Question

At ${25}^{\circ }C$, a certain elementary reaction has a forward rate constant equal to $1.00\times {10}^{-3}{s}^{-1}$ and equilibrium constant, $K_c$, equal to $4.19$. What is the rate constant for the reverse reaction?

• A. $0.23\times {10}^{-3}$ $s^{-1}$
• B. $0.23\times {10}^{-5}$ $s^{-1}$
• C. $2.23\times {10}^{-3}$ $s^{-1}$
• D. $1.23\times {10}^{-3}$ $s^{-1}$

Answer 1

The correct option is A $0.23\times {10}^{-3}$ $s^{-1}$

Given, $k_f={10}^{-3}$ $s^{-1}$
$K_c=4.19$
We know,
$K_c=\frac{k_f}{k_b}$
$\therefore k_b=0.23\times {10}^{-3}$ $s^{-1}$