chemical equilibria, Le Chatelier’s principle and Kc (A1 physical chemistry) Flashcards
explain the forward and backward reaction in a reversible reaction
forward reaction - initially reactants are used up quickly but this slows as their concentration drops
backward reaction - initially reactants are formed slowly but then speed up as the concentration of products increases
equilibrium is reached when the rate of the forward reaction equals the rate of the backward reaction, this is dynamic equilibrium and the concentration of each substance remains constant
this only occurs in closed systems (matter cannot enter or leave but energy can be exchanged with the surroundings)
this can be represented on a graph of concentration (y-axis) against time (x-axis)
state Le Chatelier’s principle
if a reaction at equilibrium is subjected to a change in pressure, temperature, or concentration, the position of equilibrium will move to counteract the change
if the change in condition results in equilibrium shifting to the left, we will make more reactants
if the change in condition results in equilibrium shifting to the right, we will form more products
this only happens in homogenous equilibria (reactants and products in the same state)
effect of changing concentration on equilibrium
if we increase the concentration of a reactant or product, the equilibrium will shift to try to reduce the concentration
if we increase the concentration of one of the reactants, equilibrium will shift to the right to use it up and reduce the concentration, therefore producing more product
if we increase the concentration of one of the products, equilibrium will shift to the left to use it up and reduce the concentration, therefore producing more reactants
if we decrease the concentration of one of the reactants, equilibrium will shift to the left to try and make up for this loss, therefore more reactants will be made
if we decrease the concentration of one of the products, equilibrium will shift to the right to try and make up for this loss, therefore more products will be made
effect of changing pressure on equilibrium
if we increase the pressure in a reaction, equilibrium will shift to try to reduce the pressure
if we increase the pressure, equilibrium will shift to the side with the fewest number (moles) of gas particles to reduce this pressure
if we decrease the pressure equilibrium will shift to the side with the most number (moles) of gas particles to increase the pressure
effect of changing temperature on equilibrium
if we increase the temperature, the equilibrium will shift to try to reduce the temperature
if we increase the temperature, equilibrium will shift in the endothermic direction to reduce the temperature
if we decrease the temperature, equilibrium will shift in the exothermic direction to increase the temperature
what factors do not affect the position of equilibrium?
catalysts - speed up the rate of the forward and back backward reaction equally, speed up the rate at which equilibrium is reached but have no effect on yield
Le Chatelier’s principle making ethanol
C₂H₄ (g) + H₂O (g) ⇌ C₂H₅OH (g) ΔH = -46 kJmol⁻¹
pressure - 60 atm
temperature - 300°C
catalyst - phosphoric acid (H₃PO₄)
since the forward reaction is exothermic, decreasing the temperature will mean equilibrium shifts to the right, producing more ethanol
however a lower temperature means lower rate of reaction, therefore 300°C is a compromise between yield and rate
as there are less moles of gas on the product side, a high pressure means equilibrium will shift to the right to reduce pressure, producing more ethanol
high pressure also increases rate
however high pressure environments are expensive so 60 atm is a compromise between yield/speed and cost
the equilibrium constant e.g. write the Kc expression and calculate the value of Kc of the following reaction, where concentration of SO₂ is 0.4moldm⁻³, concentration of O₂ is 0.2moldm⁻³, and concentration of SO₃ is 0.8moldm⁻³
2SO₂ (g) + O₂ (g) ⇌ 2SO₃ (g)
Kc can be worked out from the molar concentration of species in a reaction
products over reactants, molar values are powers, concentration is represented by square brackets
- write the Kc expression from the balanced equation
2A + B ⇌ 2C + D
Kc = [C]² [D] / [A]² [B]
Kc = [SO₃]² / [SO₂]² [O₂]
Kc = [0.8]² / [0.4]² [0.2] = 20
- work out the units by writing them all out and cancelling, then inverting the powers of the unit is on the bottom
Kc = moldm⁻³ moldm⁻³ / moldm⁻³ moldm⁻³ moldm⁻³
Kc = / moldm⁻³
Kc = mol⁻¹dm³
Kc = 20mol⁻¹dm³
equilibrium concentrations e.g. nitrogen dioxide decomposes according to this reaction
2NO₂ (g) ⇌ 2NO (g) + O₂ (g)
33.2 g of NO₂ was heated in a vessel with volume of 9.65dm³ at 450°C and 6.88 g of O₂ were found in the equilibrium mixture, calculate the value of Kc
- find the initial number of moles of reactants using mass / Mr (moles of products at the start of the reaction are 0)
- find the moles at equilibrium of given product
- use the molar ratio to work out the change in moles of the other species using this value
- divide equilibrium moles by volume given to find concentrations
- work out Kc using these concentrations
initial moles of NO₂ = 33.2g / (14+16+16) = 0.72mol
equilibrium moles of O₂ = 6.88g / (16+16) = 0.215mol
therefore change in moles for O₂ is +0.215
since there are 2 moles of NO for every mole of O₂, the change in moles of NO is +0.430
since there are 2 moles of NO₂ for every mole of O₂, and it is on the reactants side, the change in moles of NO₂ is -0.430
therefore equilibrium moles for NO₂ are 0.72 - 0.430 = 0.29
[NO₂] = 0.29 / 9.65 = 0.030moldm⁻³
[NO] = 0.430 / 9.65 = 0.045moldm⁻³
[O₂] = 0.215 / 9.65 = 0.022moldm⁻³
Kc = [0.045]² [0.02] / [0.03]²
Kc = moldm⁻³ moldm⁻³ moldm⁻³ / moldm⁻³ moldm⁻³
Kc = 0.050moldm⁻³
using Kc to find equilibrium concentrations e.g. a reaction took place and equilibrium was established when ethanol reacts with ethanoic acid. at equilibrium, the concentration of ethanoic acid was 1.8moldm⁻³ and ethanol was 3.2moldm⁻³, the value of Kc is 3.5 at 25°C, calculate the concentrations of the products at equilibrium
CH₃COOH (l) + C₂H₅OH (l) ⇌ CH₃COOC₂H₅ (l) + H₂O (l)
- write the expression and substitute in known values
- rearrange the calculation
- we assume the concentration of all products are the same, therefore square root value
Kc = [CH₃COOC₂H₅] [H₂O] / [CH₃COOH] [C₂H₅OH]
3.5 = [CH₃COOC₂H₅] [H₂O] / [1.8] [3.2]
[CH₃COOC₂H₅] [H₂O] = 3.5 * 1.8 * 3.2 = 20.16
[CH₃COOC₂H₅] = [H₂O]
√20.16 = 4.49moldm⁻³
[CH₃COOC₂H₅] = 4.49moldm⁻³
[H₂O] = 4.49moldm⁻³
factors affecting Kc
Kc is only valid for one temperature as changing the temperature will change the equilibrium concentrations
if temperature change causes equilibrium to shift right, Kc will increase i.e when product concentration increases
if temperature change causes equilibrium to shift left, Kc will decrease i.e when reactant concentration increases
the value of Kc is unaffected by any changes in concentration
adding a catalyst has no effect on the value of Kc as a catalyst speeds up the forward and reverse reaction, hence only speeding up the rate at which equilibrium is established
