1. Use pressure as concentration
   For a gaseous reaction at constant volume and temperature, pressure is proportional to concentration.
   So we can use $r = kP^n$ where $P$ is the partial pressure of acetaldehyde remaining.

2. Initial pressure
   Given initial pressure: $P_0 = 363\ \text{Torr}$

3. Pressure when 5% has reacted
   If $5\%$ has reacted, then $95\%$ remains: $P_1 = 0.95 \times 363 = 344.85\ \text{Torr}$ Given rate: $r_1 = 1.00\ \text{Torr s}^{-1}$

4. Pressure when 33% has reacted
   If $33\%$ has reacted, then $67\%$ remains: $P_2 = 0.67 \times 363 = 243.21\ \text{Torr}$ Given rate: $r_2 = 0.50\ \text{Torr s}^{-1}$

5. Apply rate law ratio
   $\frac{r_1}{r_2} = \left(\frac{P_1}{P_2}\right)^n$ Substitute values: $\frac{1.00}{0.50} = \left(\frac{344.85}{243.21}\right)^n$ $2 = (1.4178)^n$

6. Solve for $n$
   Taking logarithms: $n = \frac{\log 2}{\log 1.4178}$ Now, $\log 2 \approx 0.3010, \quad \log 1.4178 \approx 0.1510$ So, $n \approx \frac{0.3010}{0.1510} \approx 1.99 \approx 2$

7. Conclusion
   The reaction is second order.

Therefore, the correct option is: $\boxed{\text{B: }2}$