OnlineCBT
Chemistry · EQUILIBRIUM
हिन्दीMatch List-I with List-II: List-I (Chemical Reaction) List-II (Equilibrium Constant) A \( a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D} \) I \( K_c'' = (K_c)^n \) B \( c\text{C} + d\text{D} \rightleftharpoons a\text{A} + b\text{B} \) II \( K_c' = \frac{1}{K_c} \) C \( na\text{A} + nb\text{B} \rightleftharpoons nc\text{C} + nd\text{D} \) III \( K_c \) D \( \frac{1}{n}c\text{C} + \frac{1}{n}d\text{D} \rightleftharpoons \frac{1}{n}a\text{A} + \frac{1}{n}b\text{B} \) IV \( K_c''' = (K_c)^{-1/n} \) Choose the correct answer from the options given below:
Match List-I with List-II:
| List-I (Chemical Reaction) | List-II (Equilibrium Constant) | ||
|---|---|---|---|
| A | \( a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D} \) | I | \( K_c'' = (K_c)^n \) |
| B | \( c\text{C} + d\text{D} \rightleftharpoons a\text{A} + b\text{B} \) | II | \( K_c' = \frac{1}{K_c} \) |
| C | \( na\text{A} + nb\text{B} \rightleftharpoons nc\text{C} + nd\text{D} \) | III | \( K_c \) |
| D | \( \frac{1}{n}c\text{C} + \frac{1}{n}d\text{D} \rightleftharpoons \frac{1}{n}a\text{A} + \frac{1}{n}b\text{B} \) | IV | \( K_c''' = (K_c)^{-1/n} \) |
Choose the correct answer from the options given below:
Correct answer · Option 3
A-(III), B-(II), C-(I), D-(IV)
Explanation
From Table 6.4: Standard reaction has constant \( K_c \) (A-III); reversing inverts it to \( 1/K_c \) (B-II); multiplying by \( n \) raises it to power \( n \) (C-I); reversing and dividing by \( n \) gives \( (K_c)^{-1/n} \) (D-IV).
