Chapter 16, Problem 90AE

(a)

Interpretation Introduction

Interpretation: The value of the Gibb’s free energy change is to be calculated for the given reaction under the given conditions. The direction in which the reaction shifts to reach the equilibrium is to be predicted.

Concept introduction: The Le Chatelier’s principle is basically an equilibrium law. According to which the equilibrium always shifts in the direction which nullify the outside effects. The Le Chatelier’s principle states that the addition of the reactants shifts the equilibrium to the right while the addition of product shifts the equilibrium to the left at constant temperature.

Under the standard conditions the change in Gibb’s free energy is called standard Gibb’s free energy change and is denoted by $\Delta G^{\text{ο}}$.

To determine: The $\Delta G^{\text{ο}}$ value and the direction in which the reaction shifts to reach the equilibrium.

Answer to Problem 90AE

The calculated value of $\Delta G^{\text{ο}}$ is $\underset{\_}{17928.33J/mol}$.

The equilibrium shifts to the left.

Explanation of Solution

Given

The reaction is given as,

$\text{HF}\left(aq\right)\rightleftharpoons {\text{H}}^{+}\left(aq\right)+{\text{F}}^{-}\left(aq\right)$

Concentration values are given as,

$\left[\text{HF}\right]=\left[{\text{H}}^{+}\right]=\left[{\text{F}}^{-}\right]=1.0\text{M}$

The given value of $K_{\text{a}}$ is $7.2\times {10}^{-4}$.

For the given reaction the value of $\Delta G$ is calculated by t he formula,

$\Delta G=-\text{R}T\ln K_{\text{a}}$

Where,

• $\Delta G$ is the change in Gibb’s free energy.
• $\text{R}$ is the gas constant.
• $T$ is the temperature in Kelvin.
• $K_{\text{a}}$ is the acid dissociation constant.

Substitute the value in the above equation.

$\begin{array}{l}\Delta G^{\text{ο}}=-\text{R}T\ln K_{\text{a}}\\ \Delta G^{\text{ο}}=-\left(8.314\text{J/K}\cdot \text{mol}\right)\left(298\text{K}\right)\ln \left(7.2\times {10}^{-4}\right)\\ \Delta G^{\text{ο}}=\left(-2477.57\right)\left(-7.23\right)\\ \Delta G^{\text{ο}}=\underset{\_}{17928.33J/mol}\end{array}$

The direction of equilibrium shift is depend upon the following conditions,

$\begin{array}{cc}\Delta G\text{Values}& \text{Effect}\text{on}\text{equilibrium}\text{.}\\ \Delta G=\text{Negative}& \text{Equilibrium}\text{shifts}\text{to}\text{the}\text{right}\\ \Delta G=\text{Zero}& \text{No}\text{shift}\text{in}\text{quilibrium}\\ \Delta G=\text{Positive}& \text{Equilibrium}\text{shifts}\text{to}\text{the}\text{left}\end{array}$

The value of reaction quotient $Q$ for the given reaction is calculated as,

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

Where,

• $\left[{\text{H}}^{+}\right]$ is the concentration of the hydride ion.
• $\left[{\text{F}}^{\text{-}}\right]$ is the concentration of the fluoride ion.
• $\left[\text{HF}\right]$ is the concentration of the hydrogen fluoride.

Therefore, for the given condition the change in Gibb’s free energy is calculated as,

$\Delta G=\Delta G^{\text{ο}}\text{+R}T\ln Q$

Substitute the value of $Q$ in the above equation.

$\begin{array}{l}\Delta G=\Delta G^{\text{ο}}\text{+R}T\ln Q\\ \Delta G=\Delta G^{\text{ο}}\text{+R}T\ln \frac{\left[{\text{H}}^{\text{+}}\right]\left[{\text{F}}^{\text{-}}\right]}{\left[\text{HF}\right]}\end{array}$

Substitute the given values in the above equation.

