At 518$$^\circ$$C, the rate of decomposition of a sample of gaseous acetaldehyde, initially at a pressure of 363 Torr was 1.00 Torr s$$^{-1}$$ when 5% had reacted and 0.50 Torr s$$^{-1}$$ when 33% had reacted. The order of the reaction is:

We have a single-reactant gas-phase decomposition. For such a reaction the differential rate law can be written in terms of pressure as

$$-\frac{dP}{dt}=k\,P^{\,n},$$

where $$P$$ is the instantaneous pressure, $$k$$ is the rate constant and $$n$$ is the order that we have to find.

The initial pressure is given as

$$P_0=363\ \text{Torr}.$$

The problem supplies the rate at two different extents of reaction, so we first convert the percentages reacted into the corresponding instantaneous pressures.

When 5 % has reacted, the fraction of acetaldehyde that is still present is $$1-0.05=0.95$$. Hence

$$P_1=0.95\,P_0=0.95\times 363=344.85\ \text{Torr}.$$

At this stage the rate is

$$r_1=\Bigl(-\frac{dP}{dt}\Bigr)_1=1.00\ \text{Torr s}^{-1}.$$

When 33 % has reacted, the unreacted fraction is $$1-0.33=0.67$$. Therefore

$$P_2=0.67\,P_0=0.67\times 363=243.21\ \text{Torr},$$

and the rate is

$$r_2=\Bigl(-\frac{dP}{dt}\Bigr)_2=0.50\ \text{Torr s}^{-1}.$$

Because the rate law is $$r=k\,P^{\,n}$$ at any moment, we can write for the two sets of conditions

$$r_1=k\,P_1^{\,n},\qquad r_2=k\,P_2^{\,n}.$$

Dividing the first equation by the second eliminates the unknown rate constant $$k$$:

$$\frac{r_1}{r_2}=\left(\frac{P_1}{P_2}\right)^{\!n}.$$

Substituting the numerical values,

$$\frac{1.00}{0.50}=2.00=\left(\frac{344.85}{243.21}\right)^{\!n}.$$

The pressure ratio is

$$\frac{344.85}{243.21}=1.418\ (\text{approximately}).$$

So we have

$$2.00=(1.418)^{\,n}.$$

To isolate $$n$$, we take natural logarithms of both sides (any logarithm base would work):

$$\ln 2.00 = n\,\ln 1.418.$$

Using the numerical logarithms $$\ln 2.00 = 0.6931$$ and $$\ln 1.418 \approx 0.3500$$, we obtain

$$n=\frac{0.6931}{0.3500}\approx 1.98.$$

The exponent is essentially 2, which means the reaction is second order with respect to acetaldehyde.

Hence, the correct answer is Option B.