Kinetics worksheet with practice problems on writing rate laws and analyzing reaction rates.

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1. Writing generic rate laws using variables for exponents.
2. Determining rate laws from experimental data.

Let me solve each part step by step based on what’s visible in your message.

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Part 1: Write generic rate laws

For each reaction, write a rate law in the form:

> Rate = k [A]^m [B]^n ...

where m and n are exponents (orders) to be determined experimentally.

#### a. CO + NO₂ → CO₂ + NO

This is a reaction between CO and NO₂.

- The generic rate law is:

> Rate = k [CO]^m [NO₂]^n

Where:
- m = order with respect to CO
- n = order with respect to NO₂

(We don’t know m and n yet — they must be determined experimentally.)

✔ Answer:
> Rate = k [CO]^m [NO₂]^n

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#### b. 2H₂O₂ → 2H₂O + O₂

This is a decomposition reaction involving only H₂O₂.

- Only one reactant, so the rate law depends only on [H₂O₂].

> Rate = k [H₂O₂]^m

✔ Answer:
> Rate = k [H₂O₂]^m

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#### c. CH₃COCH₃ + OH⁻ → CH₃COO⁻ + CH₃OH

This is an acid-base or nucleophilic substitution reaction involving acetone (CH₃COCH₃) and hydroxide ion (OH⁻).

- Two reactants: CH₃COCH₃ and OH⁻

> Rate = k [CH₃COCH₃]^m [OH⁻]^n

✔ Answer:
> Rate = k [CH₃COCH₃]^m [OH⁻]^n

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Part 2: Use experimental data to find rate laws

We’re given tables of concentrations and rates. We'll use the method of initial rates to determine the order of reaction with respect to each reactant.

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#### a. Reaction: 2NO(g) + 2H₂(g) → N₂(g) + 2H₂O(g)

| Experiment | [NO] (M) | [H₂] (M) | Rate (M/s) |
|-----------|----------|----------|------------|
| 1 | 0.60 | 0.40 | 1.8 × 10⁻⁴ |
| 2 | 0.80 | 0.40 | 3.2 × 10⁻⁴ |
| 3 | 0.80 | 0.20 | 1.6 × 10⁻⁴ |

Assume rate law:

> Rate = k [NO]^m [H₂]^n

We’ll compare experiments to find m and n.

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Step 1: Find m (order w.r.t. NO)

Compare Exp 1 and 2, where [H₂] is constant (0.40 M), but [NO] changes:

- Exp 1: [NO] = 0.60, Rate = 1.8 × 10⁻⁴
- Exp 2: [NO] = 0.80, Rate = 3.2 × 10⁻⁴

Set up ratio:

$$
\frac{\text{Rate}_2}{\text{Rate}_1} = \frac{k [NO_2]^m [H_2]^n}{k [NO_1]^m [H_2]^n} = \left(\frac{[NO_2]}{[NO_1]}\right)^m
$$

$$
\frac{3.2 \times 10^{-4}}{1.8 \times 10^{-4}} = \left(\frac{0.80}{0.60}\right)^m
$$

$$
1.777... = (1.333)^m
$$

Take log:

$$
\log(1.777) = m \cdot \log(1.333)
\Rightarrow 0.247 = m \cdot 0.1249
\Rightarrow m ≈ 2.0
$$

✔ So, m = 2, second order in NO.

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Step 2: Find n (order w.r.t. H₂)

Compare Exp 2 and 3, where [NO] is constant (0.80 M), [H₂] changes:

- Exp 2: [H₂] = 0.40, Rate = 3.2 × 10⁻⁴
- Exp 3: [H₂] = 0.20, Rate = 1.6 × 10⁻⁴

Ratio:

$$
\frac{3.2 \times 10^{-4}}{1.6 \times 10^{-4}} = \left(\frac{0.40}{0.20}\right)^n = 2^n
\Rightarrow 2 = 2^n \Rightarrow n = 1
$$

✔ So, n = 1, first order in H₂.

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Final rate law:

> Rate = k [NO]² [H₂]¹

✔ Answer:
> Rate = k [NO]² [H₂]

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#### b. Reaction: 2NO₂(g) → 2NO(g) + O₂(g)

| Experiment | [NO₂] (M) | Rate (M/s) |
|-----------|-----------|------------|
| 1 | 0.10 | 0.050 |
| 2 | 0.20 | 0.200 |
| 3 | 0.30 | 0.450 |

Only one reactant: NO₂

Assume: Rate = k [NO₂]^m

Use Exp 1 and 2:

$$
\frac{0.200}{0.050} = \left(\frac{0.20}{0.10}\right)^m
\Rightarrow 4 = 2^m \Rightarrow m = 2
$$

Check with Exp 3:

$$
\frac{0.450}{0.050} = 9, \quad \left(\frac{0.30}{0.10}\right)^2 = 3^2 = 9 \quad ✔
$$

So, m = 2

✔ Final rate law:

> Rate = k [NO₂]²

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#### c. Reaction: 2NO(g) → N₂(g) + O₂(g)

| Experiment | [NO] (M) | Rate (M/s) |
|-----------|----------|------------|
| 1 | 0.0010 | 0.0010 |
| 2 | 0.0020 | 0.0040 |
| 3 | 0.0030 | 0.0090 |

Only one reactant: NO

Rate = k [NO]^m

Compare Exp 1 and 2:

$$
\frac{0.0040}{0.0010} = \left(\frac{0.0020}{0.0010}\right)^m
\Rightarrow 4 = 2^m \Rightarrow m = 2
$$

Check Exp 3:

$$
\frac{0.0090}{0.0010} = 9, \quad \left(\frac{0.0030}{0.0010}\right)^2 = 3^2 = 9 \quad ✔
$$

So, m = 2

✔ Final rate law:

> Rate = k [NO]²

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✔ Final Answers Summary:

#### Part 1: Generic Rate Laws
a. Rate = k [CO]^m [NO₂]^n
b. Rate = k [H₂O₂]^m
c. Rate = k [CH₃COCH₃]^m [OH⁻]^n

#### Part 2: Experimental Rate Laws
a. Rate = k [NO]² [H₂]
b. Rate = k [NO₂]²
c. Rate = k [NO]²

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If you have any additional questions or need help with calculating the rate constant (k), just let me know!