TOPIC 10 Chemical Equilibrium

Question 1

How do reversible reactions occur?

Answer 1

When there is unconvinced reactants left over in the product mixture

Question 2

When is a system said to be in equilibrium?

Answer 2

When there is no further change in concentration

Question 3

In the reversible reaction between Hydrogen and Iodine to produce Hydrogen Iodide, what happens to the reactants and therefore the forward reaction over time?

Answer 3

The concentrations decrease and so the rate of reaction decreases

Question 4

In the reversible reaction between Hydrogen and Iodine to produce Hydrogen Iodide, what happens to the product and therefore the backward reaction over time?

Answer 4

The concentration increases and so the rate of reaction increases

Question 5

When is a system said to be in dynamic equilibrium?

Answer 5

When the rates of reactions becomes equal and there is no further change in concentration

Question 6

Which 2 conditions must be met for dynamic equilibrium to be established?

Answer 6

Reaction must be reversible

Must be in a closed container

Question 7

In what position does the equilibrium shift when the concentration of the reactants is increased and decreased?

Answer 7

Increased - to the left

Decreased - to the right

Question 8

In what position does the equilibrium shift when the concentration of the products is increased and decreased?

Answer 8

Increased - to the left

Decreased - to the right

Question 9

In what position does the equilibrium shift when the pressure is increased and there is more moles in the reactants and less moles in the reactants?

Answer 9

More - to the right

Less - to the left

Question 10

In what position does the equilibrium shift in an exothermic reaction when the temperature is increased and decreased?

Answer 10

Increased - to the left

Decreased - to the right

Question 11

In what position does the equilibrium shift in an endothermic reaction when the temperature is increased and decreased?

Answer 11

Increased - to the right

Decreased - to the left

Question 12

What is the effect of adding a catalyst to an equilibrium mixture?

Answer 12

The rate of both the forward and backward reactions will increase by the same amount
The position of equilibrium is not altered
Reduces the time taken to establish equilibrium

Question 13

What is the effect of increasing the temperature of an equilibrium mixture?

Answer 13

The rates of both the forward and backward reactions will increase but the increase in the rate of the endothermic reaction will be greater

Question 14

What is the general formula for calculating Kc when the equation is :

aA + bB = cC + dD?

Answer 14

Kc= [C]^c [D]^d/[A]a [B]^b

Question 15

What temperature is used for the Haber process and why?

Answer 15

450 c

Low temperature increases ammonia yield but is slow even when a catalyst is added

Question 16

What pressure is used for the Haber process and why?

Answer 16

250 atm as high pressure favours the forward reaction but pressures that are too high are expensive to maintain

Question 17

How is efficiency of the Haber process increased?

Answer 17

The unreacted nitrogen and hydrogen , after separation from ammonia , is mixed with fresh nitrogen and hydrogen and fed into the reaction chamber

Question 18

What is the catalyst used for the Haber process in industry?

Answer 18

Iron

Question 19

What is the equation for the Contact process? Include state symbols and 🔼H.

Answer 19

SO2 (g) + 1/2O2 (g) = SO3 (g)

🔼H = -96 kJmol-1

Question 20

What temperature is used for the contact process and why?

Answer 20

450 c

Low temperature = high yield but catalyst would be ineffective at low temperatures

Question 21

What pressure is used in the contact process and why?

Answer 21

2 atm because it produces a high yield of SO3 and any higher would be more expensive

Question 22

What catalyst is used in the contact process and what are the equations (2) for the mechanisms for the reaction?

Answer 22

Vanadium (V) oxide

SO2 (g) + V2O5 (s) = SO3 (g) + V2O4 (s)

V2O4 (s) + 1/2O2 (g) = V2O5 (s)