Equilibrium Question 736
Question: Consider the following equilibrium in a closed container $ N_2{O_{4(g)}}\rightarrow 2NO_2(g) $ At a fixed temperature, the volume of the raction container is halved. For this change, which of the statements holds true regarding the equilibrium constant $ (K_{p}) $ and the degree of dissociation $ (\alpha )- $
Options:
A) Neither $ K_{p} $ nor a changes
B) Both $ K_{p} $ and a change
C) $ K_{p} $ changes but a does not change
D) $ K_{p} $ does not change but a change
Show Answer
Answer:
Correct Answer: D
Solution:
For equilibrium $ \begin{matrix} {} & N_2O_4(s) & \rightarrow & 2NO_2(g) \\ Atequilibrium & (1-\alpha )mole & {} & 2\alpha mole \\ Totalmole & =1-\alpha +2\alpha =1+\alpha & {} & {} \\ \end{matrix} $ If P is total pressure at equilibrium therefore $ {P_{N_2O_4}}=\frac{1-\alpha }{1+\alpha }\times P $ and $ {P_{N_2O}}=\frac{2\alpha }{1-\alpha }\times P $
$ K_{p}=\frac{{{({P_{NO_2}})}^{2}}}{{P_{N_2O_4}}}=\frac{4{{\alpha }^{2}}\times P}{(1-\alpha )(1+\alpha )}=\frac{4{{\alpha }^{2}}P}{1-{{\alpha }^{2}}} $ If volume is halved, therefore pressure is doubled but the value of $ K_{p} $ is not effected by pressure. Thus at higher P, $ K_{p} $ will be higher, but it is constant so for maintaining the constant value of $ K_{p},\alpha $ will be lowered. Hence $ K_{p} $ does not change, but a changes.
