Question

For the decomposition of the compound, represented as
$N{H}_{2}COON{H}_4(s)\rightleftharpoons 2N{H}_3(g)+C{O}_2(g)$
the $K_p=2.9\times 10^{-5}at{m}^3$. If the reaction is started with $1$ mole of the compound, the total pressure at equilibrium would be:

• A. $7.66\times 10^{-2}atm$
• B. $38.8\times 10^{-2}atm$
• C. $5.82\times 10^{-2}atm$
• D. $1.94\times 10^{-2}atm$

Answer 1

Answer: (C)

In case of heterogeneous equilibrium. The achive mass of solid is unit so not consider in the expression of $K_p$ and $K_c$ $(NH{)}_2(COONH{)}_4(s)\rightleftharpoons 2(NH{)}_3(g)+(CO{)}_2(g)$ At equilibrium $\Rightarrow 2p$ and $p$ will be partial pressures of ammonia and carbon dioxide gas respectively.
$K_P={\left[P_{N{H}_3}\right]}^2\left[P_{C{O}_2}\right]=[2P{]}^2[P]=4P^3$
$4P^3=2\cdot 9\times 10^{-5}at{m}^3$
$P^3=0\cdot 725\times 10^{-5}at{m}^3=7\cdot 25\times 10^{-6}at{m}^3$
$P=\sqrt[3]{7\cdot 25\times 10^{-6}}=1\cdot 94\times 10^{-2}atm$
$\Sigma P={\left((P{)}_{\text{Total }}\right)}_{\text{equilibrium }}=2P+P=3P$
$=3\times 1\cdot 94\times 10^{-2}$ atm
$=5\cdot 82\times 10^{-2}atm$