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Question:

Consider the following reversible chemical reactions:
A2(g) + Br2(g) ⇌ 2AB(g) ....(1) K1
6AB(g) ⇌ 3A2(g) + 3B2 ....(2) K2
The relation between K1 and K2 is:

K1K2=3

K2=K󔼧

K2=K31

K1K2=13

Solution:

Ans(2)
A2(g) + Br2(g) ⇌ 2AB(g) (1) K1 ⇒ eq (1) ×3
6AB(g) ⇌ 3A2(g) + 3B2 (2) K2 ⇒ (1/K1)^3 = K2 ⇒ K2 = (K1)^-3