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Chemistry Question on Equilibrium

Consider the following reversible chemical reactions : A2(g)+Br2(g)<=>[K1]2AB(g){A_2(g) + Br_2 (g) <=>[K_1] 2AB(g) } ....(1) 6AB(g)<=>[K2]3A2(g)+3B2(g){ 6AB(g) <=>[K_2] 3A_2 (g) + 3B_2(g) } ...(2) The relation between K1K_1 and K2K_2 is :

A

K2=K13K_2 = K_1^{3}

B

K2=K1−3K_2 = K_1^{-3}

C

K1K2=3K_1 K_2 = 3

D

K1K2=13K_1 K_2 = \frac{1}{3}

Answer

K2=K1−3K_2 = K_1^{-3}

Explanation

Solution

A2(g)+B2(g)<=>[k1]2AB{A_2 (g) + B_2(g) <=>[k_1] 2AB} ....(1)
⇒  e(1)×3\Rightarrow \; e (1) \times 3
6  AB(g)<=>3A2(g)+3B2(g){6 \; AB (g) <=> 3A_2 (g) + 3B_2 (g)}
⇒(1k1)3=k2⇒k2=(k1)−3\Rightarrow \left(\frac{1}{k_{1}} \right)^{3} = k_{2} \Rightarrow k_{2} =\left(k_{1}\right)^{-3}