NTA Abhyas JEE Main2020ChemistryChemical EquilibriumPractice
Equilibrium constant K C for the following reaction at 800 K is, 4 N H 3 ( g ) ⇌ 1 2 N 2 ( g ) + 3 2 H 2 ( g ) The value of K P for the following reaction will be N 2 ( g ) + 3 H 2 ( g ) ⇌ 2 N H 3 ( g )
Options
- A800 R 4 2 -
- B16 × ( 800 R ) 2
- C1 4 × 800 R 2
- D800 R 1 / 2 4
Correct answer
C. 1 4 × 800 R 2
Step-by-step solution
NH 3 (g) ⇌ 1 2 N 2 (g) + 3 2 H 2 (g); K C = 4 N 2 ( g ) + 3 H 2 ( g ) ⇌ 2 N H 3 ( g ) Δ n g = 2 - 4 = - 2 So K C for the following reaction will be equal to 1 4 2 K P = K C ( R T ) Δ n g = 1 4 2 × (800 × R) − 2 = 1 4 × 800   R 2