Concentration / Time Graphs

Rate Equations Worksheet

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1. Hydrochloric Acid & Magnesium

A reaction takes place between hydrochloric acid and magnesium pieces. The results of the reaction are recorded in the table below.

Time / s	0	5	10	15	20	25	30	35
[HCl] / mol dm^-3	0.5	0.32	0.21	0.135	0.09	0.06	0.04	0.025

a) Explain how the concentration of HCl can be known at the given intervals.

A sample is taken, the reaction is quenched (in this case – just passed through a sieve to remove the magnesium) and titrated against a known concentration of NaOH to determine the concentration.

b) Plot a graph of the results on graph paper.

c) Determine the rate of reaction at the start of the reaction; once the concentration reached 0.4 mol dm^-3 and once the concentration reached 0.2 mol dm^-3.

Time 0: 0.26 / 5.5 = 0.047 mol dm^-3 s^-1

At conc of 0.4: 0.28 / 8 = 0.035 mol dm^-3 s^-1

At conc of 0.2: 0.16 / 9 = 0.0148 mol dm^-3 s^-1

(Note: Values are calculated from tangents drawn on the graph).

d) Use the rate of reaction from question c to determine the overall order of reaction.

Overall order of reaction is 1. As the concentration halves, the rate halves as well. Additionally, the half lives remain constant.

e) Use the data to determine the order of reaction with respect to hydrochloric acid.

First order. This is because the rate equation doesn’t include solids. Therefore, the only reactant in the rate equation is the HCl and it therefore has the same order of reaction as the overall order.

2. Ethyl Ethanoate & Sodium Hydroxide

A reaction takes place between ethyl ethanoate and sodium hydroxide. The concentration of the sodium hydroxide is measured at regular time intervals. The results of the reaction are recorded in the table below.

Time / s	0	1	2	3	4	5	6	7	8	9
[OH^–] / mol dm^-3	1.29	1.03	0.86	0.73	0.64	0.56	0.51	0.46	0.43	0.39

a) Plot a graph of the results on graph paper.

b) Explain why the rate of reaction decreases over time.

The concentration of the reactants decreases over time, resulting in less frequent collisions and therefore less frequent successful collisions.

c) Use tangents to determine the overall order of reaction. If you are doing OCR, explain why 1/2 lives cannot be used for this graph.

See graph above. Half lives cannot be used, because there is not enough data for the second half life to be calculated.

d) The experiment was repeated with different conditions. One change was that a large excess of ethyl ethanoate was used. The results are in the graph below. Use this graph, and the previous results, to determine the rate equation for the reaction.

Rate = k [CH₃COOCH₂CH₃] [NaOH]

You know this because the order of reaction for NaOH is first order according to the second graph. You can use half lives (like below) or tangents to prove this. As the overall order is second and it is first order with respect to NaOH, it must also be first order with respect to ethyl ethanoate.

3. 2-chloro-2-methylpropane & NaOH

The reaction between 2-chloro-2-methylpropane and sodium hydroxide was observed. The reaction was completed twice, once with neither reactant in a large excess, and once with the halogenoalkane in a large excess. The other conditions of the experiments may have varied. Determine the rate equation for the reaction.

Experiment 1: No large excess

Time / s	0	10	20	30	40	50	60	70	80	90	100
[OH^–]	1.5	1.2	0.95	0.75	0.6	0.46	0.37	0.3	0.24	0.2	0.15

Experiment 2: Large Excess of 2-chloro-2-methylpropane

Time / s	0	10	20	30	40	50	60	70	80	90	100
[OH^–]	2.25	2.1	1.94	1.8	1.64	1.51	1.35	1.22	1.08	0.95	0.83

The reaction is first order overall according to the first graph. It is 0 order with respect to NaOH according to the second graph.

Therefore the rate equation is: Rate = k [2-methyl-2-chloropropane]

Mr Cole Chemistry