Chemical EquilibriumHard
Question
For the reaction 2NO(g) + Cl2(g) $\rightleftharpoons$2NOCl(g), NO and Cl2 are initially taken in mole ratio of 2: 1. The total pressure at equilibrium is found to be 1 atm. If the moles of NOCl are one-fourth of that of Cl2 at equilibrium, the value of KP for the reaction is
Options
A.13/36
B.13/256
C.13/512
D.13/128
Solution
2NO(g) + Cl2(g) $\rightleftharpoons$ 2NOCl(g)
Initial partial pressure 2P0 P0 0
Equ. partial pressure 2P0 – 2x P0 – x 2x
From question, (2P0 – 2x) + (P0 – x) + 2x = 1
⇒ 3P0 – x = 1 (1)
and $2x = \frac{1}{4}\left( P_{0} - x \right)$ (2)
From (1) and (2), P0 = 9x and x = 1/26
$\therefore K_{p} = \frac{(2x)^{2}}{\left( 2P_{0} - 2x \right)^{2}\left( P_{0} - x \right)} = \frac{13}{256}\text{at}\text{m}^{- 1}$
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