0 votes
24 views
ago in Physical Chemistry by (2.7k points)
What is the order of the reaction if the rate constant of a chemical reaction changes by a factor of 100 when the concentration of reactant is increased by 10 times?

1 Answer

0 votes
ago by (2.0k points)
The order of a reaction can be determined by analyzing how the rate of the reaction changes with the concentration of the reactants. In this case, the rate constant  k  changes by a factor of 100 when the concentration of the reactant is increased by 10 times.Let's denote the initial concentration of the reactant as [A] and the initial rate constant as k1. When the concentration of the reactant is increased by 10 times, the new concentration is 10[A] and the new rate constant is k2 = 100k1.The rate law for a reaction can be written as:Rate = k[A]^nwhere n is the order of the reaction with respect to the reactant A.For the initial concentration and rate constant, we have:Rate1 = k1[A]^nFor the increased concentration and rate constant, we have:Rate2 = k2 10[A] ^n = 100k1 10[A] ^nNow, we can set up a ratio of the two rates:Rate2/Rate1 =  100k1 10[A] ^n  /  k1[A]^n Since Rate2 is 100 times Rate1, we have:100 =  100k1 10[A] ^n  /  k1[A]^n Simplifying the equation, we get:1 =  10^n To find the order of the reaction  n , we can take the logarithm of both sides:log 1  = log 10^n 0 = n * log 10 Since log 10  = 1, we get:0 = nSo, the order of the reaction is 0. However, this result is contradictory to the given information that the rate constant changes by a factor of 100 when the concentration of the reactant is increased by 10 times. It seems there might be an error or misunderstanding in the problem statement. Please double-check the information provided and ensure it is accurate.

Related questions

Welcome to Sarvan Science Q&A, where you can ask questions and receive answers from other members of the community.
...