Q. At 3000 K, the equilibrium partial pressure of $CO_{2},CO\ and\ O_{2}$ are 0.6, 0.4 and 0.2 atm respectively. $K_{p}$ for the reaction, $2CO_{2} \rightleftharpoons 2CO + O_{2}$ is

A) 0.088
B) 0.0533
C) 0.133
D) 0.177
Explanation

(a)
$2CO_{2} \rightleftharpoons 2CO\ \ \ \ \ \ \ + \ \ \ \ O_{2}$
At equili. 0.6 atm 0.4 atm 0.2 atm
$\therefore$ $K_{p} = \frac{p_{CO}^{2} \times p_{O_{2}}}{{p_{CO}^{2}}_{2}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{(0.4)^{2}(0.2)}{(0.6)^{2}} = 0.088$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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