Most A Level H2 Chemistry students have no problem determining the half life of reactants in a reaction. A good proportion, however, might have issues with half lives of products. Let’s take a look at a past year preliminary question
RI, 2006 P3; DHS, 2012 P3
The kinetics of the hydrolysis of the ester, CH3CH2CO2CH3, may be investigated by the following method
CH3CH2CO2CH3 + H2O → CH3CH2CO2H + CH3OH
In a 1 dm3 mixture, 0.35 mol of the ester was hydrolysed by heating with water and using hydrochloric acid as catalyst. The following results were obtained.
|
Time /s |
Concentration of CH3CH2CO2H/moldm-3 |
|
0 |
0 |
|
340 |
0.105 |
|
680 |
0.185 |
|
1080 |
0.243 |
|
1440 |
0.278 |
a) By using a graphical method, determine the half-life of the reaction and hence the order of reaction with respect to the ester.
b) It has been found that the hydrolysis reaction is 1st order with respect to hydrochloric acid.
i) Write down the rate equation of this reaction and state the units of the rate constant.
ii)What will be the half-life of the reaction if the experiment is repeated with the concentration of hydrochloric acid doubled? Explain your answer
Discussion
We have learnt in the previous blog that for different orders of reaction with respect to a reactant, the half live with respect to that reactant can vary differently with time, ie
|
|
Half-Life |
Variation with time |
|
Zero-order |
[Reactant]i/2k |
Decreases |
|
First order |
ln 2 /k |
Constant |
|
Second order |
1/(k[Reactant]i) |
Increases |
where [Reactant]i = reactant’s initial concentratin
The variation of half life with time (as shown above) applies to both reactants and products.
So how do we determine the half life of a reactant, based on product data?
Just follow these steps and you are on your way to getting full marks for this type of question, as shown in DHS and RI
Step 1
Determine the maximum product concentration at the end of the reaction
ie the maximum concentration of CH3CH2CO2H
= moles of ester reacted at the end of the reaction / volume
= 0.35 mol/dm3
Step 2
|
Time lapsed |
[CH3CH2CO2CH3] |
[CH3CH2CO2H] |
|
0 |
0.35 |
0 |
|
1st half-life |
(0.5)x 0.35 |
(0.5)x0.35 |
|
2nd half-life |
(0.25)x0.35 |
(0.75)x0.35 |
|
3rd half-life |
(0.125)x0.35 |
(0.875)x0.35 |
Do you notice a trend?
nth half-life lapsed
[reactant] = (0.5)n ( initial concentration of reactant)
[product] = (1- 0.5n)(final concentration of product)
Step 3
One main difference in taking half-lives of reactants as compared with products, is that
Reactants’ half-lives can be take FROM ANY POINT IN TIME IN THE COURSE OF THE REACTION
Products’ half-lives MUST BE TAKEN SEQUENTIALLY FROM 0 to ½(max product formed);
from ½(max product formed) to 3/4 (max product formed)
Step 4
After determining at least 2 half lives (as required by Cambridge), compare to the table in step 1, and deduce the order of reaction with respect to the ester. Most of the questions that appear in exams are 1st order with respect to the only reactant in the reaction (ester in the case of the question in this blog)
ONE FINAL NOTE
If there is only reactant, then one product curve is needed to find the half life with respect to one product. What happens when there are 2 or more reactants? Then 2 or more product curves will be needed. In those situations, the half life method only allow the determination of the order with respect to 1 reactant. How about the other reactant?
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