For the decomposition of the compound represented as NH2COONH4( s)⇌2NH3( g)+CO2( g) the Kp=2.9×10-5 atm3. If the reaction is started with 1 mol of the compound, the total pressure at equilibrium would be

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For the decomposition of the compound represented as ${NH}_{2} {COONH}_{4} (s) ⇌ 2 {NH}_{3} (g) + {CO}_{2} (g)$ the $K_{p} = 2.9 \times 10^{- 5} {atm}^{3}$ . If the reaction is started with 1 mol of the compound, the total pressure at equilibrium would be

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$5.82 \times 10^{- 2} atm$

$38.8 \times 10^{- 2} atm$

$7.66 \times 10^{- 2} atm$

$1.94 \times 10^{- 2} atm$

answer is B.

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Detailed Solution

$\begin{array}{l} {NH}_{2} {COONH}_{4} (s) ⇌ 2 {NH}_{3} (g) + {CO}_{2} (g) \\ K_{p} = {(p_{{NH}_{3}})}^{2} (p_{{CO}_{2}}) = (2 p)^{2} (p) = 4 p^{3} \\ p = {(\frac{K_{p}}{4})}^{1 / 3} = {(\frac{2.9 \times 10^{- 5} {atm}^{3}}{4})}^{1 / 3} = 0.01935 atm \\ p_{total} = p_{{NH}_{3}} + p_{{CO}_{2}} = 2 p + p = 3 p = 3 \times 0.01935 atm = 5.81 \times 10^{- 2} atm \end{array}$