Question

In the reversible reaction shown below, \(k_f\) is the forward reaction rate constant and \(k_r\) is the reverse reaction rate constant.
The CORRECT expression representing the reaction equilibrium constant is ______.
\[A + B \underset{k_r}{\overset{k_f}{\rightleftharpoons}} C + D\]

• A. \(k_f + k_r\)
• B. \(k_f \times k_r\)
• C. \(\dfrac{k_f}{k_r}\)
• D. \(\dfrac{k_r}{k_f}\)

Hint:

At equilibrium, forward rate equals reverse rate; rearrange kf[A][B] = kr[C][D] to see that K = [C][D]/([A][B]) = kf/kr.

Answer 1

The Correct Option is C

Step 1: Write the rate expressions for the forward and reverse reactions.
For the reversible reaction \(A + B \rightleftharpoons C + D\), assuming elementary steps, the forward rate is \(\text{Rate}_f = k_f [A][B]\) and the reverse rate is \(\text{Rate}_r = k_r [C][D]\).

Step 2: Apply the condition of chemical equilibrium.
At equilibrium, the forward and reverse reactions proceed at exactly the same rate, so the concentrations of all species stop changing. This means \[\text{Rate}_f = \text{Rate}_r \implies k_f[A][B] = k_r[C][D]\]

Step 3: Rearrange to isolate the concentration ratio. \[\frac{[C][D]}{[A][B]} = \frac{k_f}{k_r}\]

Step 4: Recognize the definition of the equilibrium constant.
By definition, the equilibrium constant for this reaction is \(K = \dfrac{[C][D]}{[A][B]}\). Comparing with the result of Step 3, we get \[K = \frac{k_f}{k_r}\]

Step 5: Match with the options.
This is option (C). Option (D) is simply the reciprocal, which would represent the equilibrium constant of the reverse reaction \(C+D \rightleftharpoons A+B\) instead. Options (A) and (B), an addition and a product of the two rate constants, have no basis in the derivation and do not represent any standard equilibrium quantity.