Equilibrium Question 486
Question: If for $ H_2 {(g)}+\frac{1}{2}{S_{2(S)}} $ ⇌ $ H_2{S {(g)}} $ and $ {H_{2(g)}}+B{r_{2(g)}} $ ⇌ $ 2HB{r {(g)}} $ The equilibrium constants are K1 and K2 respectively, the reaction $ Br_2 {(g)}+H_2{S {(g)}} $ ⇌ $ 2HB{r {(g)}}+\frac{1}{2}{S_{2(S)}} $ would have equilibrium constant [MP PMT 2001]
Options:
A) $ K_1\ \times \ K_2 $
B) $ K_1/K_2 $
C) $ K_2/K_1 $
D) $ K_2^{2}/K_1 $
Show Answer
Answer:
Correct Answer: C
Solution:
$ K_1=\frac{[H_2S]}{[H_2]{{[S_2]}^{{1}/{2};}}} $ ; $ K_2=\frac{{{[HBr]}^{2}}}{[H_2][Br_2]} $
$ K_3=\frac{{{[HBr]}^{2}}\times {{[S_2]}^{{1}/{2};}}}{[Br_2],\times [H_2S]} $ ; $ \frac{K_2}{K_1}=K_3 $
BETA
