3.1.6- Chemical equilibria, Le Chatelier’s principle and Kc (PAPER 1+2)

Question 1

What makes a reversible reaction at equilibrium?

Answer 1

Carried out in a closed system so reactants and products cannot enter or leave.

-Forward and reverse reactions occur at the same rate.
-The net concentrations of the reactants and products remain constant.

Question 2

Why is it dynamic equilibrium?

Answer 2

The forward and reverse reactions are ongoing.

Question 3

Draw a graph of time ( x axis) against concentration (y axis) to show reactants and products.

Answer 3

Reactants start at the top and quickly decrease before being reformed again gradually- slow as their concentration decreases.

Products start at the bottom and increase in concentration gradually, increasing the rate the reactants are reformed at.

Question 4

What is Le Chatelier’s principle?

Answer 4

If the conditions under which an equilibrium exists are changed, the position of equilibrium shifts in a way to oppose the change in conditions.

Question 5

What happens if temperature is increased?
N2 + 3H2 ⇌ 2NH3 -76 kJmol-1

Answer 5

Forward reaction is exothermic so the equilibrium shifts to the left in the endothermic direction in order to oppose the increase in temperature. This decreases the equilibrium yield of NH3.

Question 6

What happens if temperature is decreased?N2 + 3H2 ⇌ 2NH3 -76 kJmol-1

Answer 6

Forward reaction is exothermic so the equilibrium shifts to the right in the exothermic direction in order to oppose the decrease in temperature. The equilibrium yield of NH3 increases.

Question 7

What happens if pressure is increased?
CO2 + 3H2 ⇌ CH3OH + H20 -49 kJmol-1

Answer 7

(look at big numbers before atoms)
There are 4 moles of gases on the left side but only 2 moles of gases on the right side, so the left side (the reactants) will exert a greater pressure relative to the products.

There are more molecules on the left so the equilibrium shifts to the right to oppose the increase in pressure, increasing the yield of CH3OH.

Question 8

What happens if pressure is increased?
H2 + I2 ⇌ 2HI

Answer 8

There are equal numbers of moles of gas on each side of the equation so the equilibrium position will not be affected.

The rate at which equilibrium is reached will increase as the molecules of gas will be closer together so there will be a higher number of collisions so a higher number of successful collisions.

Question 9

What happens if the concentration of CH3OH is increased by adding more moles?

CH3OH + HCOOH ⇌ CH3OCHO + H2O

Answer 9

If the concentration of CH3OH is increased, the equilibrium moves to the right side to counteract the change and decrease the concentration by producing more CH3OCHO by using up the CH3OH.

Question 10

What happens if the concentration of HCOOH is decreased by removing moles?

CH3OH + HCOOH ⇌ CH3OCHO + H2O

Answer 10

If the concentration of HCOOH is decreased, the equilibrium would move to the left side to counteract the change and produce more HCOOH.

Question 11

What do reactions have to be in order to predict a change in temp, pressure or conc on equilibrium?

Answer 11

Homogenous reactions where both the reactants and products are in the same state.

Question 12

How do catalysts affect reversible reactions?

Answer 12

Catalysts have no effect on the position of equilibria as they speed up the rate of the forward and reverse reaction equally.

A catalyst speeds up the rate at which equilibrium is reached but has no effect on the yield of either products or reactants.

Question 12

What are the conditions for the industrial manufacture of ethanol from steam and ethene?

Answer 12

C2H4 + H2O ⇌ C2H5OH -ve kJmol-1

The presence of a phosphoric acid catalyst on silica support
Temperature 300C
Pressure around 60-70atm

Question 12

What are the compromises regarding pressure with the manufacture of ethanol?

Answer 12

A high pressure is favoured as equilibrium will move to the side with the fewest molecules- which is the right side, increasing the yield of ethanol. A high pressure also increases the rate of reaction.

HOWEVER: high pressure environments are expensive due due to thicker, more robust vessels and pipes needed.

Compromise between yield/ speed and cost!

Question 12

What are the compromises regarding temperature with the manufacture of ethanol?

Answer 12

A low temperature is favoured as the forward reaction is exothermic so decreasing the temperature means equilibria will shift to the right producing more ethanol.

HOWEVER: a lower temperature means lower rate of reaction.

300C is a compromise between yield and rate!

