Chapter 5, Problem 5.43E

(a)

Interpretation Introduction

Interpretation:

For the given reaction ${\text{CO(g) + H}}_{\text{2}}\text{O(g) }\to {\text{ CO}}_{\text{2}}{\text{(g) + H}}_{\text{2}}\text{(g)}$, when K =1, the standard Gibbs free energy of the reaction has to be calculated.

Explanation of Solution

The given reaction is shown below,

  ${\text{CO(g) + H}}_{\text{2}}\text{O(g) }\to {\text{ CO}}_{\text{2}}{\text{(g) + H}}_{\text{2}}\text{(g)}$

The standard Gibbs free energy of the reaction can be expressed as follow,

  $\begin{array}{l}\Delta {\text{G}}^{\text{o}}\text{ = - RT In K}\\ \\ \text{K= 1}\\ \Delta {\text{G}}^{\text{o}}\text{ = - RT In (1)}\\ \Delta {\text{G}}^{\text{o}}\text{ = - RT (0)}\\ \Delta {\text{G}}^{\text{o}}\text{ = }0\end{array}$

The standard Gibbs free energy of the reaction is $0$.

(b)

Interpretation Introduction

Interpretation:

For the given reaction ${\text{CO(g) + H}}_{\text{2}}\text{O(g) }\to {\text{ CO}}_{\text{2}}{\text{(g) + H}}_{\text{2}}\text{(g)}$, when K =1, the temperature of the reaction has to be calculated.

Explanation of Solution

The given reaction is shown below,

  ${\text{CO(g) + H}}_{\text{2}}\text{O(g) }\to {\text{ CO}}_{\text{2}}{\text{(g) + H}}_{\text{2}}\text{(g)}$

The temperature of the reaction can be expressed as follow,

  $\begin{array}{l}\Delta {\text{G}}^{\text{o}}\text{ = - RT In K}\\ \\ Where,\\ \Delta {\text{G}}^{\text{o}}\text{ = }\Delta H^{\text{o}}-T\Delta S^{\text{o}}\\ \\ \text{K= 1}\\ \Delta {\text{G}}^{\text{o}}\text{ = - RT In (1)}\\ \Delta {\text{G}}^{\text{o}}\text{ = - RT (0)}\\ \Delta {\text{G}}^{\text{o}}\text{ = }0\end{array}$

The standard Gibbs free energy of the reaction is $0$. The enthalpy of the reaction is calculated as follows,

$\begin{array}{l}{\text{ΔH}}^{\text{o}}\text{=}\text{Product}\text{-}\text{Reactant}\\ {\text{ΔH}}^{\text{o}}\text{=}\left[-393.51kJ\cdot mo{l}^{-1}+0kJ\cdot mo{l}^{-1}\right]-\left[-110.53kJ\cdot mo{l}^{-1}+\left(-241.82kJ\cdot mo{l}^{-1}\right)\right]\\ {\text{ΔH}}^{\text{o}}\text{=}-41.16kJ\cdot mo{l}^{-1}\end{array}$

The entropy of the reaction is calculated as follows,

$\begin{array}{l}{\text{ΔS}}^{\text{o}}\text{=}\text{Product}\text{-}\text{Reactant}\\ {\text{ΔS}}^{\text{o}}\text{=}\left[-130.68J\cdot K^{-1}\cdot mo{l}^{-1}+213.74J\cdot K^{-1}\cdot mo{l}^{-1}\right]-\left[197.67J\cdot K^{-1}\cdot mo{l}^{-1}+188.83J\cdot K^{-1}\cdot mo{l}^{-1}\right]\\ {\text{ΔS}}^{\text{o}}\text{=}-42.08J\cdot K^{-1}\cdot mo{l}^{-1}\end{array}$

Therefore,

  $\begin{array}{l}\Delta {\text{G}}^{\text{o}}\text{ = }\Delta H^{\text{o}}-T\Delta S^{\text{o}}\\ \text{0 = }\Delta H^{\text{o}}-T\Delta S^{\text{o}}\\ T=\frac{\Delta H^{\text{o}}}{\Delta S^{\text{o}}}\\ T=\frac{-41.16kJ\cdot mo{l}^{-1}}{-42.08J\cdot K^{-1}\cdot mo{l}^{-1}}\\ T=\frac{-41.16kJ\cdot mo{l}^{-1}\text{( 1 000 J }{\text{. kJ}}^{\text{-1}}\text{ )}}{-42.08J\cdot K^{-1}\cdot mo{l}^{-1}}\\ T=978K\end{array}$

The temperature of the reaction is $978K$.

(c)

Interpretation Introduction

Interpretation:

When a cylinder is filled with $CO\left(g\right)at10.00bar,H_{2}O\left(g\right)at10.00bar,H_2\left(g\right)at5.00bar,andC{O}_2\left(g\right)at5.00bar$, the partial pressure of the gases has be calculated.

Explanation of Solution

The given reaction is shown below,

  ${\text{CO(g) + H}}_{\text{2}}\text{O(g) }\to {\text{ CO}}_{\text{2}}{\text{(g) + H}}_{\text{2}}\text{(g)}$

The reaction can be expressed as follow,

ICE table:

               $\text{CO(g) }\text{+}{\text{ H}}_{\text{2}}\text{O(g)}\to {\text{CO}}_{\text{2}}\text{(g) }\text{+}{\text{ H}}_{\text{2}}\text{(g)}$

Initial concentration	$10.00bar$	$10.00bar$	$5.00bar$	$5.00bar$
Change	$-\alpha$	$-\alpha$	$+\alpha$	$+\alpha$
At equilibrium	$10.0-\alpha$	$10.0-\alpha$	$5.0+\alpha$	$5.0+\alpha$

