Kp

Question 1

Def.

Total pressure.

Answer 1

• Sum of all the pressures of the individual gases (ie. all partial pressures.)

Question 2

SO₂Cl₂(g) –(reversible reaction sign) –> SO₂(g) + Cl₂(g)
At equilibrium, reaction vessel 263kPa of Cl₂. Total pressure in vessel = 714kPa. Calculate partial pressure of SO₂Cl₂ at equilbrium.

Answer 2

• SO₂ has 1:1 molar ratio with Cl₂ because they both on products side!
• Partial pressure of Cl₂= pp of SO2.
• So, 714 - 263-263 = 188kPa.

Question 3

How would you notate partial pressure of SO₂Cl₂?

Answer 3

• p(subscript SO₂Cl₂)

Question 4

True or False

SO₂Cl₂(g) –(reversible reaction sign) –> SO₂(g) + Cl₂(g).
We would say SO₂Cl₂,SO₂, Cl₂ are in molar ratio of 1:1:1.

Answer 4

• False.
• We can’t say this because we can’t equate products/ reactants.
• SO₂ + Cl₂ are in molar ratio of 1:1.

Question 5

How to calculate mole fraction of gas?

Answer 5

• number of moles of gas/ total no of moles of gas in mixture

Question 6

How do you calculate partial pressure using mole fractions?

Answer 6

• Partial pressure = mole fraction x total pressure of mixture

Question 7

3 mol SO₂Cl₂ heated, equilibrium mixture = 1.75mol Cl₂. Total pressure = 714kPa. Partial pressure of SO₂Cl₂ ?

Answer 7

• Work out equilbrium moles.
• Mole fraction: 1.25/ 4.75 x 714 = 188kPa.

Question 8

How is equilbrium constant Kp deduced?

Answer 8

• Deduced from equation for reversible reaction in gas phase.

Question 9

What is Kp?

Answer 9

• Kp is the equilibrium constant calculated from partial pressures for a system at constant temperature.

Question 10

Construct an equation for Kp using this:
N₂(g) + 3H₂(g) –(reversible reaction sign) –> 2NH₃(g)

Answer 10

Kp= (pNH₃)²/ (pN₂)(pH₂)³
Note: we use CURLY brackets here, unlike Kp!

Question 11

2A +B –(reversible) –> C +D
Partial pressures. A = 324kPa, B= 162kPa, C/D = 182kPa. Calculate value of Kp!

Answer 11

-1.95x10⁻³kPa⁻¹

Question 12

True or False

The value of Kp calculated from equation will always remain the same.

Answer 12

• False!
• Kp = temperature dependent constant.
• If temperature changes equilibrium pressures change and so Kp changes.

Question 13

What effect does temperature have on value of Kp?

Answer 13

• If temp change causes equilibrium to shift right, Kp will increase.
• If temp change causes equilbrium to shift left, Kp will decrease.

Question 14

2SO₂+O₂– (reversible reaction) –> 2SO₃.
∆H= -197kJmol⁻¹.
What effect will increase/ decrease in temp have on value of Kp?

Answer 14

• Increase: equilbrium shifts to left, endothermic direction, and Kp decreases.
• Decrease: equilbirium shifts to right, exothermic direction, and Kp will increase.

Question 15

True or False

A change in pressure will change the value of Kp.

Answer 15

• False.
• Kp is unaffected by changes in pressure.

Question 16

True or False

Adding a catalyst will change value of Kp.

Answer 16

• False.
• No effect on Kp.
• Speeds up forward and reverse reaction at same rate.
• Increases rate of which equilibrium is reached.