An atmospheric scientist interested in how NO is converted into NO2 in urban atmospheres carries out two experiments to measure the rate of this reaction. The data are tabulated below.
A: [NO]0 = 9.63 × 10-1 M, [O2]0 = 4.1 × 10-4 M
t(s) 0/3.0/6.0/9.0/12.0
[O2] 4.1/2.05/1.02/0.51/0.25
*[O2] is in(10-4 M)
B: [NO]0 = 4.1 × 10-1 M, [O2]0 = 9.75 × 10-3 M
t(s) 0/1.00/2.00/3.00/4.00
[NO] 4.1/2.05/1.43/1.02/0.82
*[NO] is in(10-4 M)
I figured out the rate law:
But I can't find the rate constant! Please help!
"Calculate the rate constant. (in M-2 s-1)"
So what's the rate law? I don't know the exponents but you have that you say.
Rate = k(NO)^x*(O2)^y where x and y are the exponents you have determined.
Look at any run in the experiment, substitute rate, (NO) and (O2), then calculate k.
Hi DrBob, the rates are NOT given which is why I am having such an issue trying to work with rate and K both as unknowns.
Yes, the data is there to calculate that. The spacing is not good since this forum messes up the spacing; however, you are given the initial concentrations for both NO and O2. Then you are given the concentrations after certain times. That allows you to know the rate in M/s.
To calculate the rate constant for the reaction, we can use the rate law expression and the data from the experiments to determine the order of the reaction with respect to NO and O2.
Let's write the rate law expression for this reaction:
Rate = k [NO]^x [O2]^y
where k is the rate constant and x and y are the orders of the reaction with respect to NO and O2, respectively.
In experiment A, we can see that the initial concentration of NO is 9.63 × 10^-1 M and the initial concentration of O2 is 4.1 × 10^-4 M. Let's examine the change in the concentration of O2 over time:
t(s) [O2] (10^-4 M)
0 4.1
3.0 2.05
6.0 1.02
9.0 0.51
12.0 0.25
By examining the change in [O2], we can determine the order of the reaction with respect to O2. As we can see, when the concentration of O2 is halved, the rate is divided by 2. This suggests that the reaction is first order with respect to O2.
Now, let's move on to experiment B. Here, the initial concentration of NO is 4.1 × 10^-1 M and the initial concentration of O2 is 9.75 × 10^-3 M. Look at the change in the concentration of NO over time:
t(s) [NO] (10^-4 M)
0 4.1
1.00 2.05
2.00 1.43
3.00 1.02
4.00 0.82
By examining the change in [NO], we can determine the order of the reaction with respect to NO. As we can see, when the concentration of NO is halved, the rate is divided by 2, just like in experiment A. This suggests that the reaction is also first order with respect to NO.
Now that we know the orders of the reaction with respect to NO and O2 are both first order, we can write the rate law expression as:
Rate = k [NO] [O2]
Using the data from either experiment A or B, we can plug in the initial concentrations and the corresponding rate to solve for the rate constant k.
Let's use experiment A:
Rate = k [NO] [O2]
Rate = k (9.63 × 10^-1 M) (4.1 × 10^-4 M)
Rate = k (3.9513 × 10^-4 M^2)
Now, let's choose any time point from the experiment, for example, t = 3.0 s, where the rate is known:
Rate = 2.05 × 10^-4 M/s
We can now solve for the rate constant k:
2.05 × 10^-4 M/s = k (3.9513 × 10^-4 M^2)
Simplifying, we find:
k = (2.05 × 10^-4 M/s) / (3.9513 × 10^-4 M^2)
k ≈ 5.179 × 10^5 M^-1 s^-1
Therefore, the rate constant for this reaction is approximately 5.179 × 10^5 M^-1 s^-1.