Chapter 17, Problem 17.79QE

(a)

Interpretation Introduction

Interpretation:

The given statement ‘The direction of spontaneous reaction depends entirely on ${\text{ΔH}}^{\text{o}}$’ has to be identified as true or false.

Explanation of Solution

The given reaction as follows,

  $\text{CO}\left(\text{g}\right)\text{+}\text{2}{\text{H}}_{\text{2}}\left(\text{g}\right)\to {\text{CH}}_{\text{3}}\text{OH}\left(\text{g}\right)$

For the forward reaction to be spontaneous the standard Gibbs free energy should be negative. At low temperature, the sign depends on standard enthalpy change and at high temperature the sign depends on standard entropy change.

Therefore, the given statement is false as the direction of reaction spontaneity depends on temperature.

(b)

Interpretation Introduction

Interpretation:

The given statement ‘The reaction is spontaneous at $\text{298K}$’ has to be identified as true or false.

Explanation of Solution

The given reaction as follows,

  $\text{CO}\left(\text{g}\right)\text{+}\text{2}{\text{H}}_{\text{2}}\left(\text{g}\right)\to {\text{CH}}_{\text{3}}\text{OH}\left(\text{g}\right)$

  $\begin{array}{l}\text{CO}\left(\text{g}\right)\text{+}\text{2}{\text{H}}_{\text{2}}\left(\text{g}\right)\to {\text{CH}}_{\text{3}}\text{OH}\left(\text{g}\right)\\ \Delta G_f^{\circ }\left(\text{kJ/mol}\right)-137.150-162.01\end{array}$

  $\begin{array}{c}{\text{ΔG}}^{\text{°}}{}_{\text{rxn}}\text{=}{\displaystyle \sum {\text{nΔG}}_{\text{f}}{}^{\text{°}}\left(\text{Products}\right)}-{\displaystyle \sum {\text{nΔG}}_{\text{f}}{}^{\text{°}}\left(\text{Reactants}\right)}\\ \\ \text{=}\left(1\times -162.01\text{kJ/mol}\right)-\left(\text{1×}-137.15\text{kJ/mol}\right)\\ \\ =-24.86kJ/mol.\end{array}$

The calculated value of $\Delta G°$ is $-24.86kJ/mol.$

The negative value reveals the reaction is spontaneous. Hence the given statement is true.

(c)

Interpretation Introduction

Interpretation:

The given statement ‘The reaction proceeds spontaneously at all temperatures greater than $\text{298K}$’ has to be identified as true or false.

Explanation of Solution

The given reaction as follows,

  $\text{CO}\left(\text{g}\right)\text{+}\text{2}{\text{H}}_{\text{2}}\left(\text{g}\right)\to {\text{CH}}_{\text{3}}\text{OH}\left(\text{g}\right)$

  $\begin{array}{l}\text{CO}\left(\text{g}\right)\text{+}\text{2}{\text{H}}_{\text{2}}\left(\text{g}\right)\to {\text{CH}}_{\text{3}}\text{OH}\left(\text{g}\right)\\ {\text{ΔH}}_{rxn}^{\text{o}}\left(\text{kJ/mol}\right)-110.520-200.66\end{array}$

  $\begin{array}{c}{\text{ΔH}}_{\text{rxn}}^{\text{o}}\text{=}{\displaystyle \sum \left(\text{n×}{\text{ΔH}}_f^{\circ }\text{products}\right)}-{\displaystyle \sum \left(\text{n×}{\text{ΔH}}_f^{\circ }\text{ reactants}\right)}\\ \\ \text{=}\left(1\text{×}-200.66\text{kJ/mol}\right)-\left(1\text{×}-110.52\text{kJ/mol}\right)\\ \\ =-90.14kJ/mol.\end{array}$

Hence, value of $\Delta {\text{H}}_{rxn}^{\circ }$ for the process is $-90.14kJ/mol.$

• The value of $\Delta {\text{S}}_f^{\circ }$ is calculated by standard entropy change as follows,

  $\begin{array}{l}\text{CO}\left(\text{g}\right)\text{+}\text{2}{\text{H}}_{\text{2}}\left(\text{g}\right)\to {\text{CH}}_{\text{3}}\text{OH}\left(\text{g}\right)\\ {\text{ΔS}}_{\text{f}}^{\text{o}}\left(\text{J/mol}\text{.K}\right)197.56130.57239.70\end{array}$

  $\begin{array}{c}{\text{ΔS}}^{\circ }\text{=}{\displaystyle \sum \left(\text{n×}{\text{ΔS}}^{\circ }\text{products}\right)}-{\displaystyle \sum \left(\text{n×}{\text{ΔS}}^{\circ }\text{ reactants}\right)}\\ \\ \text{=}\left(1\text{×}239.70\text{J/mol}\text{.K}\right)-\left(1\text{×}197.56\text{J/mol}\text{.K}\right)-\left(2\text{×}130.57\text{J/mol}\text{.K}\right)\\ \\ =-219J/mol.K.\end{array}$

  $\begin{array}{c}\text{ΔG =}\text{ΔH}\text{-}\text{T}\text{ΔS}\\ \\ \text{0}\text{=}\text{ΔH}\text{-}\text{T}\text{ΔS}\left(\text{at}\text{equilibrium}\right)\text{or}\\ \\ \text{T}\text{=}\frac{{\text{ΔH}}^{\text{o}}}{{\text{ΔS}}^{\text{o}}}\\ \\ \text{=}\frac{-\text{90}\text{.14}\text{kJ/mol}}{-\text{219J/mol}\text{.K}}=411K\end{array}$

The value of $\text{ΔG}$ will be negative below the temperature $411K$ and $\text{ΔG}$ will be positive above the temperature $411K$.

