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Since reaction is first order with respect to both reactant A and B, I can write,

$$ k=\frac{1}{t} \ln{\frac{a}{a-x}} $$ where a is initial concentration of reactant and a-x is remaining concentration.

On solving the above for A and B after 100 seconds, I get $A_r = \frac{0.1}{e^{0.2}} $ and $B_r = 0.693/e^{0.2}$

The answer does not match with any options for both concentration as well as half life.

Can you please help me with this?

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  • $\begingroup$ Don't you think a reaction that involves B should somehow depend on B? In your derivation for A, it doesn't. $\endgroup$ – Ivan Neretin Aug 21 '20 at 23:16
  • $\begingroup$ @IvanNeretin The equation is that of 1st order and depends only on 1 reactant. Even though the reaction involves A and B both, they both follow 1st order kinetics so the above equation will apply to each A and B seperately $\endgroup$ – Asad Ahmad Aug 21 '20 at 23:27
  • $\begingroup$ @IvanNeretin Rate of Reaction = k [A][B] that is true $\endgroup$ – Asad Ahmad Aug 21 '20 at 23:29
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    $\begingroup$ We have one reaction. It is first order in A, so according to you, it should depend only on A. Also, it is first order in B, so it should depend only on B. How do you reconcile that? $\endgroup$ – Ivan Neretin Aug 21 '20 at 23:29
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    $\begingroup$ have a look at pseudo first order reactions $\endgroup$ – porphyrin Aug 22 '20 at 11:42
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The whole reaction isn't first order and that can be arrived at using one simple concept. The unit of rate constant $k$ is $\pu{M^{-1} s^{-1}}$ and not $\mathrm s^{-1}$.

Since reaction is first order with respect to both reactant A and B, I can write, $$ k=\frac{1}{t} \ln{\frac{a}{a-x}} $$

This would have been correct if and only if the whole reaction was a first order reaction. Not a first order reaction with respect to each reactant individually. For example:

$$\ce{A->B}\quad \text{Rate}=k[A]$$

That is not the case here. Here the reaction would be better phrased as:

$$\ce{A + B->C}\quad \text{Rate} = k[A][B]$$

Which means that this is a second order reaction and not a first order reaction.

Now, the other issue is that this second order reaction isn't of the form $\ce{2A-> B}$ where the rate would have been relatively easy to find using a generic law and hence we move to a more basic proof for a reaction $\ce{A + B -> C}$. As given above, the rate law for this expression is "Rate $= k[A][B]$" Now we draw a RICE table for this generic reaction to get:

\begin{array}{|l|c|} \hline \mathrm{R} & \mathrm A & \mathrm B &\mathrm C \\ \hline \mathrm I & a & b & 0 \\ \mathrm C & -x & -x & x \\ \mathrm E & a-x & b-x & x \\ \hline \end{array}

Hence, we can see that for a time t, the rate of reaction can be defined as:

$$ \text{Rate} = -\frac{\mathrm d[A]}{\mathrm dt}=k(a-x)(b-x)$$

Differentiating $[A]$ with respect to $t$, we get:

$$-\frac{\mathrm d[A]}{\mathrm dt} = \frac{\mathrm dx}{\mathrm dt}$$

Substituting this in the previous equation and integrating with respect to x, we get:

$$kt = \frac{1}{b-a}\ln \frac{a(b-x)}{b(a-x)}$$

Now, in the given question, we are asked to find the value of $[A]$ and $[B]$ after $\pu{100 s}$. Therefore, $t=\pu{100 s}$, $a = \pu{0.1 M}\,,b=\pu{6.93 M}$ and $k = \pu{2 \times 10^-3 M^{-1} s^{-1}}$.

Solve for x in this equation and find the value of $a-x$ and $b-x$.

This isn't a complete solution, but the rest is just manipulating the expression using mathematics and has no actual relation to kinetics.

A hint to make calculations simpler would arise from noticing that $a<<b$ and so $b-a$, $b-x$ can both be approximated to $b$.

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  • $\begingroup$ How are you stating that the first differential of [A] with respect to time is the product of the concentrations of both the reactants after some arbitrary time period? $\endgroup$ – Aditya Roychowdhury Aug 22 '20 at 11:53
  • $\begingroup$ @AdityaRoychowdhury Isn't that what the rate law states? $-\frac{1}{a} \frac{\mