Question

If kc = 7.04 × 10-2 for the reaction: 2 hbr(g) ⇌h2(g) + br2(g), what is the value of kc for the reaction: 1/2 h2(g) + 1/2 br2 ⇌hbr(g)

Answer 1

at the first reaction when 2HBr(g) ⇄ H2(g) + Br2(g)
So Kc = [H2] [Br2] / [HBr]^2
7.04X10^-2 = [H2][Br] / [HBr]^2

at the second reaction when 1/2 H2(g) + 1/2 Br2 (g) ⇄ HBr
Its Kc value will = [HBr] / [H2]^1/2*[Br2]^1/2
we will make the first formula of Kc upside down:
1/7.04X10^-2 = [HBr]^2/[H2][Br2]
and by taking the square root: 
∴ √(1/7.04X10^-2)= [HBr] / [H2]^1/2*[Br]^1/2
∴ Kc for the second reaction = √(1/7.04X10^-2) = 3.769 

Answer 2

The value of Kc for the second reaction is mathematically given as

Kc' = 3.769  

What isthe value of Kc for the second reaction?

Question Parameters:

kc = 7.04 × 10-2 for the reaction

2 hbr(g) ⇌h2(g) + br2(g)

1/2 h2(g) + 1/2 br2 ⇌hbr(g)

Generally, the equation for the reaction  is mathematically given as

2HBr(g) ⇄ H2(g) + Br2(g)

Therefore

Kc = [H2] [Br2] / [HBr]^2

7.04X10^-2 = [H2][Br] / [HBr]^2

Upon final reaction

Kc' = [HBr] / [H2]^1/2*[Br2]^1/2

Hence

$\sqrt{(1/7.04X10^-2)}= [HBr] / [H2]^1/2*[Br]^1/2}\\\\Kc' = \sqrt{(1/7.04X10^-2)$

Kc' = 3.769  

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