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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Chemistry
Chapter
Chemical Equilibrium

NTA Abhyas 2020 · Chemistry · Chemical Equilibrium

Ammonium carbonate decomposes as N H_ 2 COON H_ 4 (s) ⇌ 2N H_ 3 (g)...

Ammonium carbonate decomposes as

\text{N}{\text{H}_{2}\text{C}\text{O}\text{O}\text{N}{\text{H}_{4}\left(\text{s}\right) ⇌ 2\text{N}{\text{H}_{3}\left(\text{g}\right) + \text{C}{\text{O}_{2} ( \text{g} )

For the reaction, {\text{K}_{\text{p}} = 2.9 \times {10}^{- 5}\text{a}\text{t}{\textm^{3}. If we start with 1 mole of the compound, the total pressure at equilibrium would be

Options

  1. A0.766 atm
  2. B0.0582 atm
  3. C0.0388 atm
  4. D0.0194 atm

Correct answer

B. 0.0582 atm

Step-by-step solution

\text{N}{\text{H}_{2}\text{C}\text{O}\text{O}\text{N}{\text{H}_{4}\left(\text{s}\right) ⇌ 2\text{N}{\text{H}_{3}\left(\text{g}\right) + \text{C}{\text{O}_{2} ( \text{g} )

At equilibrium if partial pressure of \text{C}{\text{O}_{2}\text{is} \text{p} then that of \text{N}{\text{H}_{3} is 2p2\text{p}

So, Kp=pNH32×pCO2{K}_{p} = {p}_{\text{NH} 3}^{2} \times {p}_{\text{CO} 2}

Kp=(2p)2×p=4p3{K}_{p} = {\left(2 \text{p}\right)}^{2} \times \text{p} = 4 p^{3}

2.9×10−5=4p3\text{2.9} \times {10}^{- 5} = 4 p^{3}

p=1.935×10−2\text{p} = \text{1.935} \times {10}^{- 2}

hence, total pressure p=3p\text{p} = 3\text{p}

=5.81×10−2=0.0581= 5.81 \times {10}^{- 2} = 0.0581 atm.
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