$\begin{array}{l}\Delta G=\Delta G^{\text{ο}}\text{+R}T\ln \frac{\left[{\text{H}}^{\text{+}}\right]\left[{\text{F}}^{\text{-}}\right]}{\left[\text{HF}\right]}\\ \Delta G=17928.33+\left(8.314\right)\left(298\right)\ln \frac{\left(1\right)\left(1\right)}{\left(1\right)}\\ \Delta G=17928.33+0\\ \Delta G=17928.33\end{array}$

The calculated value of $\Delta G$ is positive. Therefore, the equilibrium shifts to the left.

(b)

Interpretation Introduction

Interpretation: The value of the Gibb’s free energy change is to be calculated for the given reaction under the given conditions. The direction in which the reaction shifts to reach the equilibrium is to be predicted.

Concept introduction: The Le Chatelier’s principle is basically an equilibrium law. According to which the equilibrium always shifts in the direction which nullify the outside effects. The Le Chatelier’s principle states that the addition of the reactants shifts the equilibrium to the right while the addition of product shifts the equilibrium to the left at constant temperature.

Under the standard conditions the change in Gibb’s free energy is called standard Gibb’s free energy change and is denoted by $\Delta G^{\text{ο}}$.

To determine: The $\Delta G^{\text{ο}}$ value and the direction in which the reaction shifts to reach the equilibrium.

Answer to Problem 90AE

In the given reaction equilibrium shifts to the left.

Explanation of Solution

Given

For the given reaction the concentration values are given as,

$\begin{array}{l}\left[\text{HF}\right]=0.98\\ \left[{\text{H}}^{+}\right]=2.7\times {10}^{-2}\text{M}\\ \left[{\text{F}}^{-}\right]=2.7\times {10}^{-2}\text{M}\end{array}$

The effect on equilibrium for the given reaction is calculated as,

$\begin{array}{l}\Delta G=\Delta G^{\text{ο}}\text{+R}T\ln \frac{\left[{\text{H}}^{\text{+}}\right]\left[{\text{F}}^{\text{-}}\right]}{\left[\text{HF}\right]}\\ \Delta G=17928.33+\left(8.314\right)\left(298\right)\ln \frac{\left(2.7\times {10}^{-2}\right)\left(2.7\times {10}^{-2}\right)}{\left(0.98\right)}\\ \Delta G=17928.33-17860.45\\ \Delta G=+67.87\end{array}$

The calculated value of $\Delta G$ is positive. Therefore, the equilibrium shifts to the left.

(c)

Interpretation Introduction

Interpretation: The value of the Gibb’s free energy change is to be calculated for the given reaction under the given conditions. The direction in which the reaction shifts to reach the equilibrium is to be predicted.

Concept introduction: The Le Chatelier’s principle is basically an equilibrium law. According to which the equilibrium always shifts in the direction which nullify the outside effects. The Le Chatelier’s principle states that the addition of the reactants shifts the equilibrium to the right while the addition of product shifts the equilibrium to the left at constant temperature.

Under the standard conditions the change in Gibb’s free energy is called standard Gibb’s free energy change and is denoted by $\Delta G^{\text{ο}}$.

To determine: The direction in which the reaction shifts to reach the equilibrium.

Answer to Problem 90AE

In the given reaction equilibrium shifts to the right.

Explanation of Solution

Given

For the given reaction the concentration values are given as,

$\begin{array}{l}\left[\text{HF}\right]=1.0\times {10}^{-5}\text{M}\\ \left[{\text{H}}^{+}\right]=1.0\times {10}^{-5}\text{M}\\ \left[{\text{F}}^{-}\right]=1.0\times {10}^{-5}\text{M}\end{array}$

The effect on equilibrium for the given reaction is calculated as,

$\begin{array}{l}\Delta G=\Delta G^{\text{ο}}\text{+R}T\ln \frac{\left[{\text{H}}^{\text{+}}\right]\left[{\text{F}}^{\text{-}}\right]}{\left[\text{HF}\right]}\\ \Delta G=17928.33+\left(8.314\right)\left(298\right)\ln \frac{\left(1.0\times {10}^{-5}\right)\left(1.0\times {10}^{-5}\right)}{\left(1.0\times {10}^{-5}\right)}\\ \Delta G=17928.33-28524.07\\ \Delta G=-10595.74\text{J}\end{array}$

The calculated value of $\Delta G$ is negative. Therefore, the equilibrium shifts to the right.