Question 12

Write the expression for Kc- products/reactants.
2A + B ⇌ 2C + D

Answer 12

Kc= (C)^2 (D)
—————-
(A)^2 (B)

Question 13

Write the expression for Kc and calculate Kc for:
2SO2 + O2 ⇌ 2SO3

SO2= 0.4moldm-3, O2= 0.2moldm-3, SO3= 0.8moldm-3

Answer 13

(SO2) ^2 x (O2)

(0.4)^2 x (0.2)

Kc= 20

Question 14

Work out the units for this Kc

Answer 14

(moldm-3) (moldm-3) (moldm-3)

cancel out 2 on top and 2 on bottom,
moldm-3 becomes mol-1dm3

Question 15

Use an ICE table to work out the conc of each species at equilibrium: PRODUCTS INCREASE, REACTANTS DECREASE!

2NO2 ⇌ 2NO + O2

33.2g NO2 heated in a vessel with volume 9.65dm3 at 450C

6.88g O2 found in equilibrium mixture.

Answer 15

Moles NO2= mass/mr: 33.2/14+16+16= 0.72
Moles O2= mass/mr: 6.88/ 16+16= 0.215 at equilibrium

I 0.72 0 0
C -2(0.215) +2(0.215) +0.215
E 0.29 0.430 0.215

Question 16

Use the ICE table to work out the value for Kc

Answer 16

Work out concentrations of each species
NO2= 0.29/9.65=0.030
NO= 0.430/9.65= 0.045
O2= 0.215/9.65= 0.022

Sub these into Kc equation

(0.030)^2

=0.050moldm-3

Question 17

Use Kc to find equilibrium concentration

At equilibrium, concentration of ethanoic acid was 1.8moldm-3, ethanol was 3.2moldm-3.

Kc value is 3.5 at 25C

Calculate the concentrations of the products at equilibrium.

CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O

Answer 17

1) Write the expression for Kc
(CH3COOC2H5) (H2O)
———————————
(CH3COOH) (C2H5OH)

2) Substitute in the values that you know
3.5= (CH3COOC2H5) (H2O)
————————————
(1.8) (3.2)

3) Rearrange the equation for what you do not know
(CH3COOC2H5) (H2O) = 3.5 x 1.8 x 3.2
(CH3COOC2H5) (H2O) = 20.16

4) They will have the same concentrations so square root 20.16 to give you concentration of each one: conc CH3COOC2H5=4.49moldm-3, conc H20= 4.49moldm-3

Question 18

In an exothermic reaction, how does an increase in temp. affect size of Kc?

Answer 18

Increase in temp. causes equilibrium to move to the left, increasing the amount of reactants and decreasing the amount of products. This decreases Kc.
smaller number/ bigger number

Question 19

In an exothermic reaction, how does a decrease in temp. affect size of Kc?

Answer 19

Decrease in temp. causes equilibrium to move to the right, decreasing the amount of reactants and increasing the amount of products. This increases Kc.
bigger number/ smaller number.

Question 20

In an endothermic reaction, how does an increase in temp. affect size of Kc?

Answer 20

Increase in temp. causes equilibrium to move to the right, increasing the amount of products and decreasing the amount of reactants. This increases Kc.
bigger number/smaller number (products/ reactants)

Question 21

In an endothermic reaction, how does a decrease in temp. affect size of Kc?

Answer 21

Decrease in temp, causes equilibrium to move to the left, increasing the amount of reactants and decreasing the amount of products. This decreases Kc.
smaller number/bigger number

Question 22

What does it mean if Kc is greater than 1?

Answer 22

The concentration of products is greater than the concentration of reactants so equilibrium lies to the right.

Question 23

What does it mean if Kc is less than 1?

Answer 23

The concentration of reactants is greater than the concentration of products so equilibrium lies to the left.

Question 24

How does changing concentration affect the value of Kc?

Answer 24

The value of Kc is unaffected by changes in concentration as the ratio of products to reactants is restored by equilibrium.

Question 25

How do catalysts affect the value of Kc?

Answer 25

A catalyst does not change the value of Kc as it increases the rate of both the forward and reverse reaction, it increases the rate at which equilibrium is reached.

Question 26

What is the only factor that affects the value of Kc?

Answer 26

Temperature