Question

The correct relation between equilibrium constants, \(K_c\) (forward direction) and \(K_c'\) (reverse direction) of the reaction given below, at the same temperature, is

• A. \(K_c=K_c'\)
• B. \(K_c=\dfrac{1}{K_c'}\)
• C. \(K_c=-K_c'\)
• D. \(K_c=\sqrt{K_c'}\)

Hint:

When a chemical reaction is reversed, the new equilibrium constant becomes the reciprocal of the original equilibrium constant.

Answer 1

The Correct Option is B

Step 1: Write the forward reaction.
\[ H_2(g)+I_2(g)\rightleftharpoons 2HI(g) \] For the forward reaction, the equilibrium constant is
\[ K_c=\frac{[HI]^2}{[H_2][I_2]} \]

Step 2: Write the reverse reaction.
The reverse reaction is
\[ 2HI(g)\rightleftharpoons H_2(g)+I_2(g) \] For this reverse reaction, the equilibrium constant is denoted by \(K_c'\).

Step 3: Write expression for \(K_c'\).
\[ K_c'=\frac{[H_2][I_2]}{[HI]^2} \]

Step 4: Compare the two expressions.
Observe that the numerator of one expression becomes the denominator of the other.
\[ K_c=\frac{[HI]^2}{[H_2][I_2]} \] and
\[ K_c'=\frac{[H_2][I_2]}{[HI]^2} \]

Step 5: Multiply the two equilibrium constants.
\[ K_c\times K_c' = \frac{[HI]^2}{[H_2][I_2]} \times \frac{[H_2][I_2]}{[HI]^2} \] \[ K_c\times K_c'=1 \]

Step 6: Rearrange the relation.
\[ K_c=\frac{1}{K_c'} \]

Step 7: Final conclusion.
Hence, the equilibrium constant of the reverse reaction is the reciprocal of the forward reaction equilibrium constant.
\[ \boxed{K_c=\frac{1}{K_c'}} \]
Therefore, the correct answer is option (B).