Chapter 14.2, Problem 1RCF

For which of the following reactions is K_c equal to K_P?

(a)     ${\text{4NH}}_3(g)+{\text{5O}}_2(g)\rightleftharpoons \text{4NO}(g)+{\text{6H}}_2\text{O}(g)$

(b)     ${\text{2H}}_2{\text{O}}_2(aq)\rightleftharpoons {\text{2H}}_{\text{2}}\text{O}(l)+{\text{O}}_{\text{2}}(g)$

(c)     ${\text{PCl}}_3(g)+{\text{3NH}}_3(g)\rightleftharpoons \text{3HCl}(g)+\text{P}{({\text{NH}}_2)}_3(g)$

Interpretation Introduction

Interpretation:

For the given reactions, value of equilibrium constant ${\text{K}}_{\text{c}}$ is equal to ${\text{K}}_{\text{P}}$ has to be predicted.

Concept introduction:

Law of mass action: The rate of chemical reaction is directly proportional to the product of concentrations of reactant to products.

$\begin{array}{l}\text{aA}\text{+}\text{bB}\to \text{cC}\text{+}\text{dD}\\ \\ {\text{K}}_{eq}\text{=}\frac{{\left[\text{C}\right]}^{\text{c}}{\left[\text{D}\right]}^{\text{d}}}{{\left[\text{A}\right]}^{\text{a}}{\left[\text{B}\right]}^{\text{b}}}\text{for}\text{aqueous}\\ \\ {\text{K}}_{\text{eq}}\text{=}\frac{{\left({\text{P}}_{\text{C}}\right)}^{\text{c}}{\left({\text{P}}_{\text{D}}\right)}^{\text{d}}}{{\left({\text{P}}_{\text{A}}\right)}^{\text{a}}{\left({\text{P}}_{\text{B}}\right)}^{\text{b}}}\text{for}\text{gases}\\ \end{array}$

Relation between $K_c$ and$K_P$:

$\begin{array}{l}{\text{K}}_{\text{P}}\text{=}{\text{K}}_{\text{C}}{\left(\text{RT}\right)}^{\text{Δn}}\\ \text{where,}\\ \text{Δn}\text{is}\text{moles}\text{of}\text{products}\text{-}\text{moles}\text{of}\text{rectants}\\ \text{R}\text{is}\text{Gas}\text{Constant}\\ \text{T}\text{is}\text{Temperature}\text{in}\text{Kelvin}\end{array}$

Explanation of Solution

Reason for correct option:

• For the given reaction (c),

${\text{PCl}}_{\text{3}}\left(\text{g}\right){\text{ + 3NH}}_3\left(\text{g}\right)\underset{}{\rightleftharpoons }\text{3HCl}\left(\text{g}\right)\text{+}\text{P}{\left({\text{NH}}_2\right)}_3\left(\text{g}\right)$

The ratio of mass action can be written as,

$\begin{array}{l}{\text{PCl}}_{\text{3}}\left(\text{g}\right){\text{ + 3NH}}_3\left(\text{g}\right)\underset{}{\rightleftharpoons }\text{3HCl}\left(\text{g}\right)\text{+}\text{P}{\left({\text{NH}}_2\right)}_3\left(\text{g}\right)\\ {\text{K}}_{\text{P}}\text{=}\frac{{\left({\text{P}}_{\text{HCl}}\right)}^3\left({\text{P}}_{\text{P}{\left({\text{NH}}_2\right)}_3}\right)}{{\left({\text{P}}_{{\text{PCl}}_{\text{3}}}\right)}^1{\left({\text{P}}_{{\text{NH}}_3}\right)}^3}{\text{K}}_{\text{C}}\text{=}\frac{{\left[\text{HCl}\right]}^3\left[\text{P}{\left({\text{NH}}_2\right)}_3\right]}{{\left[{\text{PCl}}_{\text{3}}\right]}^1{\left[{\text{NH}}_3\right]}^3}\end{array}$

As the relation known,

 ${\text{K}}_{\text{P}}\text{=}{\text{K}}_{\text{C}}{\left(\text{RT}\right)}^{\text{Δn}}$

Considering T and R remains constant, the change in moles varies,

Therefore for reaction (c),

Change in moles = $\text{moles}\text{of}\text{products}\text{-}\text{moles}\text{of}\text{rectants}$

$\begin{array}{l}\Delta n=\left(3+1\right)-\left(3+1\right)\\ =0\end{array}$

The relation becomes, $\begin{array}{l}{\text{K}}_{\text{P}}\text{=}{\text{K}}_{\text{C}}{\left(\text{RT}\right)}^0\\ K_P=K_C\end{array}$

Therefore, $K_P=K_C$

Hence, the correct option is (c).

