How Fast? How Far? Test Mark Scheme

1.	(a)	(i)	Curve downwards starting at t = 0 	2
with slope gradually levelling off with no increase 
(don’t worry about hitting the x axis)
 
(ii)	Tangent shown at start 	1
(iii)	Half-life is constant	1
OR: draw tangents and then plot a 2nd graph of tangent
or rate against concentration, which is a straight line
through the origin. 
 
(iv)	Straight line through origin 	1

(b)	4 times [KI], rate increases by 4 ,
so order = 1 with respect to KI  independent marks	2
(c)	(i)	rate/r = k[C2H4Br2] [KI]  or ecf from (b)	1
(ii)	
= 0.3(0)        units: dm3 mol–1 s–1 
units dependent on rate equation in (i).
Mark independently.	3
[11]

2.	(i)	H2: Exp 2 has 2.5 times [H2] as Exp 1
and rate increases by 2.5 (1),
	so order = 1 with respect to H2 (1)	2
	NO: Exp 3 has 3 x [NO] as Exp 2;
and rate has increased by 9 = 32 (1),
	so order = 2 with respect to NO (1)	2
QWC	At least two complete sentences where the meaning is clear.	1
 
(ii)	rate = k[NO]2 [H2] (1)	1
 
(iii)	k =  (1)
	= 1300 (1)	units: dm6 mol–2 s–1 (1)
allow 1 mark for 7.69 × 10–4 or 1.3 × 10x (x not 3)	3
[9]

3.	(i)	slow step (1)	1
(ii)	(CH3)2C=CH2 + H2O → (CH3)3COH (1)	1
(iii)	H+ is a catalyst (1)
	H+ used in first step and formed in second step/
regenerated/ not used up (1)	2
(iv)	rate = k [(CH3)2C=CH2] [H+] (1)
common error will be use of H2O instead of H+	1
[5]

4.	(a)	(i)	O3: 1
and C2H4 (1)	1
(ii)	2 (1)	1
(iii)	rate = k[O3] [C2H4] (1)	1

(b)	(i)	measure gradient/tangent (1)
at t = 0/start of reaction (1)	2
(ii)	k =  (1)
k =  = 2 × 103 (1) dm3 mol–1 s–1 (1)	3
 
(iii)	2 mol CH2O forms for every 0.5 mol O2 /
stoichiometry of CH2O : O2 is not 1:1 (1)	1
(iv)	rate increases (1)
k increases (1)	2
[11]

5.	(a)	Kc =  (1)	1
 
(b)	(i)	H2		I2	HI
0.30		0.20	0
0.14		0.04	0.32
	(1)		(1)	2
 
(ii)	Kc =  = 18.28571429 (1)
= 18 (to 2 sig figs) (1)
no units (1)
(or ecf based on answers to (i) and/or (a))	3
 
(c)	Kc is constant (1)
	Composition of mixture is the same (1)	2
[8]

6.	(a)	Kc =  (1)	1

(b)	(i)	PCl5 > 0.3 mol dm–3 ; PCl3 and Cl2 < 0.3 mol dm–3 (1)	1
(ii)	At start, system is out of equilibrium with too much PCl3
and Cl2 and not enough PCL5 /
 = 0.3 is greater than Kc = 0.245 mol dm–3 (1)	1
 
(c)	(i)	Kc does not change as temperature is the same (1)	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)	2
 
(d)	(i)	Kc decreases (as more reactants than products)(1)	1
 
(ii)	Forward reaction is exothermic/
reverse reaction is endothermic (1)
equilibrium → left to oppose increase in energy/
because Kc decreases (1)	2
[9]