1. Tardigrade
  2. Question
  3. Chemistry
  4. For the decomposition reaction NH2COONH4 (s) leftharpoons 2NH3(g) + CO2(g) The Kp = 2.9 × 10-5 atm3. The total pressure of gases at equilibrium when 1 mole of NH2COONH4 (s) was taken to start with would be

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Q. For the decomposition reaction
$NH_{2}COONH_{4} (s) \rightleftharpoons 2NH_{3}(g) + CO_{2}(g)$ The $Kp = 2.9 \times 10^{-5} atm^{3}$. The total pressure of gases at equilibrium when $1$ mole of $NH_{2}COONH_{4} (s)$ was taken to start with would be

Equilibrium

Solution:

$\underset{1}{{2}COOH_{4}(s)} \rightleftharpoons \underset{2}{2HH_{3}} +\underset{1}{CO_{2}(g)}$
$K_{p} = 2.9 \times 10^{-5}atm^{3}$
If the P is the total pressure at equilibrium
$K_{p} ={(\frac {2p}{3})^{2}} (\frac{P}{3})$
$\therefore P^{3} =\frac{27 \times2.9\times10^{-5}}{4} =4.69575$
$P=\sqrt[3]{1.6575} =0.058$