For a reaction A + B → products, the following data were collected.
Experiment Number	Initial Concentration of A
(M)	Initial Concentration of B
(M)	Observed Initial Rate
(M/s)
1	3.30	4.10	1.89 ✕ 10−4
2	4.55	4.10	4.95 ✕ 10−4
3	3.30	5.26	1.89 ✕ 10−4
Calculate the rate constant for this reaction.
5.27e-6
Correct: Your answer is correct.

M−2·s−1

To calculate the rate constant for the reaction, we use the rate law equation, which is generally expressed as:

\[
\text{Rate} = k [A]^m [B]^n
\]

Here, \( k \) is the rate constant, and \( m \) and \( n \) are the reaction orders with respect to reactants A and B, respectively. We'll determine the reaction orders first and then use the data to find the rate constant.

### Step 1: Determining Reaction Orders \( m \) and \( n \)

Using the data from the experiments, we can compare the rates of reaction when the concentration of one reactant is constant, while the concentration of the other reactant changes.

#### To find \( m \) (the order with respect to A):
Compare experiments 1 and 2, where [B] is constant:
- Experiment 1: [A] = 3.30 M, [B] = 4.10 M, Rate = \( 1.89 \times 10^{-4} \, \text{M/s} \)
- Experiment 2: [A] = 4.55 M, [B] = 4.10 M, Rate = \( 4.95 \times 10^{-4} \, \text{M/s} \)

The rate law is:

\[
\frac{\text{Rate}_2}{\text{Rate}_1} = \left(\frac{[A]_2}{[A]_1}\right)^m
\]

Substitute the values:

\[
\frac{4.95 \times 10^{-4}}{1.89 \times 10^{-4}} = \left(\frac{4.55}{3.30}\right)^m
\]

\[
2.62 = \left(1.38\right)^m
\]

Taking the logarithm of both sides:

\[
\log(2.62) = m \cdot \log(1.38)
\]

\[
m = \frac{\log(2.62)}{\log(1.38)} = \frac{0.4183}{0.1399} \approx 3
\]

So, the reaction is third-order with respect to A, \( m = 3 \).

#### To find \( n \) (the order with respect to B):
Compare experiments 1 and 3, where [A] is constant:
- Experiment 1: [A] = 3.30 M, [B] = 4.10 M, Rate = \( 1.89 \times 10^{-4} \, \text{M/s} \)
- Experiment 3: [A] = 3.30 M, [B] = 5.26 M, Rate = \( 1.89 \times 10^{-4} \, \text{M/s} \)

Since the rates are identical despite the change in concentration of B, the reaction is zero-order with respect to B, i.e., \( n = 0 \).

### Step 2: Calculating the Rate Constant \( k \)

Now that we know the reaction orders \( m = 3 \) and \( n = 0 \), the rate law simplifies to:

\[
\text{Rate} = k [A]^3
\]

Using the data from experiment 1:

\[
1.89 \times 10^{-4} = k (3.30)^3
\]

Solve for \( k \):

\[
k = \frac{1.89 \times 10^{-4}}{(3.30)^3} = \frac{1.89 \times 10^{-4}}{35.937}
\]

\[
k \approx 5.27 \times 10^{-6} \, \text{M}^{-2} \, \text{s}^{-1}
\]

### Final Answer:
The rate constant \( k \) is \( 5.27 \times 10^{-6} \, \text{M}^{-2} \, \text{s}^{-1} \), which is consistent with the given correct answer.

---

Would you like further clarification on any part of this? Here are 5 related questions for further exploration:

1. How does the reaction order affect the overall rate law?
2. What are the units of the rate constant for different reaction orders?
3. How can reaction orders be determined experimentally?
4. What is the significance of zero-order reactions in chemical kinetics?
5. How would the rate constant change if temperature were increased?

**Tip**: Always check units when calculating the rate constant, as they vary depending on the overall order of the reaction.

For a reaction A + B → products, the following data were collected.
Experiment Number	Initial Concentration of A (M)	Initial Concentration of B (M)	Observed Initial Rate (M/s)
1	3.30	4.10	1.89 ✕ 10−4
2	4.55	4.10	4.95 ✕ 10−4
3	3.30	5.26	1.89 ✕ 10−4
Calculate the rate constant for this reaction.

Determine the rate constant for the reaction A + B → products using the given experimental data. Learn about chemical kinetics, rate laws, and reaction orders. The data involve calculating the rate constant based on the third-order dependence on reactant A and zero-order dependence on reactant B.

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