(d)

Interpretation Introduction

Interpretation: The value of the Gibb’s free energy change is to be calculated for the given reaction under the given conditions. The direction in which the reaction shifts to reach the equilibrium is to be predicted.

Concept introduction: The Le Chatelier’s principle is basically an equilibrium law. According to which the equilibrium always shifts in the direction which nullify the outside effects. The Le Chatelier’s principle states that the addition of the reactants shifts the equilibrium to the right while the addition of product shifts the equilibrium to the left at constant temperature.

Under the standard conditions the change in Gibb’s free energy is called standard Gibb’s free energy change and is denoted by $\Delta G^{\text{ο}}$.

To determine: The direction in which the reaction shifts to reach the equilibrium.

Answer to Problem 90AE

In the given reaction the equilibrium is not shifted.

Explanation of Solution

Given

For the given reaction the concentration values are given as,

$\begin{array}{l}\left[\text{HF}\right]=0.27\text{M}\\ \left[{\text{H}}^{+}\right]=7.2\times {10}^{-4}\text{M}\\ \left[{\text{F}}^{-}\right]=0.27\text{M}\end{array}$

The effect on equilibrium for the given reaction is calculated as,

$\begin{array}{l}\Delta G=\Delta G^{\text{ο}}\text{+R}T\ln \frac{\left[{\text{H}}^{\text{+}}\right]\left[{\text{F}}^{\text{-}}\right]}{\left[\text{HF}\right]}\\ \Delta G=17928.33+\left(8.314\right)\left(298\right)\ln \frac{\left(0.27\right)\left(2.7\times {10}^{-4}\right)}{\left(0.27\right)}\\ \Delta G=17928.33-17928.33\\ \Delta G=0\end{array}$

The calculated value of $\Delta G$ is zero. Therefore, there is no shift in the equilibrium.

(e)

Interpretation Introduction

Interpretation: The value of the Gibb’s free energy change is to be calculated for the given reaction under the given conditions. The direction in which the reaction shifts to reach the equilibrium is to be predicted.

Concept introduction: The Le Chatelier’s principle is basically an equilibrium law. According to which the equilibrium always shifts in the direction which nullify the outside effects. The Le Chatelier’s principle states that the addition of the reactants shifts the equilibrium to the right while the addition of product shifts the equilibrium to the left at constant temperature.

Under the standard conditions the change in Gibb’s free energy is called standard Gibb’s free energy change and is denoted by $\Delta G^{\text{ο}}$.

To determine: The direction in which the reaction shifts to reach the equilibrium.

Answer to Problem 90AE

In the given reaction equilibrium shifts to the left.

Explanation of Solution

Given

For the given reaction the concentration values are given as,

$\begin{array}{l}\left[\text{HF}\right]=0.52\\ \left[{\text{H}}^{+}\right]=1.0\times {10}^{-3}\text{M}\\ \left[{\text{F}}^{-}\right]=0.67\text{M}\end{array}$

The effect on equilibrium for the given reaction is calculated as,

$\begin{array}{l}\Delta G=\Delta G^{\text{ο}}\text{+R}T\ln \frac{\left[{\text{H}}^{\text{+}}\right]\left[{\text{F}}^{\text{-}}\right]}{\left[\text{HF}\right]}\\ \Delta G=17928.33+\left(8.314\right)\left(298\right)\ln \frac{\left(0.67\right)\left(1.0\times {10}^{-3}\right)}{\left(0.52\right)}\\ \Delta G=17928.33-16487.5\\ \Delta G=+1441.5\end{array}$

The calculated value of $\Delta G$ is positive. Therefore, the equilibrium shifts to the left.