Chapter 5, Problem 5.55E

(a)

Interpretation Introduction

Interpretation:

When Gibbs free energies of formation of $CO\left(g\right)and{H}_{2}O\left(g\right)at900Kare\text{ -}191.28kJ\cdot mo{l}^{-1}and\text{ -}198.08kJ\cdot mo{l}^{-1}$, K has to be calculated.

Concept introduction:

The equilibrium constant can be calculated by using following formula,

    $\begin{array}{l}\Delta G_r^o=-RT\ln K\\ \Delta G_r^o=Gibbsfreeenergy\\ R=\text{Gas}\text{constant}\\ \text{T}\text{=}\text{Temperature}\end{array}$

Explanation of Solution

The given reaction is shown below,

  $\text{C (s)}\text{+}{\text{H}}_{\text{2}}\text{O}\text{(g)}\to \text{CO(g) }\text{ +}{\text{3H}}_{\text{2}}\text{(g)}$

The equilibrium constant can be calculated by using following formula,

    $\begin{array}{l}\Delta G_r^o=-RT\ln K\\ \ln K=-\frac{\Delta G_r^o}{RT}\\ TheGibbsfreeenergiesare\text{-}191.28kJ\cdot mo{l}^{-1}and\text{ -}198.08kJ\cdot mo{l}^{-1}.\\ \\ \Delta G_r^o=\left(\text{-}191.28kJ\cdot mo{l}^{-1}\right)-\left(\text{-}198.08kJ\cdot mo{l}^{-1}\right)\\ \Delta G_r^o=6.8kJ\cdot mo{l}^{-1}\\ \Delta G_r^o=6800J\cdot mo{l}^{-1}\end{array}$

Therefore,

    $\begin{array}{l}\ln K=-\frac{\Delta G_r^o}{RT}\\ \ln K=-\frac{6800J\cdot mo{l}^{-1}}{\left(8.314J\cdot K^{-1}\cdot mo{l}^{-1}\right)\left(900K\right)}\\ \ln K=-0.909\\ K=0.403\end{array}$

K is $0.403$.

(b)

Interpretation Introduction

Interpretation:

When the mass of graphite and water are $5.20kgand125g$ placed into a $10.0L$ container then it was heated to $900K$. The equilibrium concentrations has to be calculated

Explanation of Solution

The given reaction is shown below,

  $\text{C (s)}\text{+}{\text{H}}_{\text{2}}\text{O}\text{(g)}\to \text{CO(g) }\text{ +}{\text{3H}}_{\text{2}}\text{(g)}$

The reaction can be expressed as follow,

The reaction tells that graphite reaction with water gives carbon monoxide and three molecules of hydrogen.

Number of moles of carbon is calculated as follows,

  $\begin{array}{l}\text{ Moles =}\frac{\text{Mass}}{\text{Molar mass}}\\ \text{ Moles = }\frac{\text{5200}\text{g}}{\text{12}\text{.011}\text{g/mol}}\\ \text{ Moles = 433}\text{mol}\end{array}$

Number of moles of water is calculated as follows,

  $\begin{array}{l}\text{ Moles =}\frac{\text{Mass}}{\text{Molar mass}}\\ \text{ Moles = }\frac{125\text{g}}{\text{18}\text{.016}\text{g/mol}}\\ \text{ Moles = 6}\text{.94}\text{mol}\end{array}$

According to the mole calculation, water is the limiting reagent.

Therefore, the concentration water is calculated as follows,

  $\begin{array}{l}\text{Concentration}\text{ of}\text{water =}\frac{\text{Mumber}\text{of}\text{moles}}{\text{Volume}\text{in}\text{liter}}\\ \text{Concentration}\text{ of}\text{water =}\frac{\text{6}\text{.94}\text{mol}}{\text{10}\text{L}}\\ \text{Concentration}\text{ of}\text{water =}\text{0}\text{.694}\text{M}\end{array}$

ICE table:

    $\text{C (s)}\text{+}{\text{H}}_{\text{2}}\text{O}\text{(g)}\to \text{CO(g) }\text{ +}{\text{3H}}_{\text{2}}\text{(g)}$

Initial concentration	$0.694$	0	0
Change	$-\alpha$	$+\alpha$	$+\alpha$
At equilibrium	$0.694-\alpha$	$+\alpha$	$+\alpha$

Kc of the reaction is calculated as follows,

  $\begin{array}{l}\text{Kc =K×}{\left(\text{RT}\right)}^{\text{-1}}\\ \text{Kc =}\frac{\left(\text{0}\text{.403}\right)}{\left(\text{8}{\text{.314J×K}}^{\text{-1}}{\text{×mol}}^{\text{-1}}\right)\left(\text{900}\text{K}\right)}\\ \text{Kc =}\frac{\left(\text{0}\text{.403}\right)}{\left(\text{8}{\text{.314×10}}^{\text{-3}}{\text{K}}^{\text{-1}}{\text{×mol}}^{\text{-1}}\right)\left(\text{900}\text{K}\right)}\\ \text{Kc =}\text{0}\text{.05386}\end{array}$

Where,

  $\begin{array}{l}\text{Kc =}\frac{\left[\text{CO}\right]\left[{\text{H}}_{\text{2}}\right]}{\left[{\text{H}}_{\text{2}}\text{O}\right]}\\ \text{0}\text{.05386 =}\frac{\left(\text{x}\right)\left(\text{x}\right)}{\left(\text{0}\text{.694}\text{-x}\right)}\\ {\text{x}}^{\text{2}}\text{+ 0}\text{.05386x}\text{- 0}\text{.03738}\text{=0,}\\ \text{x=}\text{0}\text{.336}\text{M}\end{array}$

The equilibrium concentration of $H_{2}and\text{ CO}$ is given below,

  $\begin{array}{l}\left[{\text{H}}_2\right]\text{ =}0.336\\ \left[CO\right]\text{=}0.336\end{array}$

The equilibrium concentration of $H_2\text{O}$ is given below,

  $\begin{array}{l}\left[{\text{H}}_{2}O\right]\text{ =}0.694-x\\ \left[{\text{H}}_{2}O\right]\text{=}0.694-0.336\\ \left[{\text{H}}_{2}O\right]\text{=}0.358\end{array}$