Therefore, the reaction proceeds spontaneously at all temperatures below than $411K.$

Hence, the given statement is false.

(d)

Interpretation Introduction

Interpretation:

The given statement ‘The reaction proceeds spontaneously at all temperatures less than $\text{298K}$’ has to be identified as true or false.

Explanation of Solution

The given reaction as follows,

  $\text{CO}\left(\text{g}\right)\text{+}\text{2}{\text{H}}_{\text{2}}\left(\text{g}\right)\to {\text{CH}}_{\text{3}}\text{OH}\left(\text{g}\right)$

  $\begin{array}{l}\text{CO}\left(\text{g}\right)\text{+}\text{2}{\text{H}}_{\text{2}}\left(\text{g}\right)\to {\text{CH}}_{\text{3}}\text{OH}\left(\text{g}\right)\\ {\text{ΔH}}_{rxn}^{\text{o}}\left(\text{kJ/mol}\right)-110.520-200.66\end{array}$

  $\begin{array}{c}{\text{ΔH}}_{\text{rxn}}^{\text{o}}\text{=}{\displaystyle \sum \left(\text{n×}{\text{ΔH}}_f^{\circ }\text{products}\right)}-{\displaystyle \sum \left(\text{n×}{\text{ΔH}}_f^{\circ }\text{ reactants}\right)}\\ \\ \text{=}\left(1\text{×}-200.66\text{kJ/mol}\right)-\left(1\text{×}-110.52\text{kJ/mol}\right)\\ \\ =-90.14kJ/mol.\end{array}$

Hence, value of $\Delta {\text{H}}_{rxn}^{\circ }$ for the process is $-90.14kJ/mol.$

• The value of $\Delta {\text{S}}_f^{\circ }$ is calculated by standard entropy change as follows,

  $\begin{array}{l}\text{CO}\left(\text{g}\right)\text{+}\text{2}{\text{H}}_{\text{2}}\left(\text{g}\right)\to {\text{CH}}_{\text{3}}\text{OH}\left(\text{g}\right)\\ {\text{ΔS}}_{\text{f}}^{\text{o}}\left(\text{J/mol}\text{.K}\right)197.56130.57239.70\end{array}$

  $\begin{array}{c}{\text{ΔS}}^{\circ }\text{=}{\displaystyle \sum \left(\text{n×}{\text{ΔS}}^{\circ }\text{products}\right)}-{\displaystyle \sum \left(\text{n×}{\text{ΔS}}^{\circ }\text{ reactants}\right)}\\ \\ \text{=}\left(1\text{×}239.70\text{J/mol}\text{.K}\right)-\left(1\text{×}197.56\text{J/mol}\text{.K}\right)-\left(2\text{×}130.57\text{J/mol}\text{.K}\right)\\ \\ =-219J/mol.K.\end{array}$

  $\begin{array}{c}\text{ΔG =}\text{ΔH}\text{-}\text{T}\text{ΔS}\\ \\ \text{0}\text{=}\text{ΔH}\text{-}\text{T}\text{ΔS}\left(\text{at}\text{equilibrium}\right)\text{or}\\ \\ \text{T}\text{=}\frac{{\text{ΔH}}^{\text{o}}}{{\text{ΔS}}^{\text{o}}}\\ \\ \text{=}\frac{-\text{90}\text{.14}\text{kJ/mol}}{-\text{219J/mol}\text{.K}}=411K\end{array}$

The value of $\text{ΔG}$ will be negative and becomes spontaneous at below the temperature $411K$ and $\text{ΔG}$ will be positive above the temperature $411K$.

Therefore, the reaction proceeds spontaneously at all temperatures below than $411K.$

Hence, the given statement is true.

(e)

Interpretation Introduction

Interpretation:

The given statement ‘The equilibrium constant increases with increasing temperature’ has to be identified as true or false.

Explanation of Solution

The given reaction as follows,

  $\text{CO}\left(\text{g}\right)\text{+}\text{2}{\text{H}}_{\text{2}}\left(\text{g}\right)\to {\text{CH}}_{\text{3}}\text{OH}\left(\text{g}\right)$

  $\begin{array}{l}\text{CO}\left(\text{g}\right)\text{+}\text{2}{\text{H}}_{\text{2}}\left(\text{g}\right)\to {\text{CH}}_{\text{3}}\text{OH}\left(\text{g}\right)\\ {\text{ΔH}}_{rxn}^{\text{o}}\left(\text{kJ/mol}\right)-110.520-200.66\end{array}$

  $\begin{array}{c}{\text{ΔH}}_{\text{rxn}}^{\text{o}}\text{=}{\displaystyle \sum \left(\text{n×}{\text{ΔH}}_f^{\circ }\text{products}\right)}-{\displaystyle \sum \left(\text{n×}{\text{ΔH}}_f^{\circ }\text{ reactants}\right)}\\ \\ \text{=}\left(1\text{×}-200.66\text{kJ/mol}\right)-\left(1\text{×}-110.52\text{kJ/mol}\right)\\ \\ =-90.14kJ/mol.\end{array}$

Hence, value of $\Delta {\text{H}}_{rxn}^{\circ }$ for the process is $-90.14kJ/mol.$

The value of enthalpy change is negative that reveals the reaction is exothermic. The equilibrium constant for exothermic reaction decreases with increasing temperature.

Hence, the given statement is false.

(f)

Interpretation Introduction

Interpretation:

The given statement ‘The reaction proceeds quickly at any spontaneous temperature’ has to be identified as true or false.

Explanation of Solution

Without knowing about equilibrium constant and temperature, the speed of reaction cannot be predicted.

Therefore, the reaction proceeds quickly at any spontaneous temperatures cannot be predicted.

Hence, the given statement is false.