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AIIMS 2012
Exam
AIIMS
Year
2012
Subject
Chemistry
Chapter
Chemical Equilibrium

AIIMS 2012 · Chemistry · Chemical Equilibrium

At 60^ and 1 ~atm, ~N_2 O_4 is 50 \% dissociated into NO_2 then K_p...

At 60∘60^{\circ} and 1 atm, N2O41 \mathrm{~atm}, \mathrm{~N}_2 \mathrm{O}_4 is 50%50 \% dissociated into NO2\mathrm{NO}_2 then KpK_p is

Options

  1. A1.33 atm1.33 \mathrm{~atm}
  2. B2 atm2 \mathrm{~atm}
  3. C2.67 atm2.67 \mathrm{~atm}
  4. D3 atm3 \mathrm{~atm}

Correct answer

B. 2 atm2 \mathrm{~atm}

Step-by-step solution

N2O4⇌2NO2\mathrm{N}_2 \mathrm{O}_4 \rightleftharpoons 2 \mathrm{NO}_2 Initially 10 At equilibrium 1−α2α\begin{array}{lcc}\text{ Initially } & 1 & 0 \\ \text{ At equilibrium } & 1-\alpha & 2 \alpha\end{array}N2O4\mathrm{N}_2 \mathrm{O}_4 is 50%50 \% dissociated, so α=12Kp=pNO22pN2O4=(2×12)2(1−12)=2 atm\alpha=\frac{1}{2} K_p=\frac{p_{\mathrm{NO}_2}^2}{p_{\mathrm{N}_2 \mathrm{O}_4}}=\frac{\left(2 \times \frac{1}{2}\right)^2}{\left(1-\frac{1}{2}\right)}=2 \mathrm{~atm}
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