KpKp. The equilibrium constant in terms of partial pressures KpKp.
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Transcript of KpKp. The equilibrium constant in terms of partial pressures KpKp.
Kp
The equilibrium constant in terms of partial pressures
Kp
Mole fraction
Partial pressures
Partial pressure, pThe contribution of a gas towards the total pressurePartial pressure = mole fraction x Total pressure
A gas mixture with a total pressure of 320 kPa contains 2 mol of N2(g) and 3 mol of O2(g).
Mole fractions
Partial pressures
Sum of partial pressures = Total pressurep(N2) + p(O2) = 128 + 192 = 320 kPa
x(N2) = = 0.4 x(O2) = = 0.6
p(O2) = x(O2)P = 0.6 x 320 = 192 kPa
p(N2) = x(N2)P = 0.4 x 320 = 128 kPa
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What is Kp
Similar to Kc but partial pressures used in place of concentration
Equilibrium: 2SO2(g) + O2(g) 2SO⇌ 3(g)
Units:
Kp = 22 )(SOp
23 )(SOp
)( 2Op
2)(kPa
)()( 2 kPakPa
1kPaKp =
Calculating Kp
Equilibrium: 2SO2(g) + O2(g) 2SO⇌ 3(g)
Partial pressures: SO2(g), 74 kPa; O2(g), 23 kPa; SO3(g), 142 kPa
Kp = 22 )(SOp
23 )(SOp
)( 2Op
Kp =2142
274 23x
x
= 0.160 kPa–1
Heterogeneous equilibria
Equilibrium contains different phases
Equilibrium: CaCO3(s) CaO(s) + CO⇌ 2(g)
Kp expression contains only gaseous species
Kp = p(CO2)
Solid species are omitted (solids have no gas pressure)
