Question
At T(K), the equilibrium constant of A_2(g) + B_2(g) C(g) is 2.7 10^ -5 . What is the equilibrium constant for 1 3 A_2(g) + 1 3 B_2(g) 1 3 C(g) at the same temperature?
JEE Main
Chemistry
Chemical Equilibrium
2026
JEE Main 2026 (04 April Shift 1)
JEE Main Chemistry Question (2026) — Solution
Options
- A. (2.7 10^ -5 )^3
- B. 6 10^ -2
- C. 2.7 10^ -5
- D. 3 10^ -2
Answer
D. 3 10^ -2
Step-by-step solution
For the reaction A_2(g) + B_2(g) C(g), the equilibrium constant is K_1 = 2.7 10^ -5 . When a reaction is multiplied by a factor n, the new equilibrium constant becomes K^n. The given reaction is multiplied by 1 3 to obtain the reaction 1 3 A_2(g) + 1 3 B_2(g) 1 3 C(g). The new equilibrium constant K_2 is given by: K_2 = (K_1)^ 1/3 K_2 = (2.7 10^ -5 )^ 1/3 K_2 = (27 10^ -6 )^ 1/3 K_2 = 3 10^ -2 Answer: 3 10^ -2
Related: Chemistry — Chemical Equilibrium · All PYQ Banks