Kp

Question 1

What are mole fractions 🟡

Answer 1

The mole fraction of a component, A, in a gaseous mixture is denoted Xa
- and is calculated by dividing amount, in moles, of A
- by the total amount of moles of gas in the mixture: 🟡
> Xa = moles of A/total moles

Mole fractions can only have values in the range O to 1.
>The sum of mole fractions of all of components in a mixture will be equal to 1.

Question 2

Whats partial pressure 🟡

Answer 2

The partial pressure (pp) of a component, eg A, in mixture of gases
- is the contribution that gas makes to total pressure of the gas mixture.
- It is also pressure if same amt of A were only substance in a container of same size.

> the partial pressure of A is denoted by Pa
- The pp of component can be calculated 🟡
- total pressure of mixture x mole fraction of component.
- pa = xa x P

Although partial pressures and concentrations are not the same thing,
> are both proportional to the amount of the substance present.

Question 3

Expression of Kp

Answer 3

Kp
For the general case of a gaseous homogeneous equilibrium:
aA(g) + bB(g) <> cC(g) + dD(g)

the equilibrium law states that
Kp= p(C)^c x p(D)^d / p(A)^a x p(B)^b
> where p(A)ª is the pp of A raised to the power of a, etc.
>
N(g) + 3H2(g) = 2NH3 (g)
Kp = p(NH3)^2 / p(N) x p(H2)^3

Calculation of Kp
> similar to those to find Kc
> The only diff is after finding number of moles at equilibrium
- instead of conc= moles/vol
- partial pressures are used where
pp = mole fraction x total pressure.

DIAGRAM EXAMPLES 1,2

Question 4

Example of units if kp

Answer 4

(Pa)2/(Pa)3 x (Pa)
= (Pa)2/(Pa)4
= 1/(Pa)2
= Pa-2

Question 5

Changes in temperature and other factors on Kp

Answer 5

• Kp is constant at particular temperature.
• The equilibrium constant, Kp, is affected by changes in temp, but not by changes in pressure,
•
• the presence of a catalyst or changes in conc in mixture.
• This is same for all equilibrium constants wch are constant at a particular temp
• and are not affected by changes in any other quantities.

..

• For the equilibrium:
N2(g) + 3H2 (g) = 2NH:g) ΔH= -92 kJmol-1
• Increasing temp favours reverse reaction and position of equilibrium will move to left.
- This will decrease pp of NH and Kp will decrease.

• Increasing pressure will favour forward reaction and Position of equilibrium will move right.
- This will have no effect on value of Kp as long as temp remains constant.

• Adding more hydrogen to the mixture will increase pp of hydrogen and will move position of equilibrium to right.
- will have no effect on the value of K, as long as the temperature remains constant.