At 200^ 0C, the equilibrium constant of the following reaction is 1.6 × 10^3.

2HBr(g)<------> H2(g) + Br2(g)

What is the amount of H2(g) at equilibrium when 1.0 × 10^-2 mol of HBr(g) is heated at

200^0C in a closed vessel?

### 1 Answer | Add Yours

Let us establish the ICE table for this reaction.

`2 HBr _(g) <=> H_2 _(g) + Br_2 _(g)`

I `1.0x10^(-2)` `0` `0`

C `-x` `+x` `+x`

E `1.0x10^(-2) -x` `x` `x`

`k = ([H_2][Br_2])/([HBr]^(2)) =1.6 x 10^3`

`l e t x =[H_2] = [Br_2]`

`k = (x * x)/(1.0x10^(-2) -x)^2 =1.6 x 10^3`

`(x^(2))/(1.0x10^(-2) -x)^2 =1.6 x 10^3`

Then we can solve for x.

`(x^(2))/(1.0x10^(-2) -x)^2 =1.6 x 10^3`

`sqrt((x^(2))/(1.0x10^(-2) -x)^2 =1.6 x 10^3)`

`(x)/(1.0x10^(-2) -x) = 40`

`x = (1.0x10^(-2) -x) * 40`

`x = 0.4 – 40x`

`x+ 40x = 0.4`

`41x = 0.4`

`x = 0.4/41`

`x = 9.756 x10^-3 = [H_2] = [Br_2]` **-> answer**

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