7.
Syngas is a mixture of carbon monoxide and hydrogen gases, used as a feedstock for the manufacture of methanol.
A dynamic equilibrium was set up between carbon monoxide, CO, hydrogen, H
2
, and methanol, CH
3
OH, in a 2.0 dm
3
sealed vessel.
The equilibrium is shown below.
CO(g) + 2H
2
(g) CH
3
OH(g)
The number of moles of each component at equilibrium is shown below component number of moles at equilibrium
CO(g)
6.20 × 10
–3
H
2
(g)
4.80 × 10
–2
CH
3
OH(g)
5.20 × 10
–5
(a) State two features of a system that is in dynamic equilibrium .
.......................................................................................................................
.
.......................................................................................................................
.
.......................................................................................................................
.
[2]
(b) (i) Write an expression for K c
for this equilibrium system.
(ii) Calculate K c
for this equilibrium. State the units.
K c
= ……………………... units:…………..…..
[4]
(c) The pressure was increased whilst keeping the temperature constant. The mixture was left to reach equilibrium.
The equilibrium position above shifted to the right.
(i) Explain why the equilibrium position shifted to the right.
[1]
(ii) What is the effect, if any, on the value of K c
?
[1]
(d) The temperature was increased whilst keeping the pressure constant. The mixture was left to reach equilibrium.
The value of K c
for the equilibrium above decreased.

(i) Explain what happened to the equilibrium position in the equilibrium.
[1]
(ii) Deduce the sign of the enthalpy change for the forward reaction shown in the equilibrium above.
Explain your reasoning.
[1]
(e) Methanol can be used as an additive to petrol.
(i) Write an equation for the complete combustion of methanol, CH
3
OH.
[1]
(ii) Suggest why methanol is added to petrol.
[1]
[Total 13 marks]
40.
When heated, phosphorus pentachloride, PC l
5
, dissociates.
PC l
5
(g) PC l
3
(g) + C l
2
(g)
A chemist placed a mixture of the three gases into a container. The initial concentration of each gas was the same: 0.30 mol dm
–3
. The container was left until equilibrium had been reached.
Under these conditions, K c
= 0.245 mol dm
–3
.
(a) Write an expression for K c
for this equilibrium.
[1]
(b) Use the value of K c
for this equilibrium to deduce whether the concentration of each gas increases, decreases or stays the same as the mixture approaches equilibrium.
(i) Show your answer by placing a tick in the appropriate cells in the table below.
PC l
PC l
C l
2
5
3 initial concentration
/ mol dm
–3
0.30 greater than
0.30 mol dm
–
3
0.30
0.30 less than
0.30 mol dm
–3 equal to
0.30 mol dm
–3
[1]
(ii) Explain your deduction.
[1]

(c) The chemist compressed the equilibrium mixture at constant temperature and allowed it to reach equilibrium under these new conditions.
(i) Explain what happens to the value of K c
.
[1]
(ii) Explain what happened to the composition of the equilibrium mixture.
[2]
(d) The chemist heated the equilibrium mixture and the equilibrium moved to the left.
(i) Explain what happens to the value of K c
.
[1]
(ii) Explain what additional information this observation reveals about the reaction.
[2]
[Total 9 marks]

MS
7.
(a) rate of forward reaction = rate of reverse reaction (1) concentrations of reactants and products are constant but they are constantly interchanging (1)
(b) (i) K c
= [CH
3
OH] / [CO] [H
2
]
2
(1)
(ii) use of K c
= [CH
3
OH] / [CO] [H
2
]
2
and moles to obtain a calculated value (1) convert moles to concentration by +2: [CO] = 3.10 × 10 –3
mol dm
–3
;
[H
2
] = 2.60 × 10 –5
mol dm
–3
; [CH
3
OH] = 2.40 × 10 –2
mol dm
–3
(1)
K c
= [2.60 × 10 –5 ] / [3.10 × 10 –3 ] [2.40 × 10 –2
]
2
= 14.6 / 14.56 (1)
If moles not converted to concentration, calculated K c
value = 3.64
(scores 1st and 3rd marks) units: dm
6
mol
–2
(1)
2
(c) (i) fewer moles of gas on right hand side (1)
(ii) None (1)
40.
(a) K c
=
[ PCl
3
][Cl
[ PCl
5
]
2
]
(1) 1
(b) (i) PCl
5
> 0.3 mol dm
–3
; PCl
3
and Cl
2
< 0.3 mol dm
–3
(1)
(ii) At start, system is out of equilibrium with too much PCl
3 and Cl
2
0 .
3

0 .
3
0 .
3
and not enough PCL
5
/
= 0.3 is greater than K c
= 0.245 mol dm
–3
(1)
1
1
1
4
1
1
(c) (i) K c
does not change as temperature is the same (1)
(ii) Fewer moles on left hand side (1) system moves to the left to compensate for increase in pressure by producing less molecules (1)
1
2

(d) (i) K c
decreases (as more reactants than products) (1)
(ii) Forward reaction is exothermic/ reverse reaction is endothermic (1) equilibrium → left to oppose increase in energy/ because K c
decreases (1)
1
2
[9]

(d) (i) moved to left hand side/reactants increase/less products (1)
(ii) ∆ H negative because high temperature favours the endothermic direction (1)
(e) (i) CH
3
OH + 1½ O
2
→ CO
2
+ 2H
2
O (1)
(ii) adds oxygen/oxygenated (1)
1
1
[13]
1
1