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JEE Main 2012Offline
Exam
JEE Main
Year
2012
Mode
Offline
Subject
Chemistry
Chapter
Chemical Equilibrium

JEE Main 2012 · Offline · Chemistry · Chemical Equilibrium

The equilibrium constant (K_ c ) for the reaction N_2(g)+O_2(~g)...

The equilibrium constant (Kc)\left(\mathrm{K}_{\mathrm{c}}\right) for the reaction N2(g)+O2( g)→2NO(g)\mathrm{N}_2(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{NO}(\mathrm{g}) at temperature T\mathrm{T} is 4×10−44 \times 10^{-4}. The value of Kc\mathrm{K}_{\mathrm{c}} for the reaction, NO(g)→1/2 N2(g)+1/2 O2(g)\mathrm{NO}(\mathrm{g}) \rightarrow{1 / 2} \mathrm{~N}_2(\mathrm{g})+{1 / 2} \mathrm{~O}_2(\mathrm{g}) at the same temperature is :

Options

  1. A0.020.02
  2. B2.5×1022.5 \times 10^2
  3. C4×10−44 \times 10^{-4}
  4. D50.050.0

Correct answer

D. 50.050.0

Step-by-step solution

N2+O2⇌2NOKC=4×10−4NO⇌12 N2+12O2 KC1=1 KCKC1=14×10−4=50\begin{array}{ll}\mathrm{N}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{NO} & \mathrm{K}_{\mathrm{C}}=4 \times 10^{-4} \\ \mathrm{NO} \rightleftharpoons \frac{1}{2} \mathrm{~N}_2+\frac{1}{2} \mathrm{O}_2 & \mathrm{~K}_{\mathrm{C}}^1=\sqrt{\frac{1}{\mathrm{~K}_{\mathrm{C}}}} \\ \mathrm{K}_{\mathrm{C}}^1=\frac{1}{\sqrt{4 \times 10^{-4}}}=50 & \end{array}
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