Topic 11 Equilibrium II

Question 1

Partial pressure of a gas in a mixture of gases.

Answer 1

The pressure that the gas would exert if it alone occupied the volume of the mixture.

Question 2

Kc

Answer 2

Uses concentrations.

Question 3

Kp

Answer 3

Uses partial pressures.

Question 4

The only factor that affects the equilibrium constant…

Answer 4

… temperature!

Question 5

For a general reaction: wA + xB <–> yC + zD
What is the expression for Kc?

Answer 5

Kc= [C]^y [D]^z / [A]^w [B]^x

Question 6

Determining the equilibrium constant.

Answer 6

Known amounts of reactants are allowed to reach equilibrium with their products. At equilibrium, the concentration of one substance is measured and the rest are determined by stoichiometry. Units of the equilibrium constant vary.

Question 7

The concentration of a solid at a given temperature.

Answer 7

Determined by its density, which has a constant value, so solid terms can be excluded from the Kc expression.

Question 8

The total pressure of a mixture=

Answer 8

The sum of the partial pressures of all the gases present in the mixture. p = pA + pB

Question 9

For a general reaction: wA (g) + xB (g) <–> yC (g) + zD (g)
What is the Kp expression?

Answer 9

Kp = (pC)^y (pD)^z / (pA)^w (pB)^x

Question 10

How can we calculate the partial pressure of a gas?

Answer 10

pA= xAP
Partial pressure = mole fraction x total pressure.

Question 11

Mole fraction of A=

Answer 11

Number of moles of A/ Total number of moles of gas

Question 12

Solids and liquids in equilibrium with gases.

Answer 12

Ignored in the Kp expression.

Question 13

For an exothermic reaction, what happens to K (Kc or Kp)?

Answer 13

K decreases as the temperature increases. The equilibrium position shifts in the endothermic direction.

Question 14

For an endothermic reaction, what happens to K (Kc or Kp)?

Answer 14

K increases as temperature increases.

Question 15

Qc

Answer 15

Reaction quotient.

Question 16

Explain the effect on the position of equilibrium of increasing the concentration of H2 at constant temperature & volume, for this reaction: H2 (g) + I2 (g) <–> 2HI(g)?

Answer 16

Kc = Qc = [HI(g)]^2 / [H2(g)] [I2(g)]
If H2 increases, Qc decreases and is no longer = Kc. To re-establish equilibrium, the equilibrium shifts to the right as the denominator of Qc must decrease to increase the numerator. The 2 values adjust until Qc = Kc and a new equilibrium system is established.

Question 17

What happens to the equilibrium constant when pressure is increased?

Answer 17

It stays the same. The equilibrium position changes to maintain a constant value for Kc or Kp.

Question 18

If the partial pressure of only one component is changed:

Answer 18

The change in equilibrium pressure to maintain Kp can be predicted in the same way as with Kc.

Question 19

If the total pressure of a gaseous system is changed, all the partial pressures will change. Example: Kp = (pNH3)^2/ (pN2) (pH2)^3

Answer 19

Replacing NH3 with c, N2 with a and H2 with b gives Kp = c^2/ab^3.
If the total pressure is doubled: Qc = 4c^2/16ab^3 (by subbing in 2c, 2a & 2b). The value of Qc is now 1/4 the value of Kp, so the equilibrium shifts right to increase the numerator and decrease the denominator until Kc = Kp.

Question 20

The effect of the catalyst on the equilibrium constant:

Answer 20

No effect. Not present in the Kp/Kc expression as it doesn’t appear in the overall equation. It increases the rate equally in both directions, so has no effect on the final concentrations.

Question 21

Thermodynamically stable

Answer 21

A substance which has less energy in an enthalpy level diagram.

Question 22

Kinetically stable/inert

Answer 22

A substance with a high activation requires lots of energy to react.

Question 23

How would you remove a gas from equilibrium?

Answer 23

Condense it!

Question 24

When doing Kp/Kc calculations, what should you draw?

Answer 24

An ICE table.

Question 25

When is it possible to calculate Kc/Kp using equilibrium moles instead of equilibrium concentrations?

Answer 25

The volume cancels out in the Kc expression as there are the same number of moles on both sides of the equation.

Question 26

How can you confirm that a mixture has had sufficient time to reach equilibrium?

Answer 26

Leave teh mixture for longer. The titre value/Kc will remain unchanged.

Question 27

Why are solids seldom included in the Kc expression?

Answer 27

The concentration of a solid is constant.

Question 28

How does increasing pressure affect Kp?

Answer 28

It doesn’t!!! The equilibrium position moves to the side with teh fewest gas moles to keep Kp constant!

Question 29

What happens to a graph of concentration when it reaches equilibrium?

Answer 29

It levels off.

Question 30

Explaining conditions: 3 things to consider.

Answer 30

Yield. Rate. Compromise.