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For the decomposition reaction ${NH}_{2} {COONH}_{4} (s) ⇌ 2 {NH}_{3} (g) + {CO}_{2} (g)$ , the $K_{p} = 2.9 \times 10^{- 5} {atm}^{3}$ . The total pressure of gases at equilibrium when $1.0 mol$ of ${NH}_{2} {COONH}_{4} (s)$ was taken to start with would be

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a

0.0194 atm

b

0.0776 atm

c

0.0388 atm

d

0.0582 atm

answer is C.

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Detailed Solution

$\underset{1.0 mol - x}{{NH}_{2} {COONH}_{4} (s)} \underset{2 p}{2 {NH}_{3} (g)} + \underset{p}{{CO}_{2} (g)}$

$K_{p} = p_{{NH}_{3}}^{2} p_{{CO}_{2}}$ i.e. $2.9 \times 10^{- 5} {atm}^{3} = (2 p)^{2} p$

This gives $p = {(\frac{2.9 \times 10^{- 5} atm}{4})}^{1 / 3} = 0.0194 atm$ and $p_{total} = 2 p + p = 3 p = 3 \times 0.0194 atm = 0.0582 atm$

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For the decomposition reaction NH2COONH4(s)⇌2NH3(g)+CO2(g), the Kp=2.9×10−5atm3. The total pressure of gases at equilibrium when 1.0mol of NH2COONH4(s) was taken to start with would be