Chapter 5, Problem 5H.5E

(a)

Interpretation Introduction

Interpretation:

The value of $K_c$ for $2NOCl\left(g\right)\rightleftharpoons 2NO\left(g\right)+C{l}_2\left(g\right)$ at $500K$ has to be calculated.

Concept Introduction:

The relationship between $Kand{K}_c$ is given below.

  $K={\left(\frac{c^{\circ }RT}{P^{\circ }}\right)}^{\Delta n_r}{K}_c$

Where, $c^{\circ }$ is the concentration, $R$ is the universal gas constant, $T$ is the absolute temperature, $P^{\circ }$ is the pressure and $\Delta n_r=n_{r,products}-n_{r,reactants}$

Answer to Problem 5H.5E

The value of $K_c$ for $2NOCl\left(g\right)\rightleftharpoons 2NO\left(g\right)+C{l}_2\left(g\right)$ at $500K$ is $4.33\times 10^{-4}.$

Explanation of Solution

The required reaction is as given below.

  $2NOCl\left(g\right)\rightleftharpoons 2NO\left(g\right)+C{l}_2\left(g\right)K=1.8\times {10}^{-2}$

The relationship between $Kand{K}_c$ is given below.

  $\begin{array}{l}K={\left(\frac{c^{\circ }RT}{P^{\circ }}\right)}^{\Delta n_r}{K}_c\\ \\ K_c={\left(\frac{c^{\circ }RT}{P^{\circ }}\right)}^{-\Delta n_r}K\\ \\ K_c={\left(\frac{T}{P^o/c^{o}R}\right)}^{-\Delta n_r}K\end{array}$

The value ${\text{P}}^{\text{o}}{\text{/c}}^{\text{o}}\text{R}$ can be calculated as given below.

  $\frac{P^{\circ }}{R{c}^{\circ }}=\frac{1bar}{\left(8.3145\times {10}^{-2}L.bar.K^{-1}mo{l}^{-1}\right)\times \left(1mol.L^{-1}\right)}=12.03K$

Now, the value of $K_c$ can be calculated as given below.

  $\begin{array}{l}\Delta n_r=n_{r,products}-n_{r,\text{reactants}}=3-2=1\\ \\ {\text{K}}_{\text{c}}\text{=}{\left(\frac{\text{T}}{{\text{P}}^{\text{o}}{\text{/c}}^{\text{o}}\text{R}}\right)}^{{\text{-Δn}}_{\text{r}}}\text{K}\\ \\ {\text{K}}_{\text{c}}\text{=}{\left(\frac{500}{12.03}\right)}^{-\left(1\right)}\times 1.8\times {10}^{-2}=4.33\times {10}^{-4}\\ \\ K_c=4.33\times 10^{-4}\end{array}$

Therefore, the value of $K_c$ for $2NOCl\left(g\right)\rightleftharpoons 2NO\left(g\right)+C{l}_2\left(g\right)$ at $500K$ is $4.33\times 10^{-4}.$

(b)

Interpretation Introduction

Interpretation:

The value of $K_c$ for $CaC{O}_3\left(s\right)\rightleftharpoons CaO\left(s\right)+C{O}_2\left(g\right)$ at $1073K$ has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 5H.5E

The value of $K_c$ for $CaC{O}_3\left(s\right)\rightleftharpoons CaO\left(s\right)+C{O}_2\left(g\right)$ at $1073K$ is $1.87.$

Explanation of Solution

The required reaction is as given below.

  $CaC{O}_3\left(s\right)\rightleftharpoons CaO\left(s\right)+C{O}_2\left(g\right)K=167$

The relationship between $Kand{K}_c$ is given below.

  $\begin{array}{l}K={\left(\frac{c^{\circ }RT}{P^{\circ }}\right)}^{\Delta n_r}{K}_c\\ \\ K_c={\left(\frac{c^{\circ }RT}{P^{\circ }}\right)}^{-\Delta n_r}K\\ \\ K_c={\left(\frac{T}{P^o/c^{o}R}\right)}^{-\Delta n_r}K\end{array}$

The value ${\text{P}}^{\text{o}}{\text{/c}}^{\text{o}}\text{R}$ can be calculated as given below.

  $\frac{P^{\circ }}{R{c}^{\circ }}=\frac{1bar}{\left(8.3145\times {10}^{-2}L.bar.K^{-1}mo{l}^{-1}\right)\times \left(1mol.L^{-1}\right)}=12.03K$

Now, the value of $K_c$ can be calculated as given below.

  $\begin{array}{l}\Delta n_r=n_{r,products}-n_{r,\text{reactants}}=1-0=1\\ \\ {\text{K}}_{\text{c}}\text{=}{\left(\frac{\text{T}}{{\text{P}}^{\text{o}}{\text{/c}}^{\text{o}}\text{R}}\right)}^{{\text{-Δn}}_{\text{r}}}\text{K}\\ \\ {\text{K}}_{\text{c}}\text{=}{\left(\frac{1073}{12.03}\right)}^{-\left(1\right)}\times 167=1.87\\ \\ K_c=1.87\end{array}$

Therefore, the value of $K_c$ for $CaC{O}_3\left(s\right)\rightleftharpoons CaO\left(s\right)+C{O}_2\left(g\right)$ at $1073K$ is $1.87.$