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Chemistry · EQUILIBRIUM
हिन्दीMatch List-I with List-II for the general reaction relationships with equilibrium constant: List-I (Chemical Reaction) List-II (Equilibrium Constant) A \( a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D} \) I \( (K_c)^n \) B \( c\text{C} + d\text{D} \rightleftharpoons a\text{A} + b\text{B} \) II \( K_c \) C \( na\text{A} + nb\text{B} \rightleftharpoons nc\text{C} + nd\text{D} \) III \( 1/K_c \) D \( \frac{1}{n}a\text{A} + \frac{1}{n}b\text{B} \rightleftharpoons \frac{1}{n}c\text{C} + \frac{1}{n}d\text{D} \) IV \( (K_c)^{1/n} \) Choose the correct answer from the options given below:
Match List-I with List-II for the general reaction relationships with equilibrium constant:
| List-I (Chemical Reaction) | List-II (Equilibrium Constant) | ||
|---|---|---|---|
| A | \( a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D} \) | I | \( (K_c)^n \) |
| B | \( c\text{C} + d\text{D} \rightleftharpoons a\text{A} + b\text{B} \) | II | \( K_c \) |
| C | \( na\text{A} + nb\text{B} \rightleftharpoons nc\text{C} + nd\text{D} \) | III | \( 1/K_c \) |
| D | \( \frac{1}{n}a\text{A} + \frac{1}{n}b\text{B} \rightleftharpoons \frac{1}{n}c\text{C} + \frac{1}{n}d\text{D} \) | IV | \( (K_c)^{1/n} \) |
Choose the correct answer from the options given below:
Correct answer · Option 4
A-(II), B-(III), C-(I), D-(IV)
Explanation
For the standard reaction, constant is \( K_c \) (II); reverse is \( 1/K_c \) (III); multiplied by \( n \) gives \( (K_c)^n \) (I); divided by \( n \) gives \( (K_c)^{1/n} \) (IV).
