Half-life of a chemical reaction at a particular concentration is 50 minutes. When the concentration of the reactant is doubled, the half-life becomes 100 minutes. What is the order of reaction?

0

\(\frac{(t_{1/2})_1}{(t_{1/2})_2}\) = [\(\frac{a_2}{a_1}\)]^n-1

\(\frac{50}{100}\) = [\(\frac{2}{1}\)]^n-1

2^-1 = 2^n-1

-1 = n - 1

n = 0

Therefore, it is a zero-order reaction.