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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?

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  • $\begingroup$ @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 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 at 1:25
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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.

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  • $\begingroup$ @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 at 21:51
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    $\begingroup$ @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 at 22:25

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