• # question_answer In a chemical equilibrium, the rate constant of the backward reaction is $7.5\times {{10}^{-4}}$ and the equilibrium constant is 1.5. So the rate constant of the forward reaction is           A) $5\times {{10}^{-4}}$ B) $2\times {{10}^{-3}}$ C) $1.125\times {{10}^{-3}}$ D) $9.0\times {{10}^{-4}}$

 [c] ${{K}_{c}}=\frac{{{K}_{f}}}{{{K}_{b}}}$ ${{K}_{f}}={{K}_{c}}\times {{K}_{b}}=1.5\times 7.5\times {{10}^{-4}}$$=1.125\times {{10}^{-3}}$