Module 5 Chapter 19
19.1 Equilibrium Constant- Kc
19.3 Controlling the position of equilibrium
19.2 Equilibrium Constant- Kp
Applies to (aq) and (g) reactions and requires concentrations
If less than 1 equilibrium lies on left (more reactants than products)
If more than 1 equilibrium lies on right (more products than reactants)
Equilibria Types
1) Homogeneous- All species have the same state
2) Heterogeneous- Species have different states
Units- work out my looking at equation and cancelling
Mole Fraction
Calculations
1) Calculate mole at equilibrium
2) Find equilibrium concentrations
3) Calculate Kc
Under constant temp and pressure the same volume of different gases contain the same number of moles of gas molecules
Sum of mole fractions of a gas mixture must equal 1
Partial Pressure
Definition- the contribution that the gas makes towards total pressure- sum of partial pressures equals total pressure
Kp
Kp is written the same way as Kc but with partial pressures instead of concentrations
Units- kPa or atm
Only include gaseous species, other species are ignored
Equation
Calculations
1) Mole Fraction
2) Partial pressure
3) Calculate Kp
Shifting Equilibrium
1) Concentration of species
2) Changing pressure
3) Changing Temperature
Equilibrium constants
K gives exact position of equilibrium
At a set temperature K is CONSTANT despite any changes to pressure, concentration or catalysts
If temperature is changed then K CHANGES
Temperature
EXOTHERMIC
ENDOTHERMIC
INCREASE- K decreases, yield decreases (less product)
DECREASE- K increases, yield increases (more product)
INCREASE- K increases, yield increases (more product)
DECREASE- K decreases, yield decreases (less product)
Explaining
Exothermic
Endothermic
2SO2(g) + O2(g) -> 2SO3(g) Enthalpy change= -197kJmol-1
Increase temp. - equilibrium to left
1) Kc/Kp decreases with increasing temp.
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2) Equation- (At equilibrium)
3) Change in temp.-system no longer in equilibrium. Ratio is greater than Kp so pp of product (SO3) must decrease and pp of reactants (SO2 and O2) must increase
4) Position of equilibrium shifts to left
5) New equilibrium where equation=Kp
1) Kc/Kp increases with increasing temp.
N2(g) + O2(g) -> 2NO(g)
Enthalpy change= +180kJmol-1
Changing other factors
Increase temp. equilibrium to right
2) Equation- (At equilibrium)
3) System not in equilibrium (change of temp)- ratio is less than Kp so pp of product (NO) must increase and pp of reactants (N2 and O2) must decrease
4) Equilibrium shift to right
5) New equilibrium where equation= Kp
As k is constant when we change factors other than temp we can use k to work out new equilibrium concentrations when a different change takes place
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1) Concentration
2) Pressure
Increasing concentration of a reactant
3) Catalyst
Increasing pressure at equilibrium
Forward and reverse increase at same rate so no change in position- equilibrium met quicker
N2O4-> 2NO2 p(N2O4)= 2atm p(NO2)=48atm Total P= 50atm
Kp= p(NO2)2/p(N2O4)= (48)2/(2)= 1152atm
Total pressure increased to 100atm the pp of both substances will initially double
p(N2O4)= 4atm p(NO2)=96atm Kp= 2304atm (out of equil.)
p(NO2) must decrease and p(N2O4) must increase until p(NO2)2/ p(N2O4)= 1152atm to get to equil. and Kp is restored
[A]= 4.0moldm-3 [B]= 0.80moldm-3 at equil.
Kc= [B]/[A]= 0.80/4.0= 0.20
Raise concentration of A to 8.0moldm-3- system not in equilibrium as [B]/[A]=0.1 (less than Kc)
To get back to equilibrium and Kc conc. of B must increase and conc. of A must decrease until [B]/[A]= 0.20