Chemistry 102: Kinetics: Predicting Initial Rate

Question

Rate laws are determined experimentally by varying the concentrations of the reactants and observing the effect on the initial rate.

Concentration and rate data are provided in the table below for the reaction shown.

Experiment [NO] [Cl_2] Initial Rate

1 0.20 0.10 0.63

2 0.20 0.30 5.70

3 0.80 0.10 2.58

4 0.40 0.20 ?

2NO(g)+Cl2​(g)2NOCl(g)

Concentration/Rate Data for Reaction

1.Predict the rate for experiment 4, given the other concentration/rate data provided (M/s).

Answer 1

Hi, so since this is a Kinetics problem, you will have to use the Kinetics Rate Equation in order to answer it. So, first we must find the rate orders in respect to each of the species involved in the reaction, so first, we can find the rate order in respect to [NO], and you can do this by making a proportion between Experiment 3 and Experiment 1, so in essence (0.80/0.20)^x = (2.58/0.63). Now you ask why I chose those two experiments specifically, and the reason is that those are the only two scenarios where the concentration of [Cl2] isn't altered, so you can find the direct effect of changing the concentration of [NO]. If you solve for x in the proportion I gave you, you get a result of x = 1, so now we know that the rate order for [NO] is first-order. We do this exact same process for [Cl2], and you should get a result of y = 2 for that, so by using these two answers, you should get a rate equation of Rate = k[NO]^1[Cl2]^2. Now with this is equation, you can find the unknown rate for Experiment 4 by solving for the rate constant, k, by using a 1 of the 3 available experiments, by substituting in the rate and respective concentrations into the rate equation. When you do this, you should get a rate constant value of 315. With this values you can finally find the rate for Experiment 4, by essentially setting up the equation like this, Rate = (315)(0.40)(0.20)^2. When you do this, you should get a rate of 5.04 M/s for Experiment 4. I hope this explanation helps.