$N X_{2} O X_{4} (g) 2 NO X_{2} (g)$ is 20\% dissociated at a total pressure of $1.0 atm$ and temperature $T$ . At the same temperature the total pressure is raised to $3.0 atm$ . The new degree of dissociation is

Answer: B

K_{p} = 4 α^{2} P / (1 - α^{2}) = 0.1667

atm. At 3 atm,

α = K_{p} / (K_{p} + 4 P) = 0.1667/12.1667 = 0.1170

. The inverse-proportion guess (

0.2/3

) is wrong because

α \propto P^{- 1/2}

only approximately.

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