19 Equilibrium Constant Kp

Question 1

Discuss with reasons the effect of increasing separately the temperature and the pressure on the yield of the products a d on the rate of reaction

Answer 1

Temperature
The forward reaction is endothermic
Increase of rate of reaction
Equilibrium would shift to the right making more product ( increasing the yield to oppose the change)

Pressure
Increase the pressure
Increase the rate of reaction
Favour the side with less moles
Reducing the yield
 Equilibrium would shift to the left hand side

Question 2

Why would industrialist not used a certain temperature / pressure

Answer 2

Pressure would be dangerous and costly

Low temperature reduces the rate of reaction

Question 3

partial pressure A

Answer 3

mole of fraction of A x total pressure

Question 4

mole fraction of gas A

Answer 4

number of moles of A/ total number of moles of gas

Question 5

3H2 (g) + N2 (g) —— 2NH3 (g) ∆H = -92kjmol-1

what would happen if you increased the temperature?

Answer 5

decrease the yield of ammonia

Question 6

3H2 (g) + N2 (g) —— 2NH3 (g) ∆H = -92kjmol-1

what would happen if you increase the pressure ?

Answer 6

shifts to side with less moles

increase the yield of ammonia

Question 7

Kp

Answer 7

The equilibrium constant Kp is deduced from the equation for a reversible reaction occurring in the gas phase.
Kp is the equilibrium constant calculated from partial pressures for a system at constant temperature.

Question 8

catalyst and Kp

Answer 8

whilst a catalyst can affect the rate of

attainment of an equilibrium, it does not affect the value of the equilibrium constant