Chapter 5, Problem 5I.16E

(a)

Interpretation Introduction

Interpretation:

The value of $K$ for the reaction $2NaHC{O}_3\left(s\right)\rightleftharpoons N{a}_{2}C{O}_3\left(s\right)+C{O}_2\left(g\right)+H_{2}O\left(g\right)$ has to be calculated.

Explanation of Solution

The required reaction is given below.

  $2NaHC{O}_3\left(s\right)\rightleftharpoons N{a}_{2}C{O}_3\left(s\right)+C{O}_2\left(g\right)+H_{2}O\left(g\right)$

The expression for equilibrium constant for the above reaction can be written as shown below,

  $K=\left(P_{C{O}_2}\right)\left(P_{H_{2}O}\right)$

The change in pressure of any solid substance present in a reaction is always taken to be one.

An equilibrium table can be set up as given below.

  $\begin{array}{ccccc}Composition\left(bar\right)& NaHC{O}_3& N{a}_{2}C{O}_3& C{O}_2& H_{2}O\\ Initialpressure& -& -& 0& 0\\ Changeinpressure& -& -& +x& +x\\ Equilibriumpressure& -& -& x& x\end{array}$

Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below.

  $K=\left(P_{C{O}_2}\right)\left(P_{H_{2}O}\right)=x^2$

Given that, the total pressure in the container is $7.68bar.$

  $\begin{array}{l}x+x=7.68bar\\ \\ x=3.84bar\end{array}$

Now, by plugging all data in the equilibrium constant expression, the value of $K$ can be calculated.

  $\begin{array}{l}K={\left(x\right)}^2={\left(3.84atm\right)}^2\\ \\ K=14.74at{m}^2\end{array}$

Therefore, the value of $K$ for the reaction $2NaHC{O}_3\left(s\right)\rightleftharpoons N{a}_{2}C{O}_3\left(s\right)+C{O}_2\left(g\right)+H_{2}O\left(g\right)$ is $14.74at{m}^2.$

(b)

Interpretation Introduction

Interpretation:

The equilibrium concentrations of $C{O}_{2}and{H}_{2}O$ has to be calculated.

Explanation of Solution

The required reaction is given below.

  $2NaHC{O}_3\left(s\right)\rightleftharpoons N{a}_{2}C{O}_3\left(s\right)+C{O}_2\left(g\right)+H_{2}O\left(g\right)$

The change in pressure of any solid substance present in a reaction is always taken to be one.

An equilibrium table can be set up as given below.

  $\begin{array}{ccccc}Composition\left(bar\right)& NaHC{O}_3& N{a}_{2}C{O}_3& C{O}_2& H_{2}O\\ Initialpressure& -& -& 1& 0\\ Changeinpressure& -& -& +x& +x\\ Equilibriumpressure& -& -& 1+x& x\end{array}$

Given that, the total pressure in the container is $7.68bar.$

  $\begin{array}{l}1+x+x=7.68bar\\ \\ x=3.34bar\end{array}$

Now, the equilibrium partial pressure of $C{O}_{2}and{H}_{2}O$ can be obtained as shown below.

  $\begin{array}{l}{P}_{C{O}_2}=1+x=1+3.34=4.34bar\\ \\ P_{H_{2}O}=x=3.34bar\end{array}$

The equilibrium concentration of $C{O}_{2}and{H}_{2}O$ can be obtained as shown below.

  $\begin{array}{l}PV=nRT\\ \\ n\left(C{O}_2\right)=\frac{PV}{RT}=\frac{\left(4.34bar\right)\times \left(2.50L\right)}{\left(8.3145\times {10}^{-2}L.bar.K^{-1}.mo{l}^{-1}\right)\left(160+273\right)K}\\ \\ \text{n}\left({\text{CO}}_{\text{2}}\right)\text{=0}\text{.30}\text{mol}\\ \\ \left[C{O}_2\right]=\frac{\text{0}\text{.30}\text{mol}}{2.50L}=0.12M\\ \\ n\left({\text{H}}_{\text{2}}\text{O}\right)=\frac{PV}{RT}=\frac{\left(3.34bar\right)\times \left(2.50L\right)}{\left(8.3145\times {10}^{-2}L.bar.K^{-1}.mo{l}^{-1}\right)\left(160+273\right)K}\\ \\ \text{n}\left({\text{H}}_{\text{2}}\text{O}\right)\text{=0}\text{.23}\text{mol}\\ \\ \left[H_{2}O\right]=\frac{\text{0}\text{.23}\text{mol}}{2.50L}=0.092M\end{array}$

Therefore, the equilibrium concentration of $C{O}_{2}and{H}_{2}O$ is $0.12Mand0.092M$ respectively.