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
Correct Answer: D
Solution :
[d] For equilibrium \[\begin{matrix} {} & {{N}_{2}}{{O}_{4}}(s) & \rightleftharpoons & 2N{{O}_{2}}(g) \\ At\,\,equilibrium & (1-\alpha )mole & {} & 2\alpha \,\,mole \\ Total\,\,mole & =1-\alpha +2\alpha =1+\alpha & {} & {} \\ \end{matrix}\] If P is total pressure at equilibrium therefore \[{{P}_{{{N}_{2}}{{O}_{4}}}}=\frac{1-\alpha }{1+\alpha }\times P\] and \[{{P}_{{{N}_{2}}O}}=\frac{2\alpha }{1-\alpha }\times P\] \[{{K}_{p}}=\frac{{{({{P}_{N{{O}_{2}}}})}^{2}}}{{{P}_{{{N}_{2}}{{O}_{4}}}}}=\frac{4{{\alpha }^{2}}\times P}{(1-\alpha )(1+\alpha )}=\frac{4{{\alpha }^{2}}P}{1-{{\alpha }^{2}}}\] 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.
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