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Rate question, help!!! (activation energy calculation)

Worms

(edited 6 years ago)

Reply 1

Plato's Trousers

Original post by Worms

A series of experiments were carried out to determine the activation energy for the oxidation of iodine ions by peroxodisulfate(VI) ions in the presebce of iron(III) ions.

S₂O₈^2-(aq) + 2I^-(aq) --> 2SO₄^2- (aq) + I²(aq)

There are a set of results from the 'clock' experiment. they were the following

Temperature, T/K - 288 - 292.5 - 299 - 308 - 315 -
Time, t for the - 10.0 - 7.0 - 5.0 - 3.5 - 2.5 -
blue colour
to appear

the questions I cannot do are the following

1) Plot a graph using the results to find a value for the activation energy in the presence of iron(III) ions

2) In the absence of iron (III) ions the activation energy for the reaction is 52.9kJmol^-1
Suggest and explanation for the effect of adding the iron (III) ions

thank you for any help, I do not expect anyone to do the question, but if anyone could point me in the right direction e.g. tell me what the put on the graph axis

Thank you in advance

Part 1)

Well... the temperature dependence of the rate constant tells you the activation energy. Here's the relevant equation.

k = A \exp \left( - \frac{E_a}{RT} \right)

Now the rate is inversely proportional to the time (t) taken to get the blue colour. Hence, your equation becomes

\dfrac{1}{t} = B \exp \left( - \frac{E_a}{RT} \right)

take logs

\ln \dfrac{1}{t} = \ln B - \frac{E_a}{RT}

- \ln t = \ln B - \frac{E_a}{RT}

\ln t = \frac{E_a}{RT} - \ln B

So, if you plot ln t against 1/T you will get a straight line with gradient E_a / R

You can just look up R (it's the gas constant). So you have E_a

Part 2)
Compare the answer you got for the E_a in part 1) with the figure they've given you without the iron (III) ions. What's your explanation of why they are different?