Edexcel – A2 Chemistry

Rate of reaction review questions

Question 1 (a)

Question 1 (b)

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.

Question 1 (c) + (d)

Question 2 (a)

Question 2 (c)

Question 3

(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.

Question 4

(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

Question 4

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.

Question 5

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.

Question 5

QUESTION 5

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