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Recently I've been going through an exercise of Chemical Kinetics and Nuclear Chemistry and right now I am stuck on one question that was previously asked in JEE(Advanced)-2013. The question says,

In the reaction,

$\ce{P + Q -> R + S}$

The time taken for $75\%$ reaction of $\ce{P}$ is twice that time taken for $50\%$ reaction of $\ce{P}$. The concentration of $\ce{Q}$ varies with reaction time as shown in the figure. What is the overall order of the reaction?

enter image description here

And here was my approach to the question,

In the question different data have been given for different reactants (i.e. $\ce{P}$ and $\ce{Q}$). For $\ce{P}$ it has been given that it take $\ce{P}$ double the time to react $75\%$ than that from $50\%$. So assuming the time taken for $50\%$ reaction to be $t$ and that of $75\%$ to be $2t$, we can easily find that order of reaction w.r.t P is $1$. As,

$$\frac{t_{\frac12}}{t_{\frac34}}=\frac{\log 2}{\log 4}=\frac12$$

Further w.r.t $Q$ there is a graph given which is something like in form $\ce{[Q]=[Q_o]-mt}$. So comparing with the equation of zero order reaction which is like $\ce{[A]=[A_o]-kt}$ (where $\ce{[A]}$ is the concentration of reactant and $k$ is rate constant), we can conclude that order of reaction w.r.t to $Q$ is $0$.

So I concluded that overall order of reaction will be

order of reaction of $P +$ order of reaction of $Q$

$\implies 1+0=1$

So was this the right approach from my side or there is any kind of error in any step?

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    $\begingroup$ The question seems self-contradictory to me. If P and Q are actually reacting stoichiometrically with each other, and [Q] decreases linearly with time, then [P] must decrease linearly with time, which contradicts the "time taken for 75% reaction of P is twice that time taken for 50% reaction of P" aspect of the question. $\endgroup$
    – DavePhD
    Jun 18, 2014 at 18:33
  • $\begingroup$ @DavePhD it's not necessary that the order of reaction w.r.t each reactant will be equal to it's stoichiometry. Check this chemguide.co.uk/physical/basicrates/orders.html $\endgroup$
    – Saharsh
    Jun 19, 2014 at 4:26
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    $\begingroup$ I agree that the order of an overall reaction with respect to a reactant does not have to be the reactant's stoichiometric coefficient. What I am saying is if P is actually reacting with Q, then the observed d[Q]/dt will be c(d[P]/dt) where c is a constant. In this question, after a certain time [Q]=0. For [P] to decrease exponentially, and therefore continue reacting for an infinite amount of time, P and Q are necessarily not reacting with each other. $\endgroup$
    – DavePhD
    Jun 19, 2014 at 18:15
  • $\begingroup$ @DavePhD: What are your thoughts on this reaction: $\ce{ArN2+ + X- -> ArX + N2}$ ? This reaction is first order with respect to $\ce{ArN2+}$ and zero order with respect to $\ce{X-}$ Source: en.wikipedia.org/wiki/Order_of_reaction#First_order $\endgroup$
    – Archer
    Mar 12, 2018 at 17:25
  • $\begingroup$ @Abcd If there is no X-, the reaction will not occur. It will appear to be first order, until [X-] is very low. Then the second order nature will be observed. It is pseudo first order. cbc.chem.arizona.edu/~salzmanr/480a/480ants/pfo3oarr/… $\endgroup$
    – DavePhD
    Mar 12, 2018 at 17:44

1 Answer 1

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As summarized in the IUPAC definition of reaction rate, the rate of change of one reactant must be directly proportional to the rate of change of each other reactant.

If the concentration of one reactant decreases linearly, the concentration of every reactant must decrease linearly (the reaction rate is a constant, and the overall reaction order is zero).

The text of the JEE question contradicts its Figure (which shows linear decrease of [Q]), by saying that the "time taken for 75% reaction of P is twice that time taken for 50% reaction of".

Instead, since [Q] decreases linearly, [P] must decrease linearly, and if 75% of an initial [P] reacts it must take 1.5 times as long as for 50% to react.

It's likely that the author of the question intends that the answer is "first order", but really there is no correct answer.

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