Answers to questions on A2 Rate of Reaction -

2. Question 1 (a)

3.  Half life 1 = 38 seconds
 Half life 2 = 38 seconds
 Half life 3 = 37 seconds
 Equal half lives so 1st order reaction with
respect to H2O2
 Rate = k [H2O2]
 k = Rate/ [H2O2] = 1.61 x 10-5s-1 from
results at 12 seconds.

4. Question 1 (c) + (d)

5. Question 2 (a)

6. Question 2 (c)

7.  (a) H2O2 + 2I- + 2H+ = I2 + 2H2O
 (b) Rate = k [H2O2] [I- ]
 (c) Overall order = 2
 (d) First step is the rate determining step
– because it involves one molecule of
hydrogen peroxide and one ion of
iodine reacting and no hydrogen ions to
confirm the rate equation.

8.  (a) Zero order with respect to X – see
experiments 1,4 and 5 – an increase in
the concentration of X does not affect
the reaction rate.
 (b) Second order with respect to Y – see
experiments 1 and 2 – doubling the
concentration of X results in a four times
faster reaction.
 Rate = k [Y]2

9.  Rate = k [Y]2
 0.0001 = k (0.1)2
 So k = 0.001 mol -1dm3s-1
 Y + Y = Y2
‡
 Y2
‡ + X = XY2
 Chemists study rate equations in order to
determine the mechanisms of reactions and
then predict factors that could make
reactions more economically viable.

10.  The sodium thiosulphate reacts with the
iodine which is initially produced and
remove it from the reaction mixture. This
allow a fixed end point of the reaction to
be determined once all the sodium
thiosulphate is used up.
 The starch solution detected the end of
the initial reaction and the point at
which iodine is no longer reacting with
the sodium thiosulpate.

11. Question 5

12.  The iron (III) ions are acting as a catalyst
and lower the activation energy.