Chapter 6, Problem 6.10P

(a)

Interpretation Introduction

Interpretation:

Equilibrium constant has to be written for given reaction and the vapour pressure of gaseous water ( ${\text{P}}_{{\text{H}}_{\text{2}}\text{O}}$) at 298K has to be calculated.

Concept introduction:

Equilibrium constant (K): In the equilibrium reaction, the ratio of concentration of the reactant and concentration of the product.  If value of K is small than 1 the reaction should be move to the left and k is more than 1 the reaction should be move to right.

Relating ${\text{ΔG}}^{\text{0}}$to K:

$\begin{array}{l}{\text{ΔG}}^{\text{0}}\text{= -RTlnK}\\ \\ {\text{K = e}}^{\left({\text{-ΔG}}^{\text{0}}\text{/RT}\right)}\end{array}$

$\begin{array}{l}{\text{ΔG}}^{\text{0}}\text{-Change in free energy}\\ \\ \text{K-Equilibrium constant}\\ \\ \text{R-Gas constant (8}\text{.314472)}\\ \\ \text{T-Temperature}\end{array}$

Answer to Problem 6.10P

$\begin{array}{l}{\text{K=P}}_{{\text{H}}_{\text{2}}\text{O}}{\text{= e}}^{{\text{-ΔG}}^{\text{0}}\text{/RT}}\\ \\ {\text{=e}}^{{\text{-(ΔH}}^{\text{0}}\text{-TΔS)/RT}}\end{array}$

The vapour pressure for given reaction is $\text{4}{\text{.7×10}}^{\text{-4}}\text{ bar}$.

Explanation of Solution

Given

Given reaction is,

$\begin{array}{l}{\text{BaCl}}_{\text{2}}{\text{.H}}_{\text{2}}{\text{O}}_{\text{(s)}}\stackrel{}{\rightleftharpoons }{\text{BaCl}}_{\text{2(s)}}\text{+}{\text{H}}_{\text{2}}{\text{O}}_{\text{(g)}}{\text{ΔH}}^{\text{0}}\text{= -63}\text{.11kJ/mol at 25°C}\\ {\text{ΔH}}^{\text{0}}\text{= +148J(K}\text{.mol) at 25°C }\end{array}$

To calculate: vapour pressure of given equation

$\text{K=}\frac{{\text{P}}_{{\text{BaCl}}_{\text{2}}}{\text{.P}}_{{\text{H}}_{\text{2}}\text{O}}}{{\text{P}}_{{\text{BaCl}}_{\text{2}}{\text{.H}}_{\text{2}}\text{O}}}$

For pure solid, the activity is equal to one.

$\begin{array}{l}{\text{K=P}}_{{\text{H}}_{\text{2}}\text{O}}{\text{= e}}^{{\text{-ΔG}}^{\text{0}}\text{/RT}}\\ \\ {\text{=e}}^{{\text{-(ΔH}}^{\text{0}}\text{-TΔS)/RT}}\end{array}$

Where,

${\text{ΔG}}^{\text{O}}{\text{= -(ΔG}}^{\text{O}}\text{-TΔS)}$

$\begin{array}{l}{\text{K= e}}^{\left\{\left(\text{63}{\text{.11×10}}^{\text{3}}\text{J/mol-(298}{\text{.15K)(148JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)\text{/}\left(\text{8}\text{.314472J/}\left(\text{K}\text{.mol}\right)\right)\left(\text{298}\text{.15K}\right)\right\}}\\ \\ \text{= 4}{\text{.7×10}}^{\text{-4}}\text{ bar}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The temperature at which gaseous water ( ${\text{P}}_{{\text{H}}_{\text{2}}\text{O}}$) over ${\text{BaCl}}_{\text{2}}{\text{.H}}_{\text{2}}{\text{O}}_{\text{(s)}}$ will be 1 bar when

${\text{ΔH}}^{\text{0}}$ and ${\text{ΔS}}^{\text{0}}$ are temperature independent has to be calculated.

Answer to Problem 6.10P

The temperature at which gaseous water ( ${\text{P}}_{{\text{H}}_{\text{2}}\text{O}}$) over ${\text{BaCl}}_{\text{2}}{\text{.H}}_{\text{2}}{\text{O}}_{\text{(s)}}$ will be 1 bar is $\text{153°C}$

Explanation of Solution

Given

Given reaction is,

$\begin{array}{l}{\text{BaCl}}_{\text{2}}{\text{.H}}_{\text{2}}{\text{O}}_{\text{(s)}}\stackrel{}{\rightleftharpoons }{\text{BaCl}}_{\text{2(s)}}\text{+}{\text{H}}_{\text{2}}{\text{O}}_{\text{(g)}}{\text{ΔH}}^{\text{0}}\text{= -63}\text{.11kJ/mol at 25°C}\\ {\text{ΔH}}^{\text{0}}\text{= +148J(K}\text{.mol) at 25°C }\end{array}$

$\text{K=}\frac{{\text{P}}_{{\text{BaCl}}_{\text{2}}}{\text{.P}}_{{\text{H}}_{\text{2}}\text{O}}}{{\text{P}}_{{\text{BaCl}}_{\text{2}}{\text{.H}}_{\text{2}}\text{O}}}$

For pure solid, the activity is equal to one.

${\text{K = P}}_{{\text{H}}_{\text{2}}\text{O}}$

$\begin{array}{l}{\text{P}}_{{\text{H}}_{\text{2}}\text{O}}{\text{=1=e}}^{{\text{-(ΔH}}^{\text{0}}\text{-TΔS)/RT}}\\ \\ {\text{ΔH}}^{\text{0}}{\text{-TΔS}}^{\text{O}}\text{must be zero}\end{array}$

$\begin{array}{l}{\text{ΔH}}^{\text{O}}{\text{-TΔS}}^{\text{O}}\text{= 0}\\ \\ \text{T=}\frac{{\text{ΔH}}^{\text{O}}}{{\text{ΔS}}^{\text{O}}}\text{= 426K=153°C}\end{array}$