A) \[0.66\times {{10}^{-9}}\,mol\,{{L}^{-1}}\]
B) \[105\times {{10}^{-5}}\,mo{{l}^{-1}}\,L\]
C) \[2.0\times {{10}^{-4}}\,{{s}^{-1}}\]
D) \[3.0\times {{10}^{5}}\,mo{{l}^{-1}}\,{{s}^{-1}}\]
Correct Answer: A
Solution :
\[\frac{dx}{dt}={{K}_{f}}\,{{[Cu\,{{(N{{H}_{3}})}_{4}}]}^{2+}}\] \[-{{K}_{b}}{{[Cu{{(N{{H}_{3}})}_{3}}{{H}_{2}}O]}^{2+}}\,[N{{H}_{3}}]\] \[\frac{{{K}_{f}}}{{{K}_{b}}}=\frac{2.0\times {{10}^{-4}}{{s}^{-1}}}{3.0\times {{10}^{5}}\,\text{L}\,\text{mo}{{\text{l}}^{-1}}\,{{\text{s}}^{-1}}}\] \[=0.66\times {{10}^{-9}}mol\,{{L}^{-1}}\]
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