Chapter 15, Problem 103AP

For which of the following reactions is $K_c\text{ equal to }{K}_p$ ? For which can we not write a $K_p$ expression?

$\begin{array}{l}\text{(a)}\\ \text{NH3(}g{\text{)+5O}}_{\text{2}}\text{(}g\text{)}\underset{}{\rightleftarrows }\text{4NO(}g{\text{)+6H}}_{\text{2}}\text{O(}g\text{)}\\ \text{(b) }\\ {\text{CaCO}}_{\text{3}}\text{(}s\text{)}\underset{}{\rightleftarrows }\text{CaO(}s{\text{)+CO}}_{\text{2}}\text{(}g\text{)}\end{array}$

$\begin{array}{l}(c)\text{ Zn(}s{\text{)+2H}}^{\text{+}}(aq)\underset{}{\rightleftarrows }{\text{Zn}}^{\text{2+}}(aq)+{\text{H}}_{\text{2}}(g)\\ (d){\text{ PCl}}_{\text{3}}\text{(}g{\text{)+3NH}}_{\text{3}}\text{(}g\text{)}\underset{}{\rightleftarrows }3\text{HCl}(g)+\text{P}{\left({\text{NH}}_{\text{2}}\right)}_{\text{3}}(g)\\ (e){\text{ NH}}_{\text{3}}(g)\text{+HCl(}g\text{)}\underset{}{\rightleftarrows }{\text{NH}}_{\text{4}}\text{Cl(}s\text{)}\end{array}$

$\begin{array}{l}\text{(f)}\\ {\text{NaHCO}}_{\text{3}}\text{(}s{\text{)+H}}^{\text{+}}\text{(}aq\text{)}\underset{}{\rightleftarrows }{\text{H}}_{\text{2}}\text{O}(l)+C{O}_2(g)+N{a}^{+}(aq)\\ (g)\\ H_2(g)+F_2(g)\underset{}{\rightleftarrows }2\text{HF}(g)\\ (h)\\ C(graphite)+C{O}_2(g)\underset{}{\rightleftarrows }2\text{CO}(g)\end{array}$

Interpretation Introduction

Interpretation:

The given reactions where $K_p\text{ and }{K}_c$ have equal values are to be determined. The reactions for which $K_p$ expressions cannot be written are also to be determined.

Concept introduction:

Equilibrium is the condition of a chemical system in which the amount of reactants consumed and the amount of products formed is equal.

The relation between $K_p$ and $K_c$ is represented as follows:

$K_p\text{= }{K}_c{\left(\text{R}T\right)}^{\text{Δn}}$

Here, $K_p$ is the equilibrium constant in terms of concentration, $K_{\text{c}}$ is the equilibrium constant in terms of concentration, $\text{R}$ is the universal gas constant, $T$ is the absolute temperature, and $\Delta n$ is the difference between the number of moles of products and the number of moles of reactants.

The values of $K_p$ and $K_c$ are equal for the reactions where there is no change in the number of moles of product and reactant, that is, $\Delta n=0$.

Answer to Problem 103AP

Solution:

a)

$K_p\ne K_c$

b)

$K_p\ne K_c$, and the expression for $K_p$ cannot be written.

c)

The equation of $K_p\text{ and }{K}_c$ is not applicable.

d)

$K_p=K_c$

e)

$K_p\ne K_c$, and the expression for $K_p$ cannot be written.

f)

The equation of $K_p\text{ and }{K}_c$ is not applicable.

g)

$K_p=K_c$.

h)

$K_p\ne K_c$, and the expression for $K_p$ cannot be written.

Explanation of Solution

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

The change in the number of moles of product and reactant is calculated as follows:

$\begin{array}{c}\Delta n=(6+4)-(4+5)\\ =10-9\\ =+1\end{array}$

Therefore, $\text{Δ}n\ne \text{ 0 }$. Hence, $K_p\ne K_c$.

Hence, the $K_p$ expressions can be written for the reaction because all the reactants are in gaseous state.

Explanation:

b) ${\text{CaCO}}_3\left(s\right)\rightleftharpoons \text{CaO}\left(s\right)+{\text{CO}}_2\left(g\right)$

The change in the number of moles of product and reactant is calculated as follows:

$\begin{array}{c}\text{Δ}n\text{ = (1)}-\text{(0)}\\ \text{ = 1}\end{array}$

Therefore, $\text{Δ}n\ne \text{ 0 }$. Hence, $K_p\ne K_c$.

Hence, the $K_p$ expressions cannot be written for the reaction because all the reactants are not in gaseous state.

Explanation:

c) $\text{Zn}\left(s\right)+2{\text{H}}^{+}\left(aq\right)\rightleftharpoons {\text{Zn}}^{+2}\left(aq\right)+{\text{H}}_2\left(g\right)$

The equation for $K_p$ and $K_c$ is not applicable for the reaction as some of the reactants and products are aqueous solutions.

Thus, the $K_p$ expressions can also be not written for the reaction.

Explanation:

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

The change in the number of moles of product and reactant is as follows:

$\begin{array}{c}\Delta n=(4)-(4)\\ =0\end{array}$

Therefore, $\Delta n=0$. Hence, $K_p=K_c$.

Hence, the $K_p$ expressions can be written for the reaction because all the reactants are in gaseous state.

Explanation:

e) ${\text{NH}}_3\left(g\right)+\text{HCl}\left(g\right)\rightleftharpoons {\text{NH}}_4\text{Cl}\left(s\right)$

The change in the number of moles of product and reactant is as follows:

$\begin{array}{c}\Delta n=(0)-(2)\\ =-2\end{array}$

Therefore, $\text{Δ}n\ne \text{ 0 }$. Hence, $K_p\ne K_c$.

Hence, the $K_p$ expression cannot be written for this reaction as the product formed is in solid state.

Explanation:

f) ${\text{NaHCO}}_3\left(s\right)+{\text{H}}^{+}\left(aq\right)\rightleftharpoons {\text{H}}_2\text{O}\left(l\right)+{\text{CO}}_2\left(g\right)+{\text{Na}}^{+}\left(aq\right)$

The equation for $K_p$ and $K_c$ is not applicable for the reaction as some of the reactants and products are aqueous solutions.

Thus, the $K_p$ expression can also be not written for this reaction.

Explanation:

g) ${\text{H}}_2\left(g\right)+{\text{F}}_2\left(g\right)\rightleftharpoons 2\text{HF}\left(g\right)$

The change in the number of moles of product and reactant is:

$\begin{array}{c}\Delta n=(2)-(2)\\ =0\end{array}$

Therefore, $\Delta n=0$. Hence, $K_p=K_c$.

$K_p$ expression can be written for this reaction as all the reactants and products are in gaseous state.

Explanation:

h) $\text{C}\left(\text{graphite}\right)+{\text{CO}}_2\left(g\right)\rightleftharpoons 2\text{CO}\left(g\right)$

The change in the number of moles of product and reactant is:

$\begin{array}{c}\Delta n=(2)-(1)\\ =1\end{array}$

Therefore, $\Delta n=1$. Hence, $K_p\ne K_c$.

Hence, the $K_{\text{p}}$ expression cannot be written for this reaction as graphite exists in solid state.