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C19 - Equilibrium - Coggle Diagram
C19 - Equilibrium
Controlling the position of equilibrium
The effect of temperature on equilibrium constants K
Endothermic
If the forward reaction is endothermic
The Eq constant increases with increasing temperature
Raising the temperature increases the Eq yield of products
N2 + O2 >< 2NO
1100K - 4 x 10^-8
700K - 5 x 10^-13
1500K - 1 x 10^-5
Kp increases with increasing temperature, Eq position shifts to the right
Explaining the Eq shift
At Eq, Kp = p(NO)^2/ p(N2) x P(O2)
If the temperature increases from 700K to 1100K
The system is no longer in Eq
The ratio p(NO)^2/ p(N2) x P(O2) is now less than Kp
Kp increases from 5 x 10^-13 to 4 x 10^-8
The Eq partial pressures that gave a Kp value of 5 x 10^-13 at 700K must change to give the new Kp value of 4 x 10^-8 at 1100K
The partial pressure of the reactants must decrease
The position of Eq shifts towards the products
The partial pressure of the product must increase
A new Eq is established where p(NO)^2/ p(N2) x P(O2) is equal to the new Kp value of 4 x 10^-8
Exothermic
If the forward reaction is exothermic
The Eq constant decreases with increasing temperature
Raising the temperature decreases the Eq yield of products
Explaining the equilibrium shift
If the temperature increases from 500K to 700K
The system is no longer in Eq
The ratio p(SO3)2/ p(SO2)2 x p(O2) is now greater than Kp
Kp decreases from 2.5 x 10^10 to 3,0 x 10^4 atm-1
The Eq partial pressures that gave a Kp value of 2.5 x 10^10 at 500K must change to give the new Kp value of 3.0 x 10^4 at 700K
The partial pressures of the reactants must increase
The position of Eq shifts towards the reactants
The partial pressure of the product SO3 must decrease
A new Eq will be reached where p(SO3)2/ p(SO2)2 x p(O2) is equal to the new Kp value of 3.0x10^4
At Eq, Kp = p(SO3)2/ p(SO2)2 x p(O2)
2SO2 + O2 >< 2SO3
700K - 3.0 x 10^4
1100 - 1.3 x 10^-1
500K - 2.5 x 10^10
Kp decreases with increasing temperature, equilibrium position shifts to left
The explanation for Kp also works for Kc
How do changes in concentration and pressure affect equilibrium constants
Eq constants and pressure changes
If the total pressure is doubled, p(N2O4) is increased to 19.2 and p(NO2) is increased to 0.48
The ratio p(NO2)2 / p(N2O4) is now greater than Kp and equal to 768atm
The system is now no longer in Eq
The partial pressures must change to return the ratio back to the Kp value of 384
The partial pressure of the reactant must increase
A new Eq position will be reached where p(NO2)2 / p(N2O4) is restored to its Kp value of 384atm
The partial pressure of the product must decrease
At a constant temperature, the Eq partial pressures are
p(N2O4) = 0.24atm
Kp = p(NO2)2 / p(N2O4) = 9.6^2/0.24 = 384atm
p(NO2) = 9.6atm
You predict a shift in the Eq position from right to left, to the side with fewer gaseous moles
Summary of Effects of increasing pressure
More moles of gaseous products, ratio > K, products decrease reactants increase, shift to the left
Same number of moles of gaseous reactants and products, ratio = K, no change, no effect on position
Fewer moles of gaseous products, ratio < K, products increase reactants decrease, shift to the right
Eq constants and concentration changes
If [N2O4] is increased to 0.02moldm-3
The system is now no longer in Eq
The ratio [NO2]2/[N2O4] is now less than Kc and is equal to 0.4^2/0.02 = 8.0moldm-3
The concentrations must change to return the ratio back to the Kc value of 16.0moldm-3
The concentration of the reactant must decrease
The concentration of the product must increase
A new Eq is established where [NO2]2/[N2O4] is restored to its Kc value of 16
At a constant temperature, the Eq concentrations are
Kc = [NO2]2/[N2O4] = 0.4^2/0.01 = 16.0moldm-3
[N2O4] = 0.01 moldm-3
[NO2] = 0.4 moldm-3
You would also predict a shift in the Eq position from left to right to decrease [N2O4]
N2O4 >< 2NO2
Le Chatelier's principle only works because Kc controls the relative concentrations of reactants and products present at Eq
