A) neither K nor a changes
B) both K and exchange
C) K changes, but a does not change
D) K does not change but a changes
Correct Answer: D
Solution :
\[{{N}_{2}}{{O}_{4}}(g)\rightleftharpoons 2N{{O}_{2}}\] Moles t = 0, 1 0 Molesa at eq. \[1-\alpha \] \[2\alpha \] \[{{K}_{P}}=\,\frac{{{\left( \frac{2\alpha }{(1+\alpha )}\times P \right)}^{2}}}{\frac{1-\alpha }{1+\alpha }\times P}\,=\frac{4{{\alpha }^{2}}}{1-{{\alpha }^{2}}}\times P\] Let total pressure at eq. = P. or \[{{K}_{P}}=\frac{4{{\alpha }^{2}}}{1-{{\alpha }^{2}}}\times P\] When volume is halved, P is double. \[\therefore \] \[\alpha \] will change as \[{{K}_{p}}\] is independent of pressure change.
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