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Step-by-Step Solution
Step 1: Understand the Given Data
• The initial pressure of acetaldehyde is 363 Torr.
• At 5% decomposition (hence 95% unreacted), the rate is 1.00 Torr s–1.
• At 33% decomposition (hence 67% unreacted), the rate is 0.5 Torr s–1.
• We assume a rate law of the form
$$
\text{Rate} = k \, [P_{t}]^n
$$
where $P_{t}$ is the partial pressure of the reactant at time t.
Step 2: Express the Pressure at Each Extent of Decomposition
• When 5% of the substance has reacted, 95% remains unreacted. Hence the pressure of unreacted acetaldehyde is:
$$
P_{5\%} = 363 \times \frac{95}{100} \text{ Torr}.
$$
• When 33% of the substance has reacted, 67% remains. Hence the pressure of unreacted acetaldehyde is:
$$
P_{33\%} = 363 \times \frac{67}{100} \text{ Torr}.
$$
Step 3: Write the Rate Expressions
• For 5% decomposition, the rate is 1.00 Torr s–1:
$$
1.00 = k \,\bigl(363 \times \frac{95}{100}\bigr)^n.
$$
• For 33% decomposition, the rate is 0.5 Torr s–1:
$$
0.5 = k \,\bigl(363 \times \frac{67}{100}\bigr)^n.
$$
Step 4: Divide One Equation by the Other
Divide the expression at 5% reaction by that at 33% reaction to eliminate k:
$$
\frac{1.00}{0.5}
= \frac{k\,\bigl(363 \times \frac{95}{100}\bigr)^n}{k\,\bigl(363 \times \frac{67}{100}\bigr)^n}
= \Bigl(\frac{95}{67}\Bigr)^n.
$$
Therefore,
$$
2 = \Bigl(\frac{95}{67}\Bigr)^n.
$$
Step 5: Simplify to Find the Order n
Numerically, $95/67 \approx 1.41$, which is close to $\sqrt{2}$. Hence:
$$
2 = (1.41)^n \approx (\sqrt{2})^n.
$$
Because $(\sqrt{2})^2 = 2$, we get:
$$
2 = 2^{n/2}.
$$
So,
$$
2^{1} = 2^{n/2} \quad\Rightarrow\quad \frac{n}{2} = 1 \quad\Rightarrow\quad n = 2.
$$
Step 6: Conclude the Order of the Reaction
The reaction is of second order ($n = 2$).
