04-12-2018, 03:14 PM

When we have an equation that is 2A-->B and we are told that it is a first order reaction, would the integrated rate law look like this: ln[A]t/[A]o = -2kt?

And the derivation looking like this:

Reaction rate = -Δ[A]/2t

-Δ[A]/2t = k[A]

-Δ[A]/t =2k[A]

therefore ln[A]t/[A]o = -2kt

So in an exam question where in the first part, we are first required to calculate the k using a table with various concentrations of the reactant and the corresponding reaction rates. And in the second part, we are required to use the calculated k to solve for the concentrations after a stated time, t we would have to multiply the k by 2 right?

And the derivation looking like this:

Reaction rate = -Δ[A]/2t

-Δ[A]/2t = k[A]

-Δ[A]/t =2k[A]

therefore ln[A]t/[A]o = -2kt

So in an exam question where in the first part, we are first required to calculate the k using a table with various concentrations of the reactant and the corresponding reaction rates. And in the second part, we are required to use the calculated k to solve for the concentrations after a stated time, t we would have to multiply the k by 2 right?