5.1

The initial concentration is 0 (of NO_{2})/concentration increases/ Curve starts at 0.
Or
Curve B has an initial concentration and is the reactant as its concentration decreases.
Tips:
 The reaction that occurs is given by the equation: 2N_{2}O_{5}(g) → 4NO_{2}(g)+O_{2}(g).
 Curve A starts from a concentration of zero therefore it must be a product because reactants have to react together for the concentration of the products to increase.
 It is very unusual for a reaction to start in a sealed container where products are already present.

1 mark

5.2

True
n mol of N_{2}O_{5} forms 2n mol of NO_{2} per unit time.
Or
The gradient of the graph for NO_{2} is twice the gradient of the graph for N_{2}O_{5}.
Tip: the reaction that occurs is given by the equation: 2N_{2}O_{5}(g) → 4NO_{2}(g)+O_{2}(g). Therefore, the mol ratio of N_{2}O_{5}: NO_{2} is 2:4 which is simplified to 1:2.

2 marks
(2 x 1 mark)

5.3.1

How to approach this question:
 Curve A gives the concentration of NO_{2 }at 400s, the [NO_{2}] is 200 x 10^{4} mol·dm^{3}.
 Do not forget to check the units of the axes as you can lose marks if you use a concentration of 200 mol·dm^{3} instead of 200 x 10^{4 }mol·dm^{3}.
 The molar mass of NO_{2} is 14 + 16 x 2 = 46g·mol^{1}, and the volume of the container is 2 dm^{3}.
Option 1
Or
Option 2

4 marks
(4 x 1 mark)
Marking criteria:
 Formula: c = m/MV or n(NO_{2}) = cV / n(NO_{2}) =m/M
 Substitute change in concentration
 Substitute M (46) and V (2)
 The final correct answer: 1,84g.

5.3.2

Option 1
Tips:
 The average rate of production of O_{2} cannot be determined directly from the graph since no curve shows the concentration of oxygen, therefore, you will need to work out the rate of change of NO_{2} or N_{2}O_{5} and use the mol ratio.
 If you use the rate of change of N_{2}O_{5, }you need to add a negative sign in the formula to make the rate of the reaction positive.
 Always remember to look at the units for the rate of the reaction to work out the correct rate.
There are more ways to answer this question. To view other options, click here.

4 marks
(4 x 1 mark)
Marking criteria:
 Substitute the change in concentration into the rate formula
 Substitute time into the rate formula
 Use mol ratio: rate(O_{2}) = ½ rate(N_{2}O_{5})/ rate(O_{2}) = ¼ rate(NO_{2}).
 Final correct answer: 1 x 10^{5} (mol·dm^{3}·s^{1})

5.4.1

Tips:
 It is essential that your diagram clearly shows both curves as the mark is given for curve Q having the same shape, a peak at the same kinetic energy, and being above curve P.
 Both curves must start at the origin and the curves should not cross over each other.

2 marks
(2 x 1 mark)
Marking criteria:
 Curve Q must be above the given curve P, and have the same shape as the given curve P, and the peaks have to correspond
 Start at the origin and do not cross curve P

5.4.2

Higher than
 When the concentration of N_{2}O_{5} is higher there are more N_{2}O_{5} particles per unit volume.
 More effective collisions per unit time/second. /Higher frequency of effective collisions.
Tip: increasing the concentration of reactant will increase the rate of the reaction. This is because there are more particles per unit volume and, therefore, more effective collisions per unit of time, leading to a higher rate of reaction.

3 marks
(3 x 1 mark)
