• # question_answer For the reaction,${{[Cu{{\left( N{{H}_{3}} \right)}_{4}}]}^{2+}}+{{H}_{2}}O\xrightarrow{\,}\,{{[Cu{{(N{{H}_{3}})}_{3}}{{H}_{2}}O]}^{2+}}$$+N{{H}_{3}}$the   net   rate   is $\frac{dx}{dt}=2.0\times {{10}^{-4}}{{s}^{-1}}$$\,{{[Cu{{(N{{H}_{3}})}_{4}}]}^{2+}}-3.0\times {{10}^{5}}\,L\,mo{{l}^{-1}}\,{{s}^{-1}}$            ${{[Cu{{(N{{H}_{3}})}_{3}}{{H}_{2}}O]}^{2+}}\,[N{{H}_{3}}]$ Then the ratio of rate constants of the forward and backward reactions is 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)  $3.0\times {{10}^{5}}\,mo{{l}^{-1}}\,{{s}^{-1}}$

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