Question

Consider the following reaction at 283 K:

2A + B → C + D

where rate = k[A][B]². An experiment was performed where [A]_o = 2.67 M and [B]_o = 0.00241 M. A plot of 1/[B] vs. time has a slope of 10.01. What will the rate of this reaction be when [A] = [B] = 0.345 M?

Answer 1

Sometimes to calculate the kinetical constant "k" we can create some conditions to emulate a pseudo order reaction.

For this particular case we know that the reaction has an order of 3.

The experiment has a concentration of one of the reactants much bigger than the other, for this case:

The key to understand this is that the concentration of A will not change drastically and the reaction will go on with an apparently a different order that will make the calculation easier.

In this case the reaction will have a pseudo second order (A is not considered to change in this part), the statement says that the graph 1/B vs Time is 10.01.

For a second order reaction this is the constant "K"

rate of original reaction = K * A * B²

pseudo second order reaction = K´ B²

K´= K * A

Since we have the value of A we can calculate real K

K = K´/ A

K = 10.01 / 2.67 = 3.749 This is the constant for the overall reaction

Now we can apply the equation

rate = 3.749 * A * B²

A = B = 0.345

rate = 3.749 * 0.345 * 0.345²

rate = 0.154 M / seg