlnQ}$

Here, R is Universal gas constant, T is temperature and Q is reaction quotient.

Answer to Problem 109AE

  $\Delta G$ = 0

There is no effect of equilibrium because of zero $\Delta G$ value.

Explanation of Solution

[HF] = [F^-] = 0.27 m, [H⁺] = $\text{7}\text{.2 }\times {\text{ 10}}^{\text{-4}}\text{m}$

  ${\text{ΔG=ΔG}}^{\text{0}}\text{+ RTlnQ}$

Or,

  ${\text{ΔG=ΔG}}^{\text{0}}\text{+ RTln}\frac{\left[{\text{F}}^{-}\right]\left[{\text{H}}^{+}\right]}{\left[\text{HF}\right]}$

Putting the values,

  $\begin{array}{l}\text{= 17929+(8}\text{.314)(298)ln}\frac{\left(0.27\right)\left(7.2\times {10}^{-4}\right)}{\left(0.27\right)}\\ =17929+(-17929.43)\\ =0\end{array}$

There is no effect of equilibrium because of zero $\Delta G$ value.

(e)

Interpretation Introduction

Interpretation:

The value of change in Gibbs free energy of the reaction needs to be determined at given condition.

[HF] = 0.52 m, [F^-] = 0.67 m

Concept Introduction :

For a reaction, the value of change in Gibbs free energy can be calculated as follows:

  ${\text{ΔG=ΔG}}^{\text{0}}\text{+ RTlnQ}$

Here, R is Universal gas constant, T is temperature and Q is reaction quotient.

Answer to Problem 109AE

  $\Delta G$ = 1441.5

The equilibrium will shift towards left because of positive $\Delta G$ value.

Explanation of Solution

Change in Gibbs free energy is calculated as follows:

  ${\text{ΔG=ΔG}}^{\text{0}}\text{+ RTln}\frac{\left[{\text{F}}^{-}\right]\left[{\text{H}}^{+}\right]}{\left[\text{HF}\right]}$

Here,

[HF] = 0.52 m, [F^-] = 0.67 m [H⁺] = $\text{1}\text{.0 }\times {\text{10}}^{\text{-3}}\text{m}$

Putting the values,

  $\begin{array}{l}\Delta G=17929+(8.314)(298)\ln \frac{\left(0.67\right)\left(1.0\times {10}^{-3}\right)}{\left(0.52\right)}\\ \Delta G=17929-16487.5\\ =1441.5\end{array}$

The equilibrium will shift towards left because of positive $\Delta G$ value.