Question:easy
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
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
Show Hint
When a chemical reaction is reversed, the new equilibrium constant becomes the reciprocal of the original equilibrium constant.
Updated On: Jun 5, 2026
- \(K_c=K_c'\)
- \(K_c=\dfrac{1}{K_c'}\)
- \(K_c=-K_c'\)
- \(K_c=\sqrt{K_c'}\)
Show Solution
The Correct Option is B
Solution and Explanation
Step 1: Write the forward constant.
For a reaction the forward $K_c$ is products over reactants, for example \[ K_c = \frac{[HI]^2}{[H_2][I_2]} \]
Step 2: Write the reverse constant.
Running the reaction backward swaps products and reactants, so \[ K_c' = \frac{[H_2][I_2]}{[HI]^2} \]
Step 3: Multiply the two.
The two expressions are upside down copies of each other, so their product is \[ K_c \times K_c' = 1 \]
Step 4: Rearrange.
Therefore $K_c = \dfrac{1}{K_c'}$.
Step 5: Answer.
\[ \boxed{K_c = \dfrac{1}{K_c'}} \]
For a reaction the forward $K_c$ is products over reactants, for example \[ K_c = \frac{[HI]^2}{[H_2][I_2]} \]
Step 2: Write the reverse constant.
Running the reaction backward swaps products and reactants, so \[ K_c' = \frac{[H_2][I_2]}{[HI]^2} \]
Step 3: Multiply the two.
The two expressions are upside down copies of each other, so their product is \[ K_c \times K_c' = 1 \]
Step 4: Rearrange.
Therefore $K_c = \dfrac{1}{K_c'}$.
Step 5: Answer.
\[ \boxed{K_c = \dfrac{1}{K_c'}} \]
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