Question

For the reaction, $A_{\left(g\right)}+2B_{\left(g\right)}\rightleftharpoons 2C_{\left(g\right)},$ the rate constants for the forward and the reverse reactions are $1\times {10}^{-4}$ and $2.5\times {10}^{-2}$ respectively. The value of equilibrium constant, K for the reaction would be

• A.
  $1\times {10}^{-4}$
• B.
  $2.5\times {10}^{-2}$
• C.
  $4\times {10}^{-3}$
• D.
  $2.5\times {10}^2$

Answer 1

The correct option is C

$4\times {10}^{-3}$
Given : $k_f=1\times {10}^{-4}$ and $k_b=2.5\times {10}^{-2}$
Apply the formula, $K=\frac{k_f}{k_b}=\frac{1\times {10}^{-4}}{2.5\times {10}^{-2}}=4\times {10}^{-3}.$