Rate Equation and Rate Law

Question

The rate of the reaction between CO and NO2, CO(g) + NO2(g) → CO2(g) +

NO(g), was studied at 540 K starting with various concentrations of CO and NO2,

and the data in the table were collected. Determine the rate equation and the

value of the rate constant, k.

Initial Concentrations (mol/L)/h

[CO], mol/L [NO2], mol/L Initial Rate

1 5.10 x 10-4 0.350 x 10-4 3.4 x 10-8

2 5.10 x 10-4 0.700 x 10-4 6.8 x 10-8

3 5.10 x 10-4 0.175 x 10-4 1.7 x 10-8

4 1.02 x 10-3 0.350 x 10-4 6.8 x 10-8

5 1.53 x 10-3 0.350 x 10-4 10.2 x 10-8

2. What are the final concentrations of KMnO4 and H2C2O4 if 10.0 mL of a 0.010 M

KMnO4 solution, 20.0 mL of a 0.30 M H2C2O4 solution, and 10.0 mL of water are

mixed together?

Answer 1

Comparing trials one and four, we can see the concentration of CO doubles while the concentration of an NO2 stays the same. We also see that the rate doubles. This tells us that the reaction is first order with respect to CO. Next, comparing trials, one and two, we see that the concentration of NO2 doubles, while the concentration of CO stays the same. We also see that the rate doubles. This tells us that the reaction is first order with respect to an NO2. Thus we can write the rate equation as follows:

Rate = k[CO][NO2]

to find the value of k, we will choose trial one and solve for k.

3.4x10^-8 = k [5.10x10^-4][0.35x10^-4]

k = 1.90 (units unknown since not given in the rate data)

For the next part…

Total volume of solution = 10 ml + 20 ml + 10 ml = 40 ml

Using V1M1 = V2M2 we can solve the problem.

[KMnO4] = (10 ml)(0.01 M)/40 ml = 0.0025 M

[H2C2O4] = (20 ml)(0.3 M) / 40 ml = 0.15 M