When the concentration of the reactant in a given reaction is halved and if the rate of reaction is halved, the order of the reaction is:

Step-by-step solution

Let the rate of the reaction be r = k[A]^n , where n is the order of the reaction. When the concentration of the reactant is halved, the new concentration is [A] 2 . The new rate is given as r 2 . Substituting these into the rate law: r 2 = k ( [A] 2 )^n Dividing the new rate equation by the original rate equation: r 2 r = k ( [A] 2 )^n k[A]^n 1 2 = ( 1 2 )^n Comparing the powers, we get n = 1 . Thus, the order of the reaction is 1 . Answer: 1