3.1.10- Equilibrium constant Kp for homogeneous systems (PAPER 1)
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What is total pressure?
The sum of all the pressures of individual gases (partial pressures)
SO2Cl2 (g) ⇌ SO2 (g) + Cl2 (g) At equilibrium reaction vessel = 263kPa Cl2 Total pressure in vessel = 714kPa Calculate partial pressure SO2Cl2 (pSO2Cl2) at equilibrium
Total pressure= p(SO2Cl2) + p(SO2) + (PCl2) 714 = p(SO2Cl2) + p(SO2) + 263 1:1 ratio SO2: Cl2 as both products, SO2= 263 714= p(SO2Cl2) + 263 + 263 (pSO2Cl2) = 188kPa
How is partial pressure of a gas calculated?
Using mole fractions- a fraction of a gas in a gas mixture
What is the equation for mole fraction of a gas?
number of moles of gas/ total n of moles of gas in the mixture
What is the equation for partial pressure?
mole fraction of gas x total pressure in mixture
What to use to calculate mole fractions? Eg 3 moles SO2Cl2 heated, equilibrium mixture contains 1.75 moles Cl2 SO2Cl2 (g) ⇌ SO2 (g) + Cl2 (g)
ICE Table- initial, change, equilibrium moles SO2Cl2 (g) ⇌ SO2 (g) + Cl2 (g) initial 3 0 0 change -1.75 +1.75 +1.75 equilibrium 1.25 1.75 1.75
Use the ICE table to work out partial pressure SO2Cl2, total pressure= 714kPa
mole fraction SO2Cl2 = 1.25/ 1.25 + 1.75 + 1.75= 1.25/ 4.75= 0.263 0.263x 714= 188kPa
Use this general equation to give expression for kp aA (g) + bB (g) ⇌ cC(g) + dD (g)
kP = p(C)^c p(D)^d/ p(A)^a p(B)^b We use partial pressures to find kp
Find value of kp and units for 2A (g) + B(g) ⇌ C(g) + D(g) Partial pressures are 324kPa for A, 162kPa for B, 182kPa for C, 182kPa for D at equilibrium
Write expression kp = (pD) x (pC) / (pB) (pA)^2 kp= 182 x 182/ 162 x (324)^2 kp= 1.95 x10^-3 Units= kPa x kPa/ kPa x kPa x kPa = 1/ kPa= kPa^-1
How does temperature affect kp?
Changing the temperature changes equilibrium pressures, changing kp If temperature change causes equilibrium to shift to right, kp INCREASES If temperature change causes equilibrium to shift to left, kp DECREASES
How will increasing and decreasing temperature affect kp of this reaction? 2SO2 (g) + O2(g) ⇌ 2SO3 (g) ΔH= -197kJmol-1
Reaction is exothermic to right Increasing temperature causes equilibrium to shift to left decreasing kp -Reactant partial pressure increases Decreasing temperature causes equilibrium to shift to right increasing kp -Product partial pressure increases
How does pressure affect kp?
Does not affect kp- partial pressure ratio of reactants to products stays the same so Kp does not change.
How do catalysts affect kp?
Do not affect kp- speeds up forward and reverse reaction so rate at which equilibrium is reached.