Kc and 6.6 The effect of changing conditions on equilibria

Question 1

What is the equation for finding Kc

Eg in equation
2A + B <–> 2C + D

Why are square brackets used

Answer 1

[A]^2 x [B]

Square brackets represent concentration

Question 2

How would you solve the question

33.2g of NO2 was heated in vessel with volume 9.65dm^3 at 450 degrees, and 6.88g of O2 were found in equilibrium mixture.
Find Kc

2NO2 <—> 2NO + O2

Answer 2

. Start by drawing ice table, where everything is in mol
. Find mol of NO2 which is 0.72
. Find mol of O2 which is 0.22
. So you know that the products initially had 0 moles so if the end mol was 0.22, then that must have been added.
. So +0.22 under O2, which means it is +0.44 under NO because of ratio
Which means -0.44 under NO2

. Then add or subtract these from the original mol
. So equilibrium mol is 0.29 for NO2, 0.43 for NO, 0.215 for O2

Then in order for it to be put into Kc, it has to be in conc so divide them all by 9.65dm^3 to get equilibrium conc

Then put into equation

Question 3

Does a catalyst impact Kc?
Does temperature impact Kc?
Does concentration impact Kc?

Answer 3

Only temperature impacts Kc, the others do not

Whether Kc increases or decreases depends on if the reaction is endothermic or exothermic

Question 4

What happens to Kc if equilibrium shifts to the right of the reaction

What happens to Kc if equilibrium shifts to the left of the reaction

Answer 4

. Kc will increase in value if equilibrium shifts to the right, which is the forward reaction to get more product

. If equilibrium shifts to the left, Kc will decrease in value so less product is made

This is because it is measured by
products / reactants

Question 5

In an exothermic reaction, what would increasing the temperature do to Kc

Answer 5

. Since forward reaction is exothermic, increasing the temp means equilibrium goes to reduce it so will shift left in the endothermic direction.

As it is shifting left, value of Kc will decrease

Question 6

In an endothermic reaction, increasing the temperature would do what to Kc?

Answer 6

So forward reaction is endothermic so increasing the temp would move equilibrium to the right

So value of Kc will increase

Question 7

Explain why changing concentration doesn’t effect Kc
Eg in reaction

C2H5OH + CH3CO2H –> CH3CO2C2H5 + H2O

Answer 7

Adding more ethanol would increase its concentration, so for equilibrium to reduce it, it will need to react with ethanoic acid to produce more ethyl ethanoate and water

So eventually a new equilibrium will be set up with relatively more of the products so equilibrium has shifted right

However, Kc will remain constant because although adding more ethanol makes the bottom of equation larger, some of it has reacted with ethanoic acid which has reduced both of their equations , and the top line value is increased as more product is made.

So it evens out to give the same value

Question 8

What does it mean if the value of Kc is much greater than 1

What does it mean if Kc is greater than 10^10

Answer 8

It means there are much more products than reactants so equilibrium position is over to the right

it means that reaction is basically completed

Question 9

What does it mean if Kc is smaller than 1

What does it mean if Kc is less than 10^-10

Answer 9

It means there are more reactants than products so equilibrium has shifted to the left

It means reaction basically didn’t occur at all

Question 10

How do catalysts not affect Kc

Answer 10

. Don’t affect Kc or position of equilibrium
. They affect the rate of forward and backward reactions equally by reducing activation energy for the reactions