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

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

1. A

0.0194 atm

2. B

0.0388 atm

3. C

0.0582 atm

4. D

0.0776 atm

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### Solution:

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

${K}_{p}={p}_{{\mathrm{NH}}_{3}}^{2}{p}_{{\mathrm{CO}}_{2}}$ i.e. $2.9×{10}^{-5}{\mathrm{atm}}^{3}=\left(2p{\right)}^{2}p$

This gives $p={\left(\frac{2.9×{10}^{-5}\mathrm{atm}}{4}\right)}^{1/3}=0.0194\mathrm{atm}$ and  