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What is the order of the reaction between A and B, if the rate of reaction is found to be tripled when the concentration of A is doubled, and the concentration of B is kept constant?

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The order of a reaction is determined by the relationship between the rate of reaction and the concentration of the reactants. In this case, we are given that the rate of reaction triples when the concentration of A is doubled, while the concentration of B is kept constant.The rate law for a reaction involving reactants A and B can be written as:Rate = k[A]^m[B]^nwhere k is the rate constant, [A] and [B] are the concentrations of A and B, and m and n are the orders of the reaction with respect to A and B, respectively.Since the concentration of B is kept constant, we can focus on the relationship between the rate and the concentration of A:Rate = k[A]^mWhen the concentration of A is doubled, the rate triples:3 * Rate = k 2[A] ^mDivide both sides by the original rate equation:3 =  2 ^mTo solve for m, take the logarithm of both sides:log 3  = m * log 2 m = log 3  / log 2 m  1.58Since the order of a reaction is typically an integer, we can round m to the nearest integer, which is 2. Therefore, the order of the reaction with respect to A is 2. Since the concentration of B is kept constant, its order is not relevant to the overall order of the reaction. Thus, the overall order of the reaction between A and B is 2.

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