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The equilibrium constant is given for one of the reactions below. Determine the value of the missing equilibrium constant.

H2(g) + Br2(g) ⇌ 2 HBr(g) Kc = 3.8 × 104
4 HBr(g) ⇌ 2 H2(g) + 2 Br2(g) Kc = ?

Respuesta :

niniabesadze
To calculate Kc of the second step, we must observe the first reaction.
If we rotate the first reaction and multiplay it by 2 we get the same thing.

So the rotation means 1/kc(first reaction)
And by multiplying it by 2, K should be squared.

Kc2=1/(kc1)^2=6.92*10^-10

The value of the missing equilibrium constant is [tex]6.9\times 10^{-10}[/tex].

Given:

The equilibrium constant of the reaction:

[tex]H_2(g) + Br_2(g) \rightleftharpoons 2 HBr(g) ,K_c = 3.8\times 10^4[/tex]...(i)

The other reaction:

[tex]4HBr(g)\rightleftharpoons 2H_2(g) + 2Br_2(g)[/tex]....(ii)

To find:

The equilibrium constant of the other reaction:

[tex]4HBr(g)\rightleftharpoons 2H_2(g) + 2Br_2(g)[/tex]

Solution:

[tex]H_2(g) + Br_2(g) \rightleftharpoons 2 HBr(g) ,K_c = 3.8\times 10^4...(i)[/tex]

The expression of the equilibrium constant of the reaction (i):

[tex]K_c=\frac{[HBr]^2}{[H_2][Br_2]}...[1]\\\\3.8\times 10^4=\frac{[HBr]^2}{[H_2][Br_2]}[/tex]

The other chemical reaction :

[tex]4HBr(g)\rightleftharpoons 2H_2(g) + 2Br_2(g)...(ii)[/tex]

The expression of the equilibrium constant of the reaction (ii):

[tex]K_c'=\frac{[H_2]^2[Br_2]^2}{[HBr]^4}\\\\K_c'=(\frac{[H_2]^1[Br_2]^1}{[HBr]^2})^2[/tex]

Using [1]:

[tex]K_c'=(\frac{1}{K_c})^2\\\\K_c'=(\frac{1}{ 3.8\times 10^4})^2\\\\=6.9\times 10^{-10}[/tex]

The value of the missing equilibrium constant is [tex]6.9\times 10^{-10}[/tex].

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