In a kinetic study of the reaction
2NO(g) + O2 --> 2NO2(g)
the following data were obtained fro the initial rates if disapperance of NO:
Initial concentraition
NO O2 Inital Rate NO
exp 1. 0.0125M 0.0253M 0.0281 M/s
exp 2. 0.0250M 0.0253M 0.0112 M/s
exp 3. 0.0125M 0.0506M 0.0561 M/s
Obtain the rate law. What is the value of the rate constant?
I got 88.9 L/(mol*s)
To obtain the rate law, we need to determine the order of the reaction with respect to each reactant (NO and O2). We can do this by comparing the initial rates in each experiment while keeping one reactant constant and varying the concentration of the other reactant.
Let's compare experiments 1 and 2, where the concentration of O2 is constant at 0.0253M. The only difference between these two experiments is the concentration of NO. We can see that doubling the concentration of NO from 0.0125M to 0.0250M decreases the initial rate from 0.0281 M/s to 0.0112 M/s. This suggests that the rate of the reaction is proportional to the concentration of NO raised to the power of 1.
Therefore, we can write the rate law for the reaction as:
Rate = k[NO]^1[O2]^m
Now, let's compare experiments 1 and 3, where the concentration of NO is constant at 0.0125M. The only difference between these two experiments is the concentration of O2. We can see that doubling the concentration of O2 from 0.0253M to 0.0506M increases the initial rate from 0.0281 M/s to 0.0561 M/s. This suggests that the rate of the reaction is proportional to the concentration of O2 raised to the power of 1.
Therefore, the rate law for the reaction becomes:
Rate = k[NO]^1[O2]^1
Simplifying the rate law, we get:
Rate = k[NO][O2]
Now, we can use any of the experimental data points to calculate the rate constant (k). Let's use the data from experiment 1:
Rate = k[NO][O2]
0.0281 M/s = k(0.0125M)(0.0253M)
Solving for k:
k = 0.0281 M/s / (0.0125M * 0.0253M)
k ≈ 88.7 M⁻²s⁻¹
Therefore, the rate law for the reaction is Rate = 88.7[NO][O2] and the value of the rate constant (k) is approximately 88.7 M⁻²s⁻¹.