For a purely reversible elementary reaction analytically defined as , the thermodynamic equilibrium — Chemical Kinetics Chemistry Question
Question
For a purely reversible elementary reaction analytically defined as $A \rightleftharpoons B$, the thermodynamic equilibrium constant $K_c$ evaluates exactly to $4.0$. If the specific forward rate constant $k_f$ is experimentally determined to be $0.08 \text{ s}^{-1}$, what is the exact numerical value of the backward rate constant $k_b$ expressed in units of $10^{-2} \text{ s}^{-1}$?
💡 Solution & Explanation
The fundamental kinetic definition of the dynamic equilibrium constant relates it to the specific rate constants via $K_c = \frac{k_f}{k_b}$. Plugging in the established values: $4.0 = \frac{0.08}{k_b}$. Rearranging to solve for $k_b$ yields $k_b = \frac{0.08}{4.0} = 0.02 \text{ s}^{-1}$. Converting this into the requested $10^{-2}$ scientific notation format results precisely in $2 \times 10^{-2} \text{ s}^{-1}$. The numeric answer is 2.