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Orders of reaction

Please can someone walk me through how to answer this question. I’m so lost…

The rate equation for a reaction between substances A, B and C is of the form:
rate = k[A]xy[C]z where x + y + z = 4
The following data were obtained in a series ofexperiments at a constant temperature.image.jpg
Idk which experiments to choose from or how to start it 😭
(edited 11 months ago)
Reply 1
This is a "please do my homework" question. DO something first. You could also post the question the right way up!
Also, this is a maths thread, try posting to Chemistry.
Original post by LaSuGi
Please can someone walk me through how to answer this question. I’m so lost…

The rate equation for a reaction between substances A, B and C is of the form:
rate = k[A]xy[C]z where x + y + z = 4
The following data were obtained in a series of experiments at a constant temperature.image.jpg
Idk which experiments to choose from or how to start it 😭


I will start you off...

Compare the rates of experiments 1 & 2
You will see that both concentration of A and the concentration of C do not change.
So, any change in the rate is due to the change in concentration of B
[B ] in experiment 1 = 0.20
[B ] in experiment 2 = 0.05

So the concentration of B in experiment 1 is 4x higher

Now look at the rate in both experiments:
Rate in experiment 1 = 8.0 x 10-5
Rate in experiment 2 = 2.0 x 10-5

So the rate in experiment 1 is 4x higher

In other words the same change in concentration and rate, therefore rate is proportional to concentration, i.e. the order is 1 with respect to B

Rate = k[A ]x[B ]1[C ]z
Reply 3
Original post by charco
I will start you off...

Compare the rates of experiments 1 & 2
You will see that both concentration of A and the concentration of C do not change.
So, any change in the rate is due to the change in concentration of B
[B ] in experiment 1 = 0.20
[B ] in experiment 2 = 0.05

So the concentration of B in experiment 1 is 4x higher

Now look at the rate in both experiments:
Rate in experiment 1 = 8.0 x 10-5
Rate in experiment 2 = 2.0 x 10-5

So the rate in experiment 1 is 4x higher

In other words the same change in concentration and rate, therefore rate is proportional to concentration, i.e. the order is 1 with respect to B

Rate = k[A ]x[B ]1[C ]z


Thank you for your help!

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