F325 How Far How Fast hw ms

1.	(i)	O3: Exp 2 has 4 times [H2] as Exp 1
and rate increases by 4 (1),
	so order = 1 with respect to O3 (1)
	C2H4: Exp 3 has 2 × [C2H4] and 2 × [O3] as Exp 2;
and rate has increased by 4 (1),
	so order = 1 with respect to C2H4 (1)
	rate = k [O3] [C2H4] (1)	5 
(ii)	use of k = rate / [O3] [C2H4] = 1.0 × 10–12 / (0.5 × 10–7 × 1.0 × 10–8)
to obtain a calculated value (1)
k = 2 × 103 (1)
units: dm3 mol–1 s–1 (1)	3 
(iii)	rate = 1.0 × 10–12 /4 = 2.5 × 10–13 (mol dm–3 s–1) (1)	1 
(iv)	rate increases and k increases (1)	1
[10]

2.	(a)	(change in) concentration/mass/volume with time	1
(b)	(i)	O2:
Exp 2 has 4 × [O2] as Exp. 1: rate increases by 4 (1),
so order = 1 with respect to O2 (1)
	NO:
Exp 3 has 3 × [NO] as Exp. 3: rate has increases by 9 (1),
so order = 2 with respect to NO (1)	4 
(ii)	rate = k[O2] [NO]2 (1)	1 
(iii)	k =  = 7.10 × 109 (1)
	units: dm6 mol–2 s–1 (1)	2
[8]

3.	From graph, constant half-life (1)
Therefore 1st order w.r.t. [CH3COCH3] (1)	2
	From table, rate doubles when [H+] doubles (1)
Therefore 1st order w.r.t. [H+] (1)	2
	From table, rate stays same when [I2] doubles (1)
Therefore zero order w.r.t. [I2] (1)
Order with no justification does not score.	2
	rate = k[H+][CH3COCH3] (1)
(from all three pieces of evidence)
 	k =  (1)
= 7.0 × 10–5 (1) dm3 mol–1 s–1 (1)	4
accept 7 × 10–5
 	rate determining step involves species in rate equation (1)
	two steps that add up to give the overall equation (1)
	The left hand side of a step that contains the species in rate-determining step (1)
i.e., for marking points 2 and 3:
CH3COCH3 + H+ → [CH3COHCH3]+
[CH3COHCH3+] + I2 → CH3COCH2I + HI + H+	3
	organises relevant information clearly and coherently,
using specialist vocabulary where appropriate
Use of the following four words/phrases:
constant, half-life, order, doubles/x2 (1)	1
[14]

4.	(i)	constant half-life (1)	1
(ii)	rate = k[N2O5] (1)	1
	Common error will be to use ‘2’ from equation.
(iii)	curve downwards getting less steep (1)
curve goes through 1200,0.30; 2400,0.15; 3600,0.075 (1)	2
(iv)	tangent shown on graph at t = 1200 s (1)	1
(v)	3.7(2) × 10–4 (1) mol dm–3 s–1 (1)
ecf possible from (ii) using [N2O5]x
(2nd order answer: 2.2(3) × 10–4)	2
[7]

 5.	(i)	The slowest step (1)	1
(ii)	2NO2 → NO + NO3 (1)
NO3 + CO → NO2 + CO2 (1)	2
(or similar stage involving intermediates)
[3]

6.	(i)	1½O2(g) → O3(g)/
O2(g) + ½O2(g) → O3(g) (1)
	NO is a catalyst (1) as it is (used up in step 1 and)
regenerated in step 2/
not used up in the overall reaction(1)
allow 1 mark for ‘O/NO2 with explanation of regeneration.’	3
 (ii)	Rate = k[NO] [O3] (1)
Species in rate equation match those reactants in the slow
step / rate determining step (1)	2
[5]

 7.	(a)	CH4 + H2O → 3H2 + CO	1
CH4 + 2H2O → 4H2 + CO2
CH4 + H2O → 2H2 + CH2O/HCHO
CH4 + 2H2O → 2H2 + CH2O2/HCOOH 
or CH4 + H2O → H2 + CH3OH 
(b)	(i)		1
[NH3]2 = (Kc × [N2] × [H2]3 ) 
           = 0.768 
(ii)	[NH3] = √0.78 = 0.876/0.88 (mol dm–3) 	3
If no powers, then rearrangement mark only.
 (c)	High pressure:
	adv: Fewer moles on r.h.s. → equilibrium moves to right 
	Greater pressure → faster rate/more frequent collisions 
	dis: Safety issues from (high) pressure 
	Expense of (high) pressure 	3
 	High temperature:
	adv: more collisions exceed activation energy/
	more successful collisions/more energetic
	collisions/molecules have more energy 
dis:  Equilibrium moves to left/reverse direction because
	(forward) reaction is exothermic 	2
 	Catalyst:
lowers activation energy/
allows reaction to take place at a lower temperature 	1
	QWC: Uses 2 words following list in the correct context:	1
exothermic/endothermic, activation energy, collisions,
equilibrium/Le Chatelier
[12]

8.	(a)	rate of forward reaction = rate of reverse reaction (1)
concentrations of reactants and products are constant but they are
constantly interchanging (1)	2
(b)	(i)	Kc = [CH3OH] / [CO] [H2]2 (1)	1
(ii)	use of Kc = [CH3OH] / [CO] [H2]2 and moles to
obtain a calculated value (1)
	convert moles to concentration by +2: [CO] = 3.10 × 10–3 mol dm–3;
[H2] = 2.60 × 10–5 mol dm–3; [CH3OH] = 2.40 × 10–2 mol dm–3 (1)
	Kc = [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 Kc value = 3.64
(scores 1st and 3rd marks)
units: dm6 mol–2 (1)	4
(c)	(i)	fewer moles of gas on right hand side (1)	1
(ii)	None (1)	1
(d)	(i)	moved to left hand side/reactants increase/less products (1)	1
(ii)	∆H negative because high temperature favours the
endothermic direction (1)	1
 
(e)	(i)	CH3OH + 1½ O2 → CO2 + 2H2O(1)	1
(ii)	adds oxygen/oxygenated (1)	1
[13]