Chapter 17, Problem 45A

(a)

Interpretation Introduction

Interpretation:

The equilibrium constant expression of given reaction is to be expressed.

Concept introduction:

The equilibrium constant is the ratio of molar concentration of products to the reactants.

Answer to Problem 45A

The equilibrium constant is given as $\text{ Keq=}\frac{\left(\left[{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\right]{\left[{\text{CO}}_{\text{2}}\right]}^{}\right)}{\left({\left[{\text{NaHCO}}_{\text{3}}\right]}^{\text{2}}\right)}$

$\text{Keq=}{\left[{\text{CO}}_{\text{2}}\right]}^{}$

Explanation of Solution

The equilibrium constant is the ratio of molar concentration of products to the reactants. It is represented by $\text{Keq}$.

The given chemical reaction is,

${\text{2NaHCO}}_{\text{3}}\left(\text{g}\right)\rightleftharpoons {\text{ Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{s}\right){\text{ + CO}}_{\text{2 }}\left(\text{g}\right)$

$\text{ Keq=}\frac{\left(\left[{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\right]{\left[{\text{CO}}_{\text{2}}\right]}^{}\right)}{\left({\left[{\text{NaHCO}}_{\text{3}}\right]}^{\text{2}}\right)}$

In case of heterogeneous reaction, the concentration of pure solids and liquid is not included. By convention these considered as unity.

So, $\text{Keq}$

$\text{= }\left[{\text{CO}}_{\text{2}}\right]$

(b)

Interpretation Introduction

Interpretation:

The equilibrium constant expression of given reaction is to be expressed..

Concept introduction:

The equilibrium constant is the ratio of molar concentration of products to the reactants.

Answer to Problem 45A

The equilibrium constant is given as $\text{Keq = }\frac{\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{6}}\right]\left(\text{l}\right)}{\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{6}}\right]\left(\text{g}\right)}$

$\text{ = }\frac{\text{1}}{\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{6}}\right]}$

Explanation of Solution

The equilibrium constant is the ratio of molar concentration of products to the reactants. It is represented by $\text{Keq}$.

The given chemical reaction is,

${\text{ C}}_{\text{6}}{\text{H}}_{\text{6 }}\left(\text{l}\right)\underset{}{\rightleftharpoons }{\text{C}}_{\text{6}}{\text{H}}_{\text{6 }}\left(\text{g}\right)$

In case of heterogeneous reaction, the concentration of pure solids and liquid is not included. By convention these considered as unity.

So, concentration of liquid benzene is unity.

Hence, $\text{Keq}$

$\text{=}\frac{\text{1}}{\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{6}}\right]}$