chemical equilibrium chapter 17

reversible reaction

A reversible reaction is one in which the reactants react to form products that then react to give the reactants back i.e. the reaction is going in both directions and does not go to completion

chemical equilibrium

Chemical equilibrium is the dynamic state where the RATE OF the forward reaction is equal to the RATE OF the reverse reaction

dynamic state

reactants are continuously forming products and the products are continuously forming reactants

Does a reaction at equilibrium cease? Explain.

• No, a reaction at equilibrium has not ceased

• System is in a dynamic state – rate of forward reaction equals rate of reverse reaction

Looking at how equilibrium is reached for a reversible reaction

Take the reaction: A + B ←→C + D

a) Initially - Reactants present, no product present

b) Reaction begins - Products formed and start reacting, rate of reverse reaction increases

c) Equilibrium reached - Rate of forward reaction = rate of reverse reaction

Important points:

1) Once equilibrium is reached the concentrations of reactants and products remain constant as long as temperature remains constant

2) A reaction being at equilibrium does not mean that the concentration of products and reactants are equal

Where: [products] greater than [reactants] – equilibrium lies on the right

Where: [reactants] greater than [products] – equilibrium lies on the left

3) In a reaction at equilibrium, the reaction does not go to completion/does not cease and there will be some of all reactants and products involved present

le chateliers principle

if a stress is applied to a system then the system re-adjusts to relieve the stress applied

What effect does a catalyst have on equilibrium? Explain.

• None - A catalyst can bring a reaction to equilibrium more quickly but does not affect the state of equilibrium (Will not affect yields or colour changes)

• A catalyst speeds up the forward reaction and the reverse reaction equally - the dynamic equilibrium position is not changed.

temperature

Increasing temperature: Equilibrium will shift to decrease temperature - in the ENDOTHERMIC DIRECTION TO USE UP HEAT

Decreasing temperature: Equilibrium will shift to increase temperature - in the EXOTHERMIC DIRECTION TO PRODUCE HEAT

Exothermic reaction - produce heat - increase temperature - △H- (forward reaction)

Endothermic reaction - uses up heat - decreases temperature - △H+ (forward reaction)

pressure

Note : Very important

(1) Changing pressure will have no effect on equilibrium unless it is a gaseous reaction at i.e. where all reactants and products are gases

(2) Changing pressure will have no effect on equilibrium if there are equal numbers of molecules on each side of the equation

Increasing pressure: Equilibrium will shift to decrease pressure – in the direction that has LESS MOLECULES

Decreasing pressure: Equilibrium will shift to increase pressure - in the direction that has MORE MOLECULES

concentration

Increasing a substance’s concentration: Equilibrium will shift to decrease its concentration - in the direction that USES IT UP

Decreasing a substance’s concentration: Equilibrium will shift to increase its concentration – in the direction that produces MORE OF IT

ideal contiodns

high pressure low temp

Give two industrial applications of Le Chatelier’s principle: 1) Manufacture of ammonia

Known as the Haber process

N2 + 3H2 ←→ 2NH3

∆H = - 92.4 kJmol-1

Notice: An iron catalyst is used

According to Le Chatelier’s principle, what conditions should be used to maximise the yield of ammonia obtained?

- High pressures

- Low temperature

Why are these conditions not used in practice during the Haber process

- High pressures - uneconomical, danger of gas leaks and explosions

- Low temperatures – rate of reaction is too slow, takes too long to come to equilibrium

Note: Compromise pressure 200 atm is used

Compromise temperature 500°C is used

Give three uses of ammonia

- Fertilisers

- Explosives

- Cleaning product

Give two industrial applications of Le Chatelier’s principle: 2) Manufacture of sulfuric acid

• Known as the Contact process

• Three reactions involved - second reaction is a reversible reaction

Reaction 1:

S + O2 → 2SO2

Reaction 2:

2SO2 + O2 ←→ 2SO3

∆H = - 196 kJmol -1

Notice: A vanadium (V2O5) catalyst is used

Reaction 3:

SO3 + H2O → H2SO4

Why is a platinum catalyst not used?

Platinum is easily ‘poisoned’ by impurities in the reactants

According to Le Chatelier’s principle, what conditions should be used to maximise the yield of sulfuric acid obtained?

- High pressures

- Low temperatures

Why are these conditions not used in practice during the contact process?

- High pressures - uneconomical, SO3 will liquify at too high pressures

- Low temperatures – rate of reaction is too slow, takes too long to come to equilibrium

Note: Compromise pressure 1-2 atm is used (just above atmospheric pressure)

Compromise temperature 450°C is used

Give three uses of sulfuric acid

- Car batteries

- Detergents

- Paints

What is meant by the equilibrium constant (KC)?

The equilibrium constant shows the mathematical relationship between the concentration

of the reactants and the products for a reaction at equilibrium

The equilibrium constant

Important points:

1) The concentration of the products are on top, and concentration of reactants are on the bottom of the expression for Kc

2) This mathematical relationship is always the case provided temperature remains constant – only changing temperature changes the value of KC

Changing concentrations i.e. adding or removing a substance does not affect KC

Changing pressure does not affect Kc

DO NOT FORGET: TEMPERATURE IS THE ONLY FACTOR THAT AFFECTS KC

Kc is always constant at constant temperature

3) Kc greater than 1; concentration of products greater than reactants – equilibrium lies on right

Kc less than 1; concentration of reactants greater than products – equilibrium lies on left

Therefore: If equilibrium shifts to right and forward reaction is favoured, Kc increases

If equilibrium shifts to left and reverse reaction is favoured, Kc decreases

An increase in Kc means the forward reaction is being favoured (more products are being made)

A decrease in Kc means the reverse reaction is being favoured (more reactants are being made

Calculations involving KC

Type 1: Calculating KC from given moles/concentrations/mass/%

Steps:

1) Write out the balanced equation for the equilibrium reaction

2) Open up an ‘ICEE’ table and fill in all ‘known’ values

Notes - must be working in moles

‘Initial moles’ of the product (s) always = 0

(Unless initial value for product is given, in which case initial moles of reactant (s) = 0)

3) Use the molar ratio in the balanced equation to complete the ‘change in moles’ line

4) Complete the ‘ICEE’ table

Note - If the number of molecules is equal on both sides of the equation, the volumes will cancel in

the Kc expression

In this situation, volume is usually not given but even if it is it can be cancelled and

5) Fill in the molarities of reactants and products into the Kc equilibrium constant expressionand calculate KC

Calculations involving KC

Type 2: Calculating moles/concentration/mass/% from a given KC

Steps:

1) Write out the balanced equation for the equilibrium reaction

2) Open up an ‘ICEE’ table and fill in all ‘known’ values

Notes - must be working in moles

‘Initial moles’ of the product (s) always = 0

(Unless initial value for product is given, in which case initial moles of reactant (s) = 0)

3) Use ‘x’ to represent moles of reactants and products and use the molar ratio in the balanced equation to complete the ‘change in moles’ line in terms of ‘x’

4) Complete the ‘ICEE’ table in terms of ‘x’

Note - If the number of molecules is equal on both sides of the equation, the volumes will cancel in the Kc expression

In this situation, volume is usually not given but even if it is it can be cancelled

5) Fill in the molarities terms of ‘x’ of reactants and products into the Kc equilibrium constant expression and allow it to equal the known value for KC

6) Work as far as a quadratic equation and use to find ‘x’

(or if your maths is good, you might be able to use algebra)

Note: two values for x will be obtained using the quadratic formula – one will be impossible i.e. a negative value or too large

7) Use ‘x’ to find the quantity asked for in the question.