Question

${NH}_{4}COO{NH}_2\left(s\right)\rightleftharpoons 2{NH}_3\left(g\right)+{CO}_2\left(g\right)$.

If equilibrium pressure is $3$ atm for the given reaction, then $K_p$ will be:

• A. $4$
• B. $27$
• C. $4/27$
• D. $1/27$

Answer 1

The correct option is A $4$

${NH}_{4}COO{NH}_2\left(s\right)\rightleftharpoons 2{NH}_3\left(g\right)+{CO}_2\left(g\right)$. If equilibrium pressure is $3$ atm for the above reaction; $K_p$ will be $4$.
Let $P_{{CO}_2}=x$ atm.

From the reaction stoichiometry,
$P_{{NH}_3}=2x$ atm.

Total pressure $=P_{{NH}_3}+P_{{CO}_2}=2x+x=3x$ atm

But equilibrium pressure is $3$ atm

$3x=3$

$x=1$
$P_{{CO}_2}=x=1$ atm.

$P_{{NH}_3}=2x=2\left(1\right)=2$ atm.
$K_p={\left[p_{{NH}_3}\right]}^2{\left[p_{{CO}_2}\right]}^{}=2^2\times 1=4$