UNIT 7 WORKSHEETS 1 Equilibrium
1. Distinguish between dynamic equilibrium and static equilibrium.
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2. Explain why wet clothes that are hung on a washing line dry best
(i) in bright sunshine.
(ii) on a windy day.
(iii) when the general humidity in the air is low.
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3. When ethanoic acid and ethanol are mixed in a closed container in the presence of sulfuric acid
(which acts as a catalyst) water and ethyl ethanoate are formed.
CH3COOH(l) + C2H5OH(l) ⇌ CH3COOC2H5(l) + H2O(l)
The reaction can be followed and the concentration of the products slowly increases. After about
one week the concentrations of all the reactants and products are found to be constant.
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(a) Explain why the concentrations of the products increase initially but then remain constant.
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(b) State and explain how the value of the reaction quotient, Q will change during the week after the
ethanoic acid and ethanol are mixed.
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(c) Suggest why the sulfuric acid speeds up the reaction but has no effect on the final concentrations of
the reactants and products.
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(d) The value for the equilibrium constant is 4 at 298 K. If equal amounts of ethanoic acid and ethanol
are added initially in a closed container at 298 K and the temperature kept constant what can be
deduced about the composition of the reaction mixture after one week has passed.
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4. Carbon monoxide reacts with hydrogen at 500 K to form gaseous methanol. The equation for the
reaction is:
CO(g) + 2H2(g) ⇌ CH3OH(g)
(a) Write the equilibrium expression for this reaction.
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(b) The value of Kc for this reaction is 14.5 at 500 K. What information does this give about the position
of equilibrium and the approximate relative concentrations of the reactants and product.
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5. Hydrogen and nitrogen react under certain conditions to form ammonia:
3H2(g) + N2(g) ⇌ 2NH3(g) ∆H = – 92 kJ
(a) The value for Kc is 152 at 500 K. Explain why the value is lower at higher temperatures and deduce
the effect that a higher temperature has on the yield of ammonia.
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(b) Explain why increasing the pressure increases the yield of ammonia.
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(c) Many text books state that the iron used as the catalyst is finely divided (or powdered). Explain why
the iron is in this form.
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6. In the high temperatures reached in an internal combustion engine the nitrogen and oxygen in the air
can combine to form nitrogen(II) oxide:
N2(g) + O2(g) ⇌ 2NO(g)
The value for the equilibrium constant, Kc, is 1.7 x 10-3 at 2300 K.
(a) Write the equilibrium expression for this reaction.
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(b) Explain the effect that increasing the pressure will have on the position of equilibrium.
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(c) This reaction is endothermic. Explain the effect that increasing the temperature above 2300 K will
have on:
(i) the yield of nitrogen(II) oxide.
(ii) the rate of the reaction.
(iii) the value of the equilibrium constant, Kc.
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Answers
1. In dynamic equilibrium the changes taking place are still proceeding but the rate of the forward
change is equal to the rate of the reverse change so the concentrations of the products and reactants
remains constant at equilibrium. In static equilibrium all movement has stopped.
2. (i) The warmth from the sun increases the temperature which increases the vapour pressure of the
water.
(ii) As the water molecules vaporise the wind blows them away so they cannot condense back on the
clothes.
(iii) When the water molecules go into the air the lack of humidity means that the air does not become
saturated with water vapour so reduces condensation.
3. (a) Initially when the concentrations of the acid and alcohol are high the rate of the forward reaction
is much faster than the rate of the reverse reaction, as products are formed the rate of the reverse
reaction increases and the rate of the forward reaction decreases. After one week equilibrium is
reached when the rate of the forward reaction is equal to the rate of the reverse reaction so the
concentration of reactants and products does not change.
(b) The value for Q will initially be very low as very little products are present. As the week progresses
the concentration of the products will increase and the concentration of the reactants will decrease so
the value of Q will continue to increase until eventually it becomes the value for Kc. At this point the
position of equilibrium has been reached.
(c) The catalyst speeds up the rate of both the forward and reverse reaction by the same amount. So
equilibrium is reached quicker but there is no effect on the position of equilibrium.
(d) The concentration of the acid and alcohol were originally equal and one mole of acid reacts with one
mole of alcohol. At equilibrium the concentration of the ester and water formed will be twice the
concentration of the remaining acid and alcohol.
Mathematically if [ester] = [water] = x at equilibrium, then [acid] = [alcohol] = ½x at equilibrium
Kc = ([ester] x [water]) / ([acid] x [alcohol]) = x2/ ¼ x2 = 4
Note that HL students could work out that if the initial amount of the acid and the alcohol in a fixed
volume was a mol each then the total amount present initially = 2a mol. Since the total amount of
particles does not change and the volume remains constant then there are a total of 3x mol of particles
at equilibrium so 3x = 2a so x = 2/3 a. Hence the concentration of the acid and the alcohol will have
reduced to 1/3 of their original value and the concentrations of the ester and water formed will each be
two thirds of the original concentration of the acid or alcohol.
4. (a)
(b) The equilibrium lies more on the product side but there will still be appreciable amounts of hydrogen
and carbon monoxide in the equilibrium mixture even though the concentration of methanol will be
greater as Kc is not extremely large.
5. (a) Because the reaction is exothermic at higher temperatures the equilibrium will move towards the
reactant side so the value of Kcwill be lower and the yield of ammonia will also be lower.
(b) Four moles of reactants are converted into two moles of product. Applying Avogadro’s law four
volumes are converted into two volumes and a higher pressure will favour the lower volume side.
(c) To increase the surface area of the catalyst which will increase the rate of the reaction as more
reactant particles can be brought closer together.
6. (a)
(b) It will have no effect as there is no change in volume (two moles of gas on each side) .
(c) (i) the yield of nitrogen(II) oxide will increase as the reaction is endothermic .
(ii) the rate of the reaction will increase as more reactant particles will have the necessary activation
energy and there will be more successful collisions.
(iii) the value of the equilibrium constant, Kc, will increase as at higher temperatures the concentration
of the product will be higher and the concentration of the reactants will be lower.