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 the reactant, when the initial rate doubles as the concentration of the reactant quadruples, is \(1/2\). When the concentration of a reactant with an order of \(-1\) doubles, the initial reaction rate decreases by a factor of \(1/2\), or the reaction rate becomes 50% of the initial rate.

Step-by-step solution

Determine the order of the reactant given the change in the reaction rate and change in concentration

To do this, we can use the formula for the rate law, which is given by: \[R = k[A]^n\] where: - R is the reaction rate - k is the rate constant - [A] is the concentration of the reactant - n is the order of the reactant We are given that the initial rate doubles when the concentration of the reactant quadruples. We can write this as: \[2R_1 = R_2\] and \[4[A_1] = [A_2]\] Plugging these into the rate law equation for the two rates, we have: \[R_1 = k[A_1]^n\] and \[R_2 = k[A_2]^n\] We can now use these equations to determine the order of the reactant (n).

Calculate the order of the reactant

To determine the order of the reactant (n), we can divide the second rate law equation by the first rate law equation: \[\frac{R_2}{R_1} = \frac{k[A_2]^n}{k[A_1]^n}\] We know that \(R_2 = 2R_1\) and \([A_2] = 4[A_1]\), so we can substitute these values into the equation: \[\frac{2R_1}{R_1} = \frac{k(4[A_1])^n}{k[A_1]^n}\] Simplify the equation: \[2 = \frac{(4)^n}{1}\] Now, we just need to solve for n: \(2 = 4^n\) Taking the logarithm of both sides: \(\log 2 = \log 4^n\) Using the power rule for logarithms: \(\log 2 = n \log 4\) Finally, solve for n: \(n = \frac{\log 2}{\log 4} = 1/2\) So, the order of the reactant is \(1/2\).

Determine the change in reaction rate for a reactant with a -1 order when its concentration doubles

Now, let's consider a reactant with a -1 order. We can use the rate law equation to find the change in the reaction rate when the concentration of the reactant increases by a factor of two: \[R’ = k([A]’)^{-1}\] Since the concentration of the reactant doubles, \([A]’ = 2[A]\). We can substitute this into the rate law equation: \[R’ = k(2[A])^{-1}\] \[R’ = k\frac{1}{2[A]}\] Now, we can divide this equation by the original rate law equation: \[\frac{R’}{R} = \frac{k\frac{1}{2[A]}}{k[A]^{-1}}\] Simplify the equation: \[\frac{R’}{R} = \frac{1}{2}\] This shows that when the concentration of the reactant with a -1 order doubles, the initial reaction rate decreases by a factor of 1/2, or the reaction rate becomes 50% of the initial rate.