MICHAEL S. answered • 11/04/16
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Given:
H2 + Br2 ↔ 2HBr, Keq = 0.74
[H2]0 = 0.1 M, initial concentration of hydrogen
[Br2]0 = 0.4 M, initial concentration of bromine
[HBr]0 = 0.8 M, initial concentration of hydrogen bromide
Find: Equilibrium Concentrations
Solution:
(i) Determine initial concentrations:
[H2]0 = 0.1 M, initial concentration of hydrogen
[Br2]0 = 0.4 M, initial concentration of bromine
[HBr]0 = 0.8 M, initial concentration of hydrogen bromide
[Br2]0 = 0.4 M, initial concentration of bromine
[HBr]0 = 0.8 M, initial concentration of hydrogen bromide
(ii) Write the equilibrium equation:
Rate of forward reaction = k1[H2][Br2]
Rate of backward reaction = k2[HBr]2
In equilibrium, the rate of the forward reaction = rate of the backward reaction.
k1[H2][Br2] = k2[HBr]2
Keq = k1/k2 = [HBr]2 / [H2][Br2]
(iii) Since we have initial concentrations of all three substances, we have to determine which way the reaction will proceed. We use the equilibrium equation to determine which way the reaction has to proceed to reach equilibrium.
K = [HBr]02 / [H2]0[Br2]0
= (0.8)2 / (0.1)(0.4)
= 16
Since K > Keq (16 > 0.74), this means that there is a higher proportion of products to reactants than what you would see at equilibrium. Therefore, the reaction will proceed towards H2 and Br2 to get to equilibrium.
(iv) Since we are proceeding in the backward, we know:
Every 2 mol HBr produces 1 mol H2 and 1 mol Br2
Every '2x' mol HBr produces 'x' mol H2 and 'x' mol Br2.
We have to subtract '2x' from [HBr]0.
We have to add 'x' to [H2]0.
We have to add 'x' to [Br2]0.
(v) At equilibrium, we will have:
[HBr] = 0.8 - 2x
[H2] = 0.1 + x
[Br2] = 0.4 + x
(vi) Set up the equilibrium equation.
Keq = [HBr]2 / [H2][Br2]
0.74 = (0.8 - 2x)2 / (0.1 + x)(0.4 + x)
(vii) Solve for 'x', then calculate the equilibrium concentrations by plugging 'x' below.
[HBr] = 0.8 - 2x =
[H2] = 0.1 + x =
[Br2] = 0.4 + x =
[H2] = 0.1 + x =
[Br2] = 0.4 + x =
