Using the information in the table, the value of the rate constant for the reaction A(g) + 3 B(g) → C(g) + 2 D(g) is

Question

Problem Table

[A]₀ (M)

[B]₀ (M)

Rate (M/min)

0.300

0.180

0.0234

0.300

0.360

0.0934

0.150

0.180

0.0234

It wouldnt let me paste the picture so the first [A] is the first three numbers, second [B] is the second set of three numbers, the grey ones.

Answer 1

Compare the multiplier for one of the concentrations to the multiplier for the rate. The power required to make them equal is the order of the reaction. Generally in these problems, you can isolate the effects of the two concentrations. In this problem the trials were chosen to make the problem harder.

Trial 2 and 1 (.18/.3)^m(.36/.3)ⁿ = (.18/.15) or ratio of A to the m power x ratio of B to the n power = ratio of rs

So you get .6^m1.2ⁿ = 1.2 (by inspection, you can see m=0, n=1 works

For Trial 3 and 1 you get (.0234/.3)^m(.0934/.3)ⁿ = .0234/.15) = .078^m.31133ⁿ = .31133

You can see that m=0, n=1 works again.

r = k[B]