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K(C) value of a gaseous reaction is 5mol...

`K_(C)` value of a gaseous reaction is 5mole/lit. If pressure is increased

A

Forward reaction is favoured

B

Backward reaction is favoured

C

Reaction is uneffected

D

`K_(C)` value increases

Text Solution

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The correct Answer is:
To solve the question regarding the effect of pressure on the equilibrium constant \( K_c \) of a gaseous reaction, we can follow these steps: ### Step 1: Understand the Concept of Equilibrium Constant The equilibrium constant \( K_c \) is a value that expresses the ratio of the concentrations of products to reactants at equilibrium for a given reaction at a specific temperature. It is defined as: \[ K_c = \frac{[products]}{[reactants]} \] ### Step 2: Analyze the Effect of Pressure on Equilibrium According to Le Chatelier's principle, if the pressure of a gaseous reaction is increased, the equilibrium will shift to favor the side with fewer moles of gas. However, this does not directly affect the value of \( K_c \). ### Step 3: Determine the Independence of \( K_c \) from Pressure The value of \( K_c \) is dependent only on temperature. Changes in pressure do not change the equilibrium constant itself; they may shift the position of equilibrium but do not affect the value of \( K_c \). ### Step 4: Conclusion Since the question states that the \( K_c \) value is 5 mole/liter and asks what happens if pressure is increased, we conclude that the reaction is unaffected in terms of the \( K_c \) value. Therefore, the correct answer is that the reaction is unaffected by the increase in pressure. ### Final Answer The correct answer is option C: the reaction is unaffected. ---
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