  $\begin{array}{l}\text{CO(g) }\text{+}{\text{ H}}_{\text{2}}\text{O(g)}\to {\text{CO}}_{\text{2}}\text{(g) }\text{+}{\text{ H}}_{\text{2}}\text{(g)}\\ \alpha \text{ =}\frac{\left[{\text{CO}}_{\text{2}}\right]\left[{\text{H}}_{\text{2}}\right]}{\left[\text{CO}\right]\left[{\text{H}}_{\text{2}}\text{O}\right]}\\ \alpha \text{ =}\frac{\left(5\right)\left(5\right)}{\left(10\right)\left(10\right)}\\ \alpha \text{ =}\text{+}\text{2}\text{.50}\text{bar}\end{array}$

The partial pressures of $CO,H_2\text{O,}{\text{CO}}_{\text{2 }}and{H}_2$ is given below,

  $\begin{array}{l}\left[CO\right]\text{ =}10.0-\alpha \\ \left[CO\right]\text{ =}10.0-0.25bar\\ \left[CO\right]\text{ =}0.75bar\end{array}$

  $\begin{array}{l}\left[H_{2}O\right]\text{ =}10.0-\alpha \\ \left[H_{2}O\right]\text{ =}10.0-0.25bar\\ \left[H_{2}O\right]\text{ =}0.75bar\end{array}$

  $\begin{array}{l}\left[C{O}_2\right]\text{ =}5.00+\alpha \\ \left[C{O}_2\right]\text{ =}5.00+0.25bar\\ \left[C{O}_2\right]\text{ =}0.75bar\end{array}$

  $\begin{array}{l}\left[H_2\right]\text{ =}5.00+\alpha \\ \left[H_2\right]\text{ =}5.00+0.25bar\\ \left[H_2\right]\text{ =}0.75bar\end{array}$

Therefore, all the pressure is equal to $\text{+}7.\text{50}\text{bar}$

(d)

Interpretation Introduction

Interpretation:

When a cylinder is filled with $\text{CO(g) at 6}{\text{.00 bar, H}}_{\text{2}}\text{O(g) at 4}{\text{.00 bar, H}}_{\text{2}}\text{(g) at 5}{\text{.00 bar, and CO}}_{\text{2}}\text{(g) at 10}\text{.00 bar,}$, the partial pressure of the gases has be calculated.

Explanation of Solution

The given reaction is shown below,

  ${\text{CO(g) + H}}_{\text{2}}\text{O(g) }\to {\text{ CO}}_{\text{2}}{\text{(g) + H}}_{\text{2}}\text{(g)}$

Q value is calculated as follows,

  $\begin{array}{l}\text{Q =}\frac{\left[{\text{CO}}_{\text{2}}\right]\left[{\text{H}}_{\text{2}}\right]}{\left[\text{CO}\right]\left[{\text{H}}_{\text{2}}\text{O}\right]}\\ \text{Q =}\frac{\left(10.0\right)\left(5.0\right)}{\left(6.0\right)\left(4.0\right)}\\ \text{Q =}\text{2}\text{.08}\end{array}$

The Q value is greater than one, hence the reaction shift produce reactants.

ICE table:

               $\text{CO(g) }\text{+}{\text{ H}}_{\text{2}}\text{O(g)}\to {\text{CO}}_{\text{2}}\text{(g) }\text{+}{\text{ H}}_{\text{2}}\text{(g)}$

Initial concentration	$6.00bar$	$4.00bar$	$10.00bar$	$5.00bar$
Change	$+\alpha$	$+\alpha$	$-\alpha$	$-\alpha$
At equilibrium	$6.00+\alpha$	$4.0+\alpha$	$10.0-\alpha$	$5.0-\alpha$

Pressure,

  $\begin{array}{l}\text{1 =}\frac{\left(\text{10}\text{.0}\text{-}\text{α}\right)\left(\text{5}\text{.0-}\text{α}\right)}{\left(\text{6}\text{.0+}\text{α}\right)\left(\text{4}\text{.0+}\text{α}\right)}\\ \left(\text{6}\text{.0+}\text{α}\right)\left(\text{4}\text{.0+}\text{α}\right)\text{=}\left(\text{10}\text{.0}\text{-}\text{α}\right)\left(\text{5}\text{.0-}\text{α}\right)\\ {\text{α}}^{\text{2}}\text{-15}\text{.0α}\text{+}\text{50}\text{.0}{\text{=α}}^{\text{2}}\text{+10}\text{.0α}\text{+}\text{24}\text{.0}\\ \text{25}\text{.0α}\text{=}\text{26}\text{.0}\\ \text{α}\text{=}\text{1}\text{.04}\text{bar}\end{array}$

The partial pressures of $CO,H_2\text{O,}{\text{CO}}_{\text{2 }}and{H}_2$ is given below,

  $\begin{array}{l}\left[CO\right]\text{ =}6.0+\alpha \\ \left[CO\right]\text{ =}6.0+1.04bar\\ \left[CO\right]\text{ =}7.04bar\end{array}$

  $\begin{array}{l}\left[H_{2}O\right]\text{ =}4.0+\alpha \\ \left[H_{2}O\right]\text{ =}4.0+1.04bar\\ \left[H_{2}O\right]\text{ =}5.04bar\end{array}$

  $\begin{array}{l}\left[C{O}_2\right]\text{ =}10.00-\alpha \\ \left[C{O}_2\right]\text{ =}10.00-1.04bar\\ \left[C{O}_2\right]\text{ =}8.96bar\end{array}$

  $\begin{array}{l}\left[H_2\right]\text{ =}5.00-\alpha \\ \left[H_2\right]\text{ =}5.00-1.04bar\\ \left[H_2\right]\text{ =}3.96bar\end{array}$

Therefore, all the pressure is calculated.