Question

Consider the following reaction: H2(g) +Br2(g) <=> 2HBr(g) Kc=2.0x10^9 at 25C

If 0.100 mol H2 and 0.200 mol of Br2 were placed in a 10.0L container at 25C, what will the equilibrium concentrations of H2, Br2 and HBr be?

Answer 1

The equation goes as

H2 + Br = 2HBr

Kc = 2*10^9

since Kc > 1, this favorus products, so expect HBr formation and H2 Br2 reaction

Calculate concentrations (initally)

[H2] = mol of H2/V = (0.1/10) = 0.01 M

[Br2] = mol of Br2/V = (0.2/10) = 0.02 M

[HBr] = 0

the extent of reaction

[H2] = 0.01 - x

[Br2] = 0.02 - x

[HBr] = 0 + 2x

now, substitute this value in Kc

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

2*10^9 = (2x)^2 / ((0.01-x) (0.02-x))

solve for x

(2*10^9)(0.01*0.02 - 0.03x + x^2) = 4x^2

(2*10^9)*(0.01*0.02 - 0.03x + x^2) = x^2

x = 0.0099999

substitute

[H2] = 0.01 - x = 0.01-0.0099999 = 0.0000001 M

[Br2] = 0.02 - x = 0.02-0.0099999 = 0.0100001 M

[HBr] = 0 + 2x = 2*0.0099999 = 0.0199998 M

NOTE = this clearly depends on the amount of decimals you use when substitute x value, but approx values are shown previously. In engineering applications, we will typically assume this is a compelte reaction, i.e. all the limiting reactant reacts to form the max amount of HBr product