P1: Equilibria

Question 1

Are some reactions reversible?

Answer 1

Yes

Question 2

What is happening in an equilibria reaction?

Answer 2

A forward and backward reaction happening at the same time.

Represented by opposing single-headed arrows

Question 3

What is a dynamic equilibria?

Answer 3

Two reactions that are opposite are going at the same rate (equal) in a closed system. (concs don’t change)

Question 4

What are the conditions for Equilibria?

Answer 4

-Can only happen in a closed system
-Can be approached from either direction
-Final equilibrium position will be the same (as long as conditions are the same)
-It is dynamic, moving in opposite directions at the same time.

Question 5

What is Le Chatelier’s Principle?

Answer 5

“If a system at equilibrium is disturbed, the equilibrium moves in the direction that tends to reduce the disturbance.”

AKA- Equilibrium opposes change

Question 6

How is equilibria reached?

Answer 6

-Reactants only at the start
-The rate of the forward reaction is much greater at first than the backwards
-The rate of forwards slows as reactants get used up, so backwards increases.
-Eventually rates are the same, equilibria is reached.

If equilibrium is met, the concs do not change (if conditions are the same)
The balance does not have to be 50/50 for equilibria

Question 7

What is the effect on equilibria if I change the pressure?

Answer 7

Only changes reactions involving gas, and will only change if there are a different number of molecules on each side of the equation.

e.g

2NO2 (g) > N2O4 (g)

Increase in forward direction, as it moves to the side with less gas.
Decrease in backwards direction.

To test:
Decrease: brown gas
Increase: pale yellow gas

Question 8

What is the effect on equilibria if I change the temperature?

Answer 8

Depends on if it exothermic or endothermic.

Exothermic:

This means it will be exothermic in the forward direction, giving out heat.

Therefore the equilibria wants to oppose this, and decreases temp in the forwards direction.
So increasing heat, we go in the backwards direction.

Endothermic:

Will be endo in the forwards direction.

The equilibria will go forward when we increase heat, to oppose the cooling of endo.
Vice versa.

Question 9

What is the effect of a catalyst on equilibria?

Answer 9

-Does not have an effect
-The forward and backward reactions are equal.
-However, does allow dynamic equilibrium to be reached more quickly, so important for industry.
-Does not effect yield

Question 10

What is the symbol for equilibrium constant?

Answer 10

Kc

Question 11

What is the equation for equilibrium constant?

Answer 11

Kc= conc products/ conc reactants

Question 12

How do you work out concentration?

Answer 12

Mol/ vol

Vol in dm3, so if in cm3, divide by 1000

Question 13

In a reversible reaction, will the actual yield always be more or less than the theoretical yield?

Answer 13

Less

Question 14

How does Concentration effect the equilibrium position?

Answer 14

If you increase the conc of the reactant, you increase the conc of the product yield, shifts to the right.

If you decrease it, shifts to the left, and the yield of product decreases.

Question 15

How would you answer a ‘Le Chatelier’s Principe’ question?

Answer 15

1. State any relevant features of the reaction e.g exothermic, left has more molecules?
2. State where the equilibrium will move, right or left?
3. State why the equilibrium position moves e.g to counteract the decrease in temp.
4. State how the change effects the yield of the product.

Question 16

What is equilibrium used for in industry?

Answer 16

1. The Haber process:
   for fertilisers
   N2 + 3H2 -> 2NH3
2. Contact process:
   for detergent, dye, explosives
   2SO2 + O2 -> 2SO3

Question 17

How do you maximise rate and yield?

Answer 17

Yield:
- Increase pressure
- Increase conc of reactants
- Decrease conc of the product
- Decrease temp

Rate:
- Increase pressure
- Increase conc of reacatnts
- Increase temp
- Add a catalyst

Question 18

What is a factor you have to think about before trying to increase yield?

Answer 18

Economic cost:

Equipment
Conditions

+safety of workers

Question 19

How do you express Kc?

Answer 19

[product]/ [reactant]

with the coefficients (number of moles) as exponents outside the bracket (to the power of)

Question 20

How do you work out the units for Kc/

Answer 20

Well technically its conc/conc
so mol dm-3/ mol dm-3

BUT if there are multiple moles, therefore an exponent in the fraction, we subtract the bottom exponents from the top e.g.:

[pro]^2 / [react 1] [react 2]^3

-add the two reactants exponents 1+3=4
- 2/4 as the fraction, but a subtraction not a divide, 2-4=-2

Units: mol^-2 dm^3

Sometimes they cancel out and there might not be any units.

Question 21

How do you calculate Kc?

Answer 21

Sub in the concs of each product and reaction to the equation
-May have to times the products with another product if there is multiple (or reactants)
-Figure out units

Question 22

How do you calculate Kc if you don’t have the concs?

Answer 22

-Use conc= mol/vol (make sure vol is dm3)
-Plug into equation

Question 23

Is there an exception to only using concs to calculate Kc?

Answer 23

Yes, if the concs are not given, and the volume is not either to work out conc, you may be able to plug in the moles into the equation instead!

ONLY if you have the same number of moles of products and reactants on each side of the equation, they are equal

so 2:2 and 5:5
but not 1:2 or 3:1

Question 24

You may be given the initial moles of the reaction and told to find the equilibrium moles (vice versa)- what method would you use?

Answer 24

An ICE table:

(the compounds in the reaction across the top of the table)

Initial moles

Change in moles

Equilibrium moles

-Fill in what we already have been given.

Question 25

What equation do you use to find the missing values in an ICE table (initial or equilibrium moles)?

Answer 25

Change in moles of what we want to find (A)=
change in moles of what we already know (B) x (multiplied)
coefficient of A / coefficient of B
(the moles)

-Gives us the change in moles of the missing values.
-Reactants are always negative values
-Products are always positive values
-Initial moles for products is always zero.

Question 26

What effects Kc?

Answer 26

Only temperature

Question 27

What has happened if the Kc value increases?

Answer 27

The equilibrium position shifts right, and more product is produced.

Question 28

What has happened if the Kc value decreases?

Answer 28

The equilibrium position shifts left, less product is produced.

Question 29

Why does the Kc value decrease if the reactants are increasing?

Answer 29

Kc= products/reactants

so if the reactants are increasing, the denominator is getting bigger, making the end value smaller (Kc).

Question 30

What happens to Kc if we increase the temperature of an Exothermic reaction?

Answer 30

Kc decreases

The forward (right) reaction is exothermic, so it is hot, if we increase that, it gets opposed to be cooler, shifts to the left where the reactants increase, decreasing Kc value.

Question 31

What happens to Kc if we decrease the temperature of an Exothermic reaction?

Answer 31

Kc increases

The forward reaction (exo) is hot, if we are cooling it down, it will oppose by trying to heat up, which is the ways its already going (right) so products increase, Kc increases.

Question 32

What happens to Kc if we increase the temperature of an Endothermic reaction?

Answer 32

Kc increases

Forward reaction (endo) is cool, so opposing the heat by cooling is going in the right (forward) direction, increasing products, increasing Kc.

Question 33

What happens to Kc if we decrease the temperature of an Endothermic reaction?

Answer 33

Kc decreases

Forward reaction (endo) is cool, if we cool it even more, it will oppose by heating, which is the opposite direction (shift left), increasing reactants, decreasing Kc.

Question 34

Why does conc not effect Kc?

Answer 34

If we add more of Reactant 1, equilibrium will combat this by reacting with Reactant 2, to produce more of Products 1 and 2, eventually reaching equilibrium again.