Hydrogen bromide and oxygen react to form bromine and water. Also, a chemist finds that…
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Hydrogen bromide and oxygen react to form bromine and water. Also, a chemist finds that at a certain temperature the equilibrium mixture of hydrogen bromide, oxygen, bromine, and water has the following composition: HBr has a pressure at equilibrium of 72.1 atm. O2 has a pressure at equilibrium of 60.3 atm. Br2 has a pressure at equilibrium of 5.35 atm. H2O has a pressure at equilibrium of 77.9 atm. Calculate the value of the equilibrium constant Kp for this reaction
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We are given the following reaction:
2 HBr (g) + O2 (g) ⇌ 2 Br2 (g) + 2 H2O (g)
And the equilibrium pressures of each component are as follows:
[HBr] = 72.1 atm [O2] = 60.3 atm [Br2] = 5.35 atm [H2O] = 77.9 atm
The equilibrium constant (Kp) can be calculated with the following equation:
Kp = ([Br2]^2 * [H2O]^2) / ([HBr]^2 * [O2])
Plugging in the known pressures:
Kp = (5.35^2 * 77.9^2) / (72.1^2 * 60.3)
Kp = (28.6225 * 6072.41) / (5196.41 * 60.3)
Kp = (173786.0651375) / (313110.723)
Kp ≈ 0.555
The value of the equilibrium constant Kp for this reaction at the given temperature is approximately 0.555.