Question

The initial rate of a reaction doubles as the concentration of one of the reactants is quadrupled. What is the order of this reactant? If a reactant has a \(-1\) order, what happens to the initial rate when the concentration of that reactant increases by a factor of two?

Short answer

The order of reactant A is 1, as the initial rate doubles when the concentration of the reactant quadruples. If a reactant has a -1 order, when the concentration of that reactant increases by a factor of two, the initial rate will be halved.

Step-by-step solution

Understand the rate law expression

The rate law expression relates the rate of reaction to the concentration of the reactants. It is given by: Rate \(= k[A]^m\) Where 'Rate' is the initial rate of the reaction, 'k' is the rate constant, \([A]\) is the concentration of reactant A, and 'm' is the order of reactant A.

Set up the given information and the rate law expressions

We are given that the initial rate doubles when the concentration of reactant A is quadrupled. Therefore, we can set up the following expressions: \(Rate_1 = k[A_1]^m\) \(Rate_2 = k[A_2]^m = 2Rate_1\) Where \([A_2]\) is four times the concentration of \([A_1]\): \([A_2] = 4[A_1]\)

Find the order of the reactant

Now, we can write the rate law expressions for the two cases and divide the second equation by the first to eliminate 'k': \(\dfrac{Rate_2}{Rate_1} = \dfrac{k[A_2]^m}{k[A_1]^m}\) \(\dfrac{2Rate_1}{Rate_1} = \dfrac{k(4[A_1])^m}{k[A_1]^m}\) \(2 = (4)^m\) To find 'm', we can take the logarithm of both sides: \(\log_4{2} = m\) \(m = 1\) The order of reactant A is 1.

Analyze the impact of a -1 order and increasing concentration by a factor of 2

Now, we need to determine what happens to the initial rate when the concentration of reactant A increases by a factor of 2 and the order is -1. Let's use the rate law expression: Rate \(= k[A]^m\) Here, \(m = -1\) and \([A'] = 2[A]\), where \([A']\) is the new concentration of reactant A. We can write the expression for the new rate: New Rate \(= k[A']^{-1}\) \(New Rate = k(2[A])^{-1}\) \(New Rate = k\dfrac{1}{2[A]}\) Since the rate is inversely proportional to the concentration, when the concentration of reactant A is doubled, the rate of the reaction will be halved.