At 300 K, N 2 O 4 (g) is 20% dissociated at a pressure of 1 atm. Now a sufficient quantity of an inert gas at constant pressure is introduced into the reaction mixture to produce an inert gas partial

Question

At 300 K, N 2 O 4 (g) is 20% dissociated at a pressure of 1 atm. Now a sufficient quantity of an inert gas at constant pressure is introduced into the reaction mixture to produce an inert gas partial pressure of 0.75 atm. What is the percentage dissociation of N 2 O 4 (g) when equilibrium is re-established?

TestHub · Accepted Answer

N 2 O 4 (g) ⇌ 2NO 2 (g) gives K p = 4α 2 P/(1-α 2 ) K p = 4(0.20) 2 (1)/(1 - 0.04) = 0.16/0.96 = 1/6 atm After adding the inert gas the total pressure stays 1 atm, so the pressure of the reacting gases is 1 - 0.75 = 0.25 atm 4α 2 (0.25)/(1-α 2 ) = 1/6 α 2 /(1-α 2 ) = 1/6 so α 2 = 1/7 α = 0.378, that is 37.8% This is option (b).

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