7.1 Equilibrium

Equilibrium

Many reactions are reversible.

N₂ + 3H₂ ⇌ 2NH₃ (ΔH = −92.4 kJmol⁻¹)

These will reach equilibrium in a closed system where nothing else can get in or out and where Temperature and Pressure are constant.

This doesn’t mean half the atoms are on the product side & half on the reactant side.

The Haber Process operates under conditions where the split is 70:30.

Imagine enclosing 1 mole of Nitrogen & 3 moles of Hydrogen in a sealed container.

• Why would the forward rate of reaction be high at the start and decrease over time?

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• Why would the backward rate of reaction be zero at the start & increase over time?

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• What would happen to concentrations of reactants and products when the two rates of reaction became equal?

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Most equilibria you encounter will have all reactants and products in the same phase – all gases or all solutions.

These are homogeneous.

But it is possible for equilibria to become established between different phases.

These are Heterogeneous.

Saturated solutions are a good example.

When the solution is saturated no more will dissolve.

• Write an equilibrium for the two processes that are now happening at the same speed.

• What is dynamic about this dynamic equilibrium?

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Since equilibria are rarely a 50:50 split between reactants and products, we need a way to measure the balance between the two sides.

The Equilibrium Constant is used – K_c

It couldn’t be much simpler.

In reality, powers rarely get as high as 5.

Common sense suggests that a lot of products will lead to a large K_c

And a lot of remaining reactant will lead to a low one.

The IB should only ask for K_c for Homogeneous Equilibria because when a solid is involved the concentration of the solid doesn’t change as it is used up.

• Write K_c expressions for the following equilibria

a. 4NH_3(g) + 5O_2(g) ⇌ 4NO_(g) + 6H₂O_(g)

Using [H₂] =2 [I₂]= 2 and [HI]=4 calculate K_C for b & c.

b. H_2(g) + I_2(g) ⇌ 2HI_(g)

• What do you notice about the values?

c. 2HI_(g) ⇌ H_2(g) + I_2(g)

Still using [H₂] =2 [I₂]= 2 and [HI]=4 calculate K_C for d.

d. ½H_2(g) + ½I_2(g) ⇌ HI_(g)

(d) and (b) are basically the same. Are K_cs the same?

1. Reversing Reaction K_{c (forwards)} = 1/K_c(backwards) and vice versa

eg For equilibrium N₂ + 3H₂ ⇌ 2NH₃ - if K_{c (forwards)} = 10, 1/K_c(backwards) = 0.1

2 Halving coefficients K_c(new) = √K_c(old)

eg For equilibrium 2N₂ + O₂ ⇌ 2NO₂ with K_{c (forwards)} = 10

Equilibrium N₂ + ½ O₂ ⇌ NO₂ will have K_{c (forwards)} = √10 = 3.16

3 Doubling coefficients K_c(new) = K_c(old)²

eg For equilibrium N₂ + ½ O₂ ⇌ NO₂ with K_{c (forwards)} = 10

Equilibrium 2N₂ + O₂ ⇌ 2NO₂ will have K_{c (forwards)} = 10² = 100

1 For the equilibrium below:

H_2(g) + I_2(g) ⇌ 2HI_(g) ΔH = -10.4 kJmol^-1

a) Write an expression for Kc

b) Will a high or a low temperature increase the yield of the reaction?

Explain your answer

c) What effect would a higher pressure have on the yield of the reaction?

d) Explain your answer

2 Predict the effect on the equilibrium position of an increase in pressure.

a) N₂O_4(g) ⇌ 2NO_2(g)

b) H_2(g) + CO_2(g) ⇌ CO_(g) + H₂O_(g)

c) CaCO_3(s) ⇌CaO_(s) + CO_2(g)

3 Predict the effect of a temperature increase on the equilibrium position of,

a) H_2(g) + CO_2(g) ⇌ CO_(g) + H₂O_(g) ∆H = + 40 kJ mol^-1

b) 2SO_2(g) + O_2(g) ⇌ 2SO_3(g) ∆H = - ive

What if your reaction hasn’t reached equilibrium yet?

The balance of [products] & [reactants] won’t equal the K_c.

But the number that the concentrations give in the K_c expression will tell you how close it is to getting to equilibrium.

If Q is GREATER than K_c then there must be more product than at equilibrium.

If Q is LESS than K_c then there must be less product than at equilibrium.

This has yet to appear in IB exams!!!!!