EQUILIBRIUM

EQUILIBRIUM

Many chemical reactions are reversible:

Reactants → Products or Reactants ← Products Reactants form Products Products form Reactants

For example, under certain conditions, one mole of the colourless gas N2O4 will decompose to form two moles of brown NO2 gas:

N2O4 2 NO2

Under other conditions, you can take 2 moles of brown NO2 gas and change it into one mole of N2O4 gas:

N2O4 2 NO2

EQUILIBRIUMIn other words, this reaction is as written can go forward or in reverse depending on the conditions

If we were to put some N2O4 in a flask, the N2O4 molecules would collide with each other some of them would break apart to form NO2

This is called a FORWARD reaction N2O4 2 NO2

EQUILIBRIUMOnce this has happened for awhile, there is a build up of NO2 molecules in the same flask Once in awhile, two NO2 molecules will collide with each other and join to form a molecule of N2O4 !

This is called a REVERSE reaction: N2O4 2 NO2

EQUILIBRIUM

2 things to notice:

• 1. as long as N2O4 present, the forward reaction will keep on happening

• 2. as long as there is NO2 present, the reverse reaction will keep on happening!

•At one particular time a molecule of N2O4 might be breaking up, and at the same time two molecules of NO2 might be joining to form another molecule of N2O4

EQUILIBRIUM

In any reversible reaction, the forward reaction and the reverse reaction are going on at the same time

N2O4 2 NO2

The double arrow means that both the forward and reversereaction are happening at the same time.

RATE OF REACTION

Rate of the forward reaction

Rate of forward reaction will be fast at first

Rate will slow down

RATE OF REACTION

Rate of reverse reaction

RATE OF REACTION

RATE OF REACTION

RATE OF REACTION

The rate of the forward reaction = the rate of the reverse reaction

N02 is being used up at the same rate that it is being formed

N2O4 2 NO2

Rate of reaction does not change because both N202 and NO2 are constant

EQUILIBRIUM

• When this happens we say the system is at EQUILIBRIUM

• We have reached a state of DYNAMIC EQUILIBRIUM

KEY POINTS

• The reaction has not stopped

• The forward and reverse reactions are still taking place but their rates are equal so there are no changes in concentrations of the reactants and products• It is happening at a microscopic level so we don’t

see the individual molecules reacting (macroscopic)

KEY POINTS

• All observable properties are constant

• include the reactants, products, total pressure, temperature

• If no changes were made to the condition a system would stay at equilibrium forever

KEY POINTS

• Temperature must remain constant

• A system at equilibrium is a closed system

• Equilibrium can be approaches form the left (reactants) or from the right (products)

https://www.youtube.com/watch?v=dUMmoPdwBy4

https://www.youtube.com/watch?v=wlD_ImYQAgQ

CALCULATIONS

Equilibrium calculations

We can figure out what the concentrations are for the products and reactants when they are at equilibrium

We use:

KeqIt tells us the ratio of products: reactants when the reaction is at equilibrium

CALCULATIONS

to figure out the concentration of the products and reactants at equilibrium we use:

Keq = [products]

[reactants]

A + B C + D

Keq = [C] [D]

[A] [B]

CALCULATIONS

Given the equilibrium system:

PCl5(g) PCl3(g) + Cl2(g)

The system is analyzed at a certain temperature and the equilibrium concentrations are as follows:

[PCl5] = 0.32 M, [PCl3] = 0.40 M and the [Cl2] = 0.40 M.

Calculate the Keq for this reaction at the temperature this was carried out.

CALCULATIONSSOLUTION:

Step 1 - Use the balanced equation to write the Keq expression:

PCl5(g) PCl3(g) + Cl2(g)

Keq = [PCl3] [Cl2]

[PCl5]

Step 2 - "Plug in" the values for the equilibrium

Keq = [0.4] [0.4]

[0.32]

Step 3 - Calculate the value of Keq :

Keq = 0.5

CALCULATIONS Given the equilibrium system:

3 PCl5(g) 2PCl3(g) + Cl2(g)

[PCl5] = 0.32 M, [PCl3] = 0.40 M and the [Cl2] = 0.40 M.

We write it as:

Keq = [PCl3]2 [Cl2]

[PCl5]3

Raise to the power of the coefficient

Keq = [0.4]2 [0.4]

[0.32]3

CALCULATIONS At 200°C, the Keq for the reaction: N2(g) + 3H2(g) 2NH3(g) is 625

If the [N2] = 0.030 M, and the [NH3] = 0.12 M, at equilibrium, calculate the equilibrium [H2].

Solution: All concentrations given are at equilibrium so: Write out the Keq expression:

Keq = [NH3]2

[N2] [H2]3

625 = [0.12]2

[0.30] [H2]3

(625) [H2] 3 (0.030) = (0.12) 2

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2 H

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