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

NTA Abhyas 2020 · Chemistry · Chemical Equilibrium

What will be the expression of K _ p for the given reaction if the...

What will be the expression of Kp{K}_{p} for the given reaction if the total pressure inside the vessel is P and degree of dissociation of the reactant is a? The reaction

N2O4⇄2NO2{\text{N}}_{2} {\text{O}}_{4} \overset{}{⇄} 2{\text{NO}}_{2}

Options

  1. A4a2P1+a2\dfrac{4{a}^{2}P}{1 + {a}^{2}}
  2. B4a2P1−a2\dfrac{4{a}^{2}P}{1 - {a}^{2}}
  3. Ca2P1−a2\dfrac{{a}^{2}P}{1 - {a}^{2}}
  4. Da21−a2\dfrac{{a}^{2}}{1 - {a}^{2}}

Correct answer

B. 4a2P1−a2\dfrac{4{a}^{2}P}{1 - {a}^{2}}

Step-by-step solution

Molesatt=0MolesatequilibriumN2O411−a⇌2NO202a\mathop{\mathop{}_{Molesatt = 0}}_{Molesatequilibrium}\mathop{{N}_{2}{O}_{4}}_{\mathop{1}_{1 - a}} ⇌ \mathop{2N{O}_{2}}_{\mathop{0}_{2a}}

Total moles at equilibrium =1−a+2a=1+a= 1 - a + 2a = 1 + a

PN2O4=(1−a1+a)P;PNO2=(2a1+a)P{\text{P}}_{{\text{N}}_{2} {\text{O}}_{4}} = ( \dfrac{1 - \text{a}}{1 + \text{a}} )\text{P}; {\text{P}}_{{\text{NO}}_{2}} = ( \dfrac{2\text{a}}{1 + \text{a}} )\text{P}

Kp=PNO2PN2O4=(2a1+a)2P2(1−a1+a)P=4a2P1−a2{K}_{p} = \dfrac{{P}_{N{O}_{2}}}{{P}_{{N}_{2}{O}_{4}}} = \dfrac{{\left(\dfrac{2a}{1 + a}\right)}^{2}{P}^{2}}{\left(\dfrac{1 - a}{1 + a}\right)P} = \dfrac{4{a}^{2}P}{1 - {a}^{2}}
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