Reason for in-correct option:

• For the given reaction (a),

${\text{4NH}}_{\text{3}}\left(\text{g}\right){\text{ + 5O}}_{\text{2}}\left(\text{g}\right)\underset{}{\rightleftharpoons }\text{NO}\left(\text{g}\right)\text{+}{\text{6H}}_{\text{2}}\text{O}\left(\text{g}\right)$

The ratio of mass action can be written as,

${\text{K}}_{\text{P}}\text{=}\frac{{\left({\text{P}}_{\text{NO}}\right)}^1}{{\left({\text{P}}_{{\text{NH}}_{\text{3}}}\right)}^4{\left({\text{P}}_{{\text{O}}_{\text{2}}}\right)}^5}{\text{K}}_{\text{C}}\text{=}\frac{{\left[\text{NO}\right]}^1}{{\left[{\text{NH}}_{\text{3}}\right]}^4{\left[{\text{O}}_{\text{2}}\right]}^5}$

Water (also act as solvents in reaction) in homogeneous equilibria does not appear in the equilibrium constant expressions.

As the relation known,

 ${\text{K}}_{\text{P}}\text{=}{\text{K}}_{\text{C}}{\left(\text{RT}\right)}^{\text{Δn}}$

Considering T and R remains constant, the change in moles varies,

Therefore for reaction (a),

Change in moles = $\text{moles}\text{of}\text{products}\text{-}\text{moles}\text{of}\text{rectants}$

$\begin{array}{l}\Delta n=\left(6+1\right)-\left(5+4\right)\\ =7-9=-2\end{array}$

The relation becomes, ${\text{K}}_{\text{P}}\text{=}{\text{K}}_{\text{C}}{\left(\text{RT}\right)}^{-2}$

Therefore, $K_P\ne K_C$

• For the given reaction (b),

${\text{2H}}_2{\text{O}}_2\left(\text{aq}\right)\underset{}{\rightleftharpoons }{\text{2H}}_{\text{2}}\text{O}\left(\text{l}\right){\text{+ O}}_{\text{2}}\left(\text{g}\right)$

The ratio of mass action can be written as,

$\begin{array}{l}{\text{2H}}_2{\text{O}}_2\left(\text{aq}\right)\underset{}{\rightleftharpoons }{\text{2H}}_{\text{2}}\text{O}\left(\text{l}\right){\text{+ O}}_{\text{2}}\left(\text{g}\right)\\ {\text{K}}_{\text{P}}\text{=}\frac{{\left({\text{P}}_{{\text{O}}_{\text{2}}}\right)}^1}{{\left({\text{P}}_{{\text{H}}_2{\text{O}}_2}\right)}^2}{\text{K}}_{\text{C}}\text{=}\frac{{\left[{\text{O}}_{\text{2}}\right]}^1}{{\left[{\text{H}}_2{\text{O}}_2\right]}^2}\end{array}$

As the relation known,

 ${\text{K}}_{\text{P}}\text{=}{\text{K}}_{\text{C}}{\left(\text{RT}\right)}^{\text{Δn}}$

Considering T and R remains constant, the change in moles varies,

Therefore for reaction (b),

Change in moles = $\text{moles}\text{of}\text{products}\text{-}\text{moles}\text{of}\text{rectants}$

$\begin{array}{l}\Delta n=\left(2+1\right)-\left(2\right)\\ =3-2=1\end{array}$

The relation becomes, ${\text{K}}_{\text{P}}\text{=}{\text{K}}_{\text{C}}{\left(\text{RT}\right)}^1$

Therefore, $K_P\ne K_C$

Hence, the in-correct options are (a and b).