For cases where [A] exactly doubles:

• if the rate does not change, the order for A is 0.
• if the rate doubles, then the order for A is 1.
• if the rate quadruples, the order for A is 2.

For more complex cases, use the general equation:

What is the reaction order with respect to A?
[A], mol/L	[B], mol/L	rate, mol/(L-min)
Trial 1	0.771	0.682	0.0893
Trial 2	1.54	0.682	0.357
Trial 3	0.771	1.36	0.357
Trial 4	0.774	0.291	0.0164
Rate = k[A]orderA[B]orderB

The problem requires determining the reaction order with respect to \([A]\) and \([B]\) using the rate law:

\[
\text{Rate} = k[A]^{\text{order}_A}[B]^{\text{order}_B}
\]

We'll analyze the given data to find the reaction orders with respect to \( [A] \) and \( [B] \).

### Step 1: Data Table Recap
| Trial | \([A]\) (mol/L) | \([B]\) (mol/L) | Rate (mol/(L·min)) |
|-------|-----------------|-----------------|--------------------|
| 1     | 0.771           | 0.682           | 0.0893             |
| 2     | 1.54            | 0.682           | 0.357              |
| 3     | 0.771           | 1.36            | 0.357              |
| 4     | 0.774           | 0.291           | 0.0164             |

### Step 2: Determine Order of Reaction with Respect to \([A]\)

To isolate the effect of \([A]\), we compare trials where \([B]\) remains constant. Comparing **Trial 1** and **Trial 2**:

- Trial 1: \([A]_1 = 0.771 \, \text{mol/L}, \, \text{Rate}_1 = 0.0893 \, \text{mol/(L·min)}\)
- Trial 2: \([A]_2 = 1.54 \, \text{mol/L}, \, \text{Rate}_2 = 0.357 \, \text{mol/(L·min)}\)
- \([B]\) remains constant at 0.682 mol/L.

Now, calculate the ratio of the rates and the concentrations:

\[
\frac{\text{Rate}_2}{\text{Rate}_1} = \frac{k[A]_2^{\text{order}_A}[B]^{\text{order}_B}}{k[A]_1^{\text{order}_A}[B]^{\text{order}_B}} = \left(\frac{[A]_2}{[A]_1}\right)^{\text{order}_A}
\]

\[
\frac{0.357}{0.0893} = \left(\frac{1.54}{0.771}\right)^{\text{order}_A}
\]

\[
4 = 2^{\text{order}_A}
\]

This gives us \(\text{order}_A = 2\).

### Step 3: Determine Order of Reaction with Respect to \([B]\)

Now, compare **Trial 1** and **Trial 3**, where \([A]\) remains constant:

- Trial 1: \([B]_1 = 0.682 \, \text{mol/L}, \, \text{Rate}_1 = 0.0893 \, \text{mol/(L·min)}\)
- Trial 3: \([B]_3 = 1.36 \, \text{mol/L}, \, \text{Rate}_3 = 0.357 \, \text{mol/(L·min)}\)
- \([A]\) remains constant at 0.771 mol/L.

Again, calculate the ratio of the rates and concentrations:

\[
\frac{\text{Rate}_3}{\text{Rate}_1} = \frac{k[A]^{\text{order}_A}[B]_3^{\text{order}_B}}{k[A]^{\text{order}_A}[B]_1^{\text{order}_B}} = \left(\frac{[B]_3}{[B]_1}\right)^{\text{order}_B}
\]

\[
\frac{0.357}{0.0893} = \left(\frac{1.36}{0.682}\right)^{\text{order}_B}
\]

\[
4 = 2^{\text{order}_B}
\]

This gives us \(\text{order}_B = 2\).

### Step 4: Conclusion

- The reaction order with respect to \([A]\) is \( 2 \).
- The reaction order with respect to \([B]\) is \( 2 \).
- Therefore, the overall rate law is:

\[
\text{Rate} = k[A]^2[B]^2
\]

Would you like more details on this or have any questions?

Here are 5 related questions you might find helpful:
1. How do you calculate the rate constant \( k \) given reaction data?
2. What happens to the reaction rate if only \([B]\) is doubled?
3. How does temperature affect the rate constant \( k \)?
4. What is the difference between zero-order and first-order reactions?
5. How would the rate law change if the reaction order with respect to \( A \) was 1 instead of 2?

**Tip:** When determining reaction orders experimentally, always look for trials where only one reactant concentration changes while the others remain constant.

For cases where [A] exactly doubles: • if the rate does not change, the order for A is 0. • if the rate doubles, then the order for A is 1. • if the rate quadruples, the order for A is 2. For more complex cases, use the general equation: What is the reaction order with respect to A? [A], mol/L [B], mol/L rate, mol/(L-min) Trial 1 0.771 0.682 0.0893 Trial 2 1.54 0.682 0.357 Trial 3 0.771 1.36 0.357 Trial 4 0.774 0.291 0.0164 Rate = k[A]orderA[B]orderB

Learn how to determine the reaction orders for [A] and [B] using the rate law: Rate = k[A]^orderA[B]^orderB. Analyze the experimental data from four trials and calculate the reaction orders using the provided formulas. Suitable for high school chemistry students.

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