Equilibrium II
Quick Notes
- Kp is the equilibrium constant for reactions involving gases, expressed in terms of partial pressures instead of concentrations.
- Expression for Kp:
- a, b, c, d = molar ratios from the balanced equation.
- Kp is only affected by temperature.
- Catalysts do not affect Kp, but they increase the rate at which equilibrium is reached.
Full Notes
Kp has been outlined in more detail here.
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The equilibrium constant, Kp, works in the same way as Kc – see Kc . It applies to homogeneous gaseous equilibria, where all reactants and products are in the gas phase and uses partial pressures instead of concentration.
For a general gas-phase reaction:
The Kp expression is:
where:
- P[A], P[B], P[C], P[D] are the partial pressures of each gas.
- a, b, c, d are the stoichiometric coefficients from the balanced equation.
Only gases are included in the Kp expression – solids and liquids are ignored.
Calculating Partial Pressure
The total pressure of a gaseous system at equilibrium is directly related to the number of moles of each gas in the mixture.
How much pressure one type of gas contributes to the total pressure is called its partial pressure.
All partial pressures of gases in a system add up to give the total pressure of the system.
The mole fraction of a gas is the moles of that gas in the mixture compared to moles of all gases. It is calculated using:
Find Kp for the following system, given the total pressure = 400 kPa and the moles of N2O4 and NO2 at equilibrium are 0.40 and 0.60, respectively.
Reaction: N2O4(g) ⇌ 2NO2(g)
- Total pressure = 400 kPa
Moles of N2O4 = 0.40
Moles of NO2 = 0.60 - Calculate mole fractions:
X(N2O4) = 0.40 / (0.40 + 0.60) = 0.40
X(NO2) = 0.60 / (0.40 + 0.60) = 0.60 - Calculate partial pressures:
P(N2O4) = 0.40 × 400 = 160 kPa
P(NO2) = 0.60 × 400 = 240 kPa - Kp expression: Kp = (P²[NO2]) / (P[N2O4)
Kp = (240²) / (160) = 360 kPa
Calculating Kc and Kp from Experimental Data
To calculate Kc or Kp using experimental data:
- Write the balanced equation and the correct expression for Kc or Kp.
- Use data given to determine equilibrium concentrations (for Kc) or partial pressures (for Kp).
- Substitute values into the expression.
- Ensure correct units:
- Kc units depend on the number of moles (mol dm⁻³).
- Kp units depend on the reaction stoichiometry (commonly atm, but can be Pa).
Example: For A ⇌ B + C (all gases), Kp = (pB)(pC)/(pA).
Effect of Temperature on Equilibrium Constant
Temperature is the only factor that changes the value of Kc or Kp.
If forward direction is exothermic:
- Increasing temperature shifts equilibrium left (towards reactants).
- Kc and Kp values decrease.
If forward direction is endothermic:
- Increasing temperature shifts equilibrium right (towards products).
- Kc and Kp values increase.
This is because temperature alters the balance of the forward and reverse reaction rates, changing the equilibrium position and therefore the constant.
What Does Not Affect the Equilibrium Constant
- Concentration changes: Affect equilibrium position, not Kc or Kp.
- Pressure changes (for gases): May shift position, but Kp remains constant unless temperature is changed.
- Catalysts: Increase rate of both forward and reverse reactions equally – no effect on Kc or Kp.
- Catalysts help the system reach equilibrium faster but do not change the equilibrium composition or the value of the constant.
Summary
- Kp is the equilibrium constant for gases, using partial pressures.
- Partial pressure = mole fraction × total pressure.
- Kp is only affected by temperature, not by concentration, pressure, or catalysts.
- Exothermic reactions: higher temperature decreases Kp.
- Endothermic reactions: higher temperature increases Kp.