A) \[8\]
B) \[0.25\]
C) \[0.125\]
D) \[32\]
Correct Answer: B
Solution :
For the reaction, \[2A+{{B}_{2}}2A{{B}_{2}}\], Equilibrium constant, \[{{K}_{P}}=\frac{P_{A{{B}_{2}}}^{2}}{P_{A}^{2}.{{P}_{{{B}_{2}}}}}=16\] ....(i) For the reaction, \[A{{B}_{2}}A+1/2{{B}_{2}},\] \[K_{P}^{}=\frac{{{p}_{A}}.p_{{{B}_{2}}}^{1/2}}{{{p}_{A{{B}_{2}}}}}\] ???(ii) On squaring Eq. (ii), we get \[{{(K_{P}^{})}^{2}}=\frac{p_{A}^{2}{{p}_{{{B}_{2}}}}}{p_{A{{B}_{2}}}^{2}}\] ???(iii) From Eq. (i) and (iii)^ we get \[{{K}_{p}}.{{(K_{p}^{})}^{2}}=1\] \[16{{(K_{p}^{})}^{2}}=1\] \[{{(K_{p}^{})}^{2}}=\frac{1}{16}\] \[K_{p}^{}=\frac{1}{4}=0.25\]You need to login to perform this action.
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