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?

  • $\begingroup$ @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. $\endgroup$ – Karsten Theis Feb 10 '20 at 21:49
  • $\begingroup$ @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? $\endgroup$ – Martin - マーチン Feb 11 '20 at 1:25

You are doing it wrong in situation when concentration of B alone was doubled but half life did not change.

In this situation, the reaction order should be 1 according to B as Half life in first order reaction doesn't depend upon initial concentration of reactants.

$$t_{1/2} = \ln2 / k $$

So the final order will be $2$ and the answer of the question will be b.

  • $\begingroup$ @orthocresol But if it is first order in B and first order in A, the integrated rate law (starting with equal concentrations of A and B) would not be an exponential function with a half-life. $\endgroup$ – Karsten Theis Feb 10 '20 at 21:51
  • 1
    $\begingroup$ @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...) $\endgroup$ – orthocresol Feb 10 '20 at 22:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.