3.1.10 - Equilibrium constant K p for homogeneous systems
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Summary notes for AQA A-Level Chemistry Unit 3.1.10 - Equilibrium constant K p for homogeneous systems (A-level only)
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3.1.10 - Equilibrium constant K p for homogeneous systems
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AQA A Level Chemistry Student Book 2
Summary notes for AQA A-Level Chemistry Unit 3.1.10 - Equilibrium constant K p for homogeneous systems (A-level only) by an Imperial College London MSci Chemistry graduate. Notes divided into the following sections: Mole Fractions, Partial Pressure, Equilibrium constant K p
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3.1.10 - Equilibrium constant K p for homogeneous systems
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Section 1 : Physical Chemistry
Equilibrium Constant Kp for Homogeneous Systems
Mole Fractions
The mole fraction of a component, A, in a gaseous mixture is denoted xA + is calculated by dividing the
amount, in moles, of A by the total amount of moles of gas in the mixture.
• For: A(g) + B(g) ⇌ C(g) …
- xA = nA / (nA + nB + nC)
• where nA, nB + nC are the amounts, in moles, of A, B + C present in the mixture.
• The sum of the mole fractions of all of the components in a mixture will be equal to 1.
Partial Pressure
The partial pressure of a component, A, in a mixture of gases is the contribution which that gas makes
to the total pressure of the gas mixture.
• The partial pressure of A is denoted p(A) + its units are Pa (or kPa = 103 Pa, or MPa = 106 Pa).
The partial pressure of A, p(A), in a mixture is given by the following expression:
p(A) = xA x P
• ∴ partial pressure = mole fraction x total pressure
Equilibrium Constant (Kp)
The equilibrium constant Kp is deduced from the equation for a reversible reaction occurring in the gas
phase. Kp is calculated from partial pressures for a system at constant temp.
For the general case of a gaseous homogeneous equilibrium:
aA(g) + bB(g) ⇌ cC(g) + dD(g)
p(C)c p(D)d where p(C) represents the partial pressure of C in the equilibrium mixture and c
Kp =
p(A)a p(B)b is the balancing no. for C. Same applies to A, B + D.
The units are in terms of partial pressures in Pa but the overall power depends on the balancing
numbers in the eq. for the reaction.
(Pa)(c+d)
∴ units of Kc =
(Pa)(a+b)
∴ if Kp has no units then it is because there are an equal no. of moles on both sides of the eq. + they
cancel each other out in the Kp expression.
E.g. for the following reaction:
H2(g) + Br2(g) ⇌ 2HBr(g)
1.00 mol of H2(g) was mixed w/ 1.00 mol of Br2(g) + the mixture allowed to attain equilibrium.
0.824 mol of HBr(g) were present in the equilibrium mixture. Calculate a value for Kp.
Kp = p(HBr)2/[p(H2)p(Br2)] = (0.412P)2/(0.294P)2 = 1.96
∴ units = (Pa)2/(Pa)2 = no units, because the partial pressures cancel out in the expression
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