# How to find the units of the rate constant?

Consider the reaction $$\ce{2A + B -> Products},$$ when the concentration of $$\ce{B}$$ alone was doubled, the half-life did not change. When the concentration of $$\ce{A}$$ alone was doubled, the rate increased by two times. The units of the rate constant is:

a) $$\mathrm{s^{-1}}$$

b) $$\mathrm{L\ mol^{-1}\ s^{-1}}$$

c) Unitless

d) $$\mathrm{mol\ L^{-1}\ s^{-1}}$$

I have tried it solving through this method:

According to me, when the concentration of $$\ce{A}$$ alone was doubled, reaction rate also increased by two time, implies that reaction is first order according to $$\ce{A}$$. In same way, the reaction should be 0th order according to $$\ce{B}$$. So the net order would be 1. So this gives me answer $$\mathrm{s^{-1}}$$ but the answer is $$\mathrm{L\ mol^{-1}\ s^{-1}}$$. Where am I going wrong?

• @Martin If the reaction were zero order in B and first order in A, it would be first order overall and have a half-life. If the reaction is first order in B and first order in A, it would not have a half-life (or in other words, the half-life would be concentration dependent) unless it is pseudo-first order because A is much more concentrated than B. I don't think the question makes sense. – Karsten Theis Feb 10 at 21:49
• @Karsten I'm not sure how I can help you here. I've just added some markup and removed the homework tag. Have I introduced an error in the question? – Martin - マーチン Feb 11 at 1:25

$$t_{1/2} = \ln2 / k$$
So the final order will be $$2$$ and the answer of the question will be b.
• @KarstenTheis, thank you, I do see your point, but my only edit was to change $K$ to $k$. I was too lazy to check for factual correctness. Please feel free to edit as you wish... (or provide your own answer if you prefer...) – orthocresol Feb 10 at 22:25