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

NTA Abhyas 2020 · Chemistry · Chemical Equilibrium

For the decomposition of the compound, represented as NH _ 2 COONH...

For the decomposition of the compound, represented as

NH2COONH4(s)⇌2NH3(g)+CO2(g){NH}_{2} {COONH}_{4} \left(\text{s}\right)⇌ 2 {NH}_{3} \left(\text{g}\right) + {CO}_{2}\left(\text{g}\right)

the Kp=2.9×10−5atm3.Ifthereactionisstartedwith1molofthecompound,thetotalpressureatequilibriumwouldbe:{K}_{p} = \text{2.9} \times1 {0}^{- 5} {atm}^{3}. If the reaction is started with 1 mol of the compound, the total pressure at equilibrium would be :

Options

  1. A7.66×10−2atm\text{7.66} × 1 {0}^{- 2} atm
  2. B38.8×10−2atm\text{38.8} × 1 {0}^{- 2} atm
  3. C5.82×10−2atm\text{5.82} \times 1 {0}^{- 2} atm
  4. D1.94×10−2atm\text{1.94} × 1 {0}^{- 2} \text{atm}

Correct answer

C. 5.82×10−2atm\text{5.82} \times 1 {0}^{- 2} atm

Step-by-step solution

In case of heterogeneous equilibrium. The achive mass of solid is unit so not consider in the expression of KpandKc{K}_{p} \text{and}{K}_{c}

NH2COONH4(s)⇌2NH3(g)+CO2(g){NH}_{2} {COONH}_{4} \left(\text{s}\right)⇌ 2 {NH}_{3} \left(\text{g}\right) + {CO}_{2}\left(\text{g}\right)

At equilibrium ⇒⇒ 2 p & p&

KP=[PNH3]2[PCO2]=[2P]2[P]=4P3{K}_{P} \text{=}{\left[{P}_{{\text{NH}}_{3}}\right]}^{2} \left[{P}_{{\text{CO}}_{2}}\right]\text{=}{\left[\text{2P}\right]}^{2} \left[\text{P}\right]\text{=}{\text{4P}}^{3}

4P3=2⋅9×10−5atm34 P^{3} = 2 \cdot9 \times1 {0}^{- 5} {atm}^{3}

P3=0⋅725×10−5atm3=7⋅25×10−6atm3P^{3} = 0 \cdot7 2 5 \times1 {0}^{- 5} {atm}^{3} = 7 \cdot2 5 \times1 {0}^{- 6} {atm}^{3}

P=7⋅25×10−63=1⋅94×10−2atm\text{P} =7 \cdot2 5 \times1 {0}^{- 6} 3 = 1 \cdot9 4 \times1 {0}^{- 2} \text{atm}

ΣP=(PTotal)equilibrium=2P+P=3PΣ \text{P} = {\left({P}_{Total}\right)}_{equilibrium} =2 \text{P} + \text{P} =3 \text{P}

3×1⋅94×10−2atm3 \times1 \cdot9 4 \times1 {0}^{- 2} \text{atm}

5⋅82×10−2atm5 \cdot8 2 \times1 {0}^{- 2} \text{atm}
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