At $$345$$ K, the half life for the decomposition of a sample of a gaseous compound initially at $$55.5$$ kPa was $$340$$ s. When the pressure was $$27.8$$ kPa, the half life was found to be $$170$$ s. The order of the reaction is ______ [integer answer]

For an $$\mathrm{n^{th}}$$ order reaction, the half-life relation is:

$$\mathrm{t_{1/2} \propto \frac{1}{P_0^{\,n-1}}}$$

Thus:

$$\mathrm{\frac{(t_{1/2})_1}{(t_{1/2})_2} = \left(\frac{P_1}{P_2}\right)^{1-n}}$$

Substituting the given values:

$$\mathrm{\frac{340}{170} = \left(\frac{55.5}{27.8}\right)^{1-n}}$$

Simplifying:

$$\mathrm{2 \approx 2^{\,1-n}}$$

For this equality to hold:

$$\mathrm{1-n = 1}$$

$$\mathrm{n = 0}$$

Thus, the reaction is:

$$\mathrm{Zero\ Order}$$

Correct Answer: $$\mathrm{0}$$