Question

For the decomposition of the compound, represented as:

$N{H}_{2}COON{H}_4\left(s\right)\rightleftharpoons 2N{H}_3\left(g\right)+C{O}_2\left(g\right)$

The $K_P=2.9\times {10}^{-5}at{m}^3$. If the reaction is started with $1mol$ of the compound, the total pressure at equilibrium would be:

• A. $1.94\times {10}^{-2}atm$
• B. $5.80\times {10}^{-2}atm$
• C. $7.66\times {10}^{-2}atm$
• D. $38.8\times {10}^{-2}atm$

Answer 1

The correct option is B $5.80\times {10}^{-2}atm$

The balanced chemical equation is

$N{H}_{2}COON{H}_4\left(s\right)\rightleftharpoons 2N{H}_3\left(g\right)+C{O}_2\left(g\right)$.

Initially 1 mole of $N{H}_{2}COON{H}_4$ is present.
At equilibrium, the number of moles of $N{H}_{2}COON{H}_4$, ammonia and carbon dioxide is $1-x$, $2x$ and $x$ respectively.

Let P be the total pressure at equilibrium.
Then the equilibrium partial pressures of ammonia and carbon dioxide are $\left(\frac{2}{3}P\right)$ atm and $\left(\frac{1}{3}P\right)$ atm respectively.
The expression for the equilibrium constant is

$K_P=P_{N{H}_3}^2\times P_{C{O}_2}$

Substitute values in th above equation.

$2.9\times {10}^{-5}=(\frac{2}{3}P{)}^2\times (\frac{1}{3}P)$

$P^3=2.9\times {10}^{-5}\times \frac{27}{4}$

$P=5.8\times {10}^{-2}atm$.

Hence, the correct option is $\left(B\right)$