Question

For the reaction; 2A + B -> D + E
the following initial rate data were collected at constant temperature. Determine the correct rate law for this reaction. All units are arbitrary.
Trial [A] [B] Rate
1 0.225 0.150 0.0217
2 0.320 0.150 0.0439
3 0.225 0.250 0.0362
4 0.225 0.150 0.01270
What is the rate order for A?
rate order for B?
Rate constant k?

Answer 1

Compare 2 trials when only one reactant is changing (in terms of concentration) this allows you to ignore the contribution of the reactant kept constant toward the rate.

for example, in trial 1 and 2, B is held \(\sf \color{red}{constant}\).
Trial [A] [B] Rate
1 0.225 \(\color{red}{0.150}\) 0.0217
2 0.320 \(\color{red}{0.150}\) 0.0439

so A is the only contributing factor.

Now determine what [A] increased by, and compare that to the rate.

i ended up getting a non-integer order, which is not typical if you're in hs. Also, trial 1 and 4 have the same concentrations but a very different rate. you may want to double check that you copied the data down correctly.

Answer 2

This is a college course. Inorganic Chemistry II. I double checked and all the data is correctly copied from my study guide.

Answer 3

Do you mean general chemistry II?

Answer 4

It is titled General Inorganic Chemistry II. I don't see how that is relevant though. If you can help me that would be great.

Answer 5

if all the data is correct, then ignore the last thing i said and carry on through with the method i posted.