Do equilibrium constants change
At a set temperature, K is constant and doesn't change despite any modifications to concentration, pressure of the presence of a catalyst
The magnitude of an equilibrium constant K indicates the extent of a chemical equilibrium
K= 100 indicates an equilibrium well in favour of the products
K = 1 indicates an equilibrium halfway between reactants and products
K = 1 x 10^-2 indicates an equilibrium well in favour of the reactants
K does change if the temperature is changed - a temperature change is the only condition that will cause K to change its value
How does a catalyst affect equilibrium constants
Catalysts speed up both the forward and reverse reactions in the Eq by the same factor
Eq is reached quicker but the Eq position isn't changed
Catalysts affect the rate of a chemical reaction but not the position of the Eq
Why does the equilibrium position shift
If the pressure is increased, the equilibrium position shifts towards the side with fewer gaseous moles
If the temperature is increased, the equilibrium position shifts in the endothermic direction
If the concentration of a species is increased, the equilibrium position shifts in the direction that reduces the concentration
The equilibrium constant Kc
Homogenous equilibria
Contains equilibrium species that all have the same state of phase and are gases
The Kc expression contains concentrations of all species
Heterogenous equilibria
The concentration of solids and liquids are essentially constant
The Kc only includes species that are gases or aqueous
Contains equilibrium species that all have different states or phases
Kc = products/reactants
Determining Kc from experimental results
Results
Amount of HCl in control = 0.0500mol
Amount of HCl and CH3COOH in Eq mixture = 0.115mol
Calculation of Kc
Use the Eq equation to determine the Eq amounts of each component
Initial Mole - 0.100 0.100 0 0.500
Change mol - -0.035 -0.035 +0.035 +0.035
Equation - CH3COOH + C2H5OH >< CH3COOC2H5 + H2O
Eq Mole - 0.065 0.065 0.035 0.535
Find the Eq concentrations in moldm-3
[C2H5OH] = 0.065/0.02 = 3.25moldm-3
[Ch3COOC2H5] = 0.035/0.02 = 1.75moldm-3
[CH3COOH] = 0.065/0.02 = 3.25moldm-3
[H2O] = 0.535/0.02 = 26.75moldm-3
The volume of the solution is 20cm3 = 0.02dm3 so Eq amounts are divided by 0.02 to give the concentrations in moldm-3
Determine the Eq amount of CH3COOH
0.115 - 0.05 = 0.065mol
Write the expression for Kc, and calculate
Kc = [CH3COOC2H5][H2O]/[CH3COOH][C2H5OH]
1.75 x 26.75/3.25 x 3.25 = 4.43
Method
Stopper both glasks and leave for a week to reach equilibrium
Carry out a titration on the Eq mixture using a standard solution of sodium hydroxide
Add 0.0500 mol of HCl to a second conical flask as a control
Repeat the titration with the control to determine the amount of acid catalyst that had been added
In a conical flask, mix together 0.100 mol CH3COOH and 0.100 mol C2H5OH
Add 0.0500 mol of HCl as an acid catalyst to the flask
The total volume of the mixture in the flask is 20.0cm^3
The amount of water in the aqueous acid catalyst is 0.500mol
The equilibrium constant Kp
Partial Pressure
The sum of partial pressures of each gas equals the total pressure
Partial pressure = mole fraction x total pressure
The partial pressure of a gas is the contribution that the gas makes towards the total pressure
Calculating Kp
Homogenous Eq
Find the partial pressures
p(H2) = 54/120 x 200 = 90atm
p(NH3) = 48/120 x 200 = 80atm
p(N2) = 18/120 x 200 = 30atm
Calculate Kp
N2 + 3H2 >< 2NH3
Kp = p(NH3)2/p(N2) x p(H2)3
Kp = 80^2/30 x 90^3 = 2.9 x 10^-4 atm-2
Find the mole fractions of nitrogen, hydrogen and ammonia
x(N2) = 18/120
x(H2) = 54/120
Total number of gas moles = 18 + 54 + 48 = 120mol
x(NH3) = 48/120
Heterogenous Eq
Write down the expression for Kp with only gaseous species
CaCO3(s) >< CaO(s) + CO2(g)
Kp = p(CO2)
Kp = 2.5 x 10^-2atm
Mole Fraction
number of moles of A/total number of moles in gas mixture
The sum of mole fractions must equal one
The mole fraction of a gas is the same as its proportion by volume to the total volume of gases in a gas mixture