Given this data in a study on how the rate of a reaction was affected by the concentration of the reactants
Initial Rate,
RUN # A], M [B, M C], M mol L-1 s-1
1 0.200 0.100 0.600 5.00
2 0.200 0.400 0.400 80.0
3 0.600 0.100 0.200 15.0
4 0.200 0.100 0.200 5.00
5 0.200 0.200 0.400 20.0
The rate constant for this reaction (all in the same units) is
a. 6667 b. 208 c. 2083 d. 139 e. 2500
help me!
The secret here is to determine the order of the reaction. For example, it is zero order in C. Do you know how to do the others? After you determine the orders, then k is determined as in your post above.
To find the rate constant for the reaction, we need to use the rate equation. The rate equation relates the rate of reaction to the concentrations of the reactants and the rate constant.
The rate equation for a reaction is usually of the form:
Rate = k[A]^m[B]^n[C]^p
Where:
- Rate is the rate of reaction
- k is the rate constant
- [A], [B], [C] are the concentrations of the reactants A, B, and C, respectively
- m, n, p are the orders of the reaction with respect to A, B, and C, respectively
Given the data, we can see that the concentration of reactants A, B, and C changes for different runs, and the rate of reaction is also recorded.
Let's analyze one of the runs (let's say Run 1) to determine the rate equation and subsequently the rate constant.
For Run 1:
[A] = 0.200 M
[B] = 0.100 M
[C] = 0.600 M
Initial Rate = 5.00 mol L^-1 s^-1
Comparing with the rate equation:
Rate = k[A]^m[B]^n[C]^p
We can see that p = 1 because the rate is directly proportional to the concentration of reactant C.
To determine the values of m and n, we can compare two runs where only one reactant concentration changes. Let's compare Run 1 and Run 4 (where only [C] changes):
Run 1: [A] = 0.200 M, [B] = 0.100 M, [C] = 0.600 M, and Initial Rate = 5.00 mol L^-1 s^-1
Run 4: [A] = 0.200 M, [B] = 0.100 M, [C] = 0.200 M, and Initial Rate = 5.00 mol L^-1 s^-1
Since the concentration of [C] is the only difference between these two runs, we can deduce that the rate is not affected by the change in [C]. Therefore, n = 0 for reactant C.
So now we have:
Rate = k[A]^m[B]^n[C]^p
Rate = k[A]^m[B]^0[C]^1
Rate = k[A]^m
Comparing Run 1 and Run 2 (where the concentration of [B] changes):
Run 1: [A] = 0.200 M, [B] = 0.100 M, [C] = 0.600 M, and Initial Rate = 5.00 mol L^-1 s^-1
Run 2: [A] = 0.200 M, [B] = 0.400 M, [C] = 0.400 M, and Initial Rate = 80.0 mol L^-1 s^-1
We can see that the concentration of [B] has doubled, and the rate has increased by a factor of 16 (80.0 / 5.00). This implies that the rate is directly proportional to the concentration of [B] to the power of 1 (m = 1).
Therefore, the rate equation becomes:
Rate = k[A]^1[B]^1[C]^0
Rate = k[A][B]
Using the values from Run 1:
Rate = k[0.200][0.100] = 5.00
k = 5.00 / (0.200 * 0.100) = 250
Now we can calculate the rate constant for any of the runs. Let's take Run 3 as an example:
Run 3: [A] = 0.600 M, [B] = 0.100 M, [C] = 0.200 M, Initial Rate = 15.0 mol L^-1 s^-1
Rate = k[A][B]
15.0 = k(0.600)(0.100)
k = 15.0 / (0.600 * 0.100) = 2500
Therefore, the rate constant for the reaction is 2500 (option e).