For the decomposition of ammonium carbamate according to the reaction $NH_2COONH_4(s) \rightleftharpoons 2NH_3(g) + CO_2(g)$, the equilibrium constant $K_p$ is $2.9 \times 10^{-2}\,\text{atm}^3$. If 1 mole of $NH_2COONH_4$ is initially present, what is the total pressure at equilib

Question

For the decomposition of ammonium carbamate according to the reaction $NH_2COONH_4(s) \rightleftharpoons 2NH_3(g) + CO_2(g)$, the equilibrium constant $K_p$ is $2.9 \times 10^{-2}\,\text{atm}^3$. If 1 mole of $NH_2COONH_4$ is initially present, what is the total pressure at equilibrium?

Accepted Answer

7.66 × 10⁻² atm. For $NH_2COONH_4(s) \rightleftharpoons 2NH_3(g) + CO_2(g)$, let the total pressure at equilibrium be $P$. Then $P_{NH_3} = \tfrac{2P}{3}$ and $P_{CO_2} = \tfrac{P}{3}$. So $K_p = (P_{NH_3})^2(P_{CO_2}) = \frac{4P^3}{27}$, which gives $P = 7.66 \times 10^{-2}\,\text{atm}$.

For the decomposition of ammonium carbamate according to the reaction \(NH_2COONH_4(s) \rightleftharpoons 2NH_3(g) + CO_2(g)\), the equilibrium constant \(K_p\) is \(2.9 \times 10^{-2}\,\text{atm}^3\). If 1 mole of \(NH_2COONH_4\) is initially present, what is the total pressure at equilibrium?

Solution

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