| \[{{A}_{2}}(g)\,\,+\,\,{{B}_{2}}\,(g)\,\,\,\,\,2AB\,(g)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.......\,(1)\] |
| \[6AB\,(g)\,\,\,\,\,3{{A}_{2}}\,(g)\,\,+3{{B}_{2}}\,(g)\,\,\,\,\,\,\,\,\,\,\,\,\,.......\,\,(2)\] |
A) \[{{K}_{2}}={{K}_{1}}^{-3}\]
B) \[{{K}_{1}}{{K}_{2}}\,\,=\,\,\frac{1}{3}\]
C) \[{{K}_{1}}{{K}_{2}}\,\,=\,\,3\]
D) \[{{K}_{2}}={{K}_{1}}^{3}\]
Correct Answer: A
Solution :
\[{{K}_{1}}=\,\frac{{{\left[ AB \right]}^{2}}}{\left[ {{A}_{2}} \right]\left[ {{B}_{2}} \right]}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\,(1)\] \[{{K}_{2}}=\,\frac{{{\left[ {{A}_{2}} \right]}^{3}}{{\left[ {{B}_{2}} \right]}^{3}}}{{{\left[ AB \right]}^{6}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\,(2)\] From Eq. (1) & (2) \[{{K}_{2}}\,=\,{{\left( \frac{1}{{{k}_{1}}} \right)}^{3}}\] \[{{K}_{2}}\,=\,{{K}_{{{1}^{-3}}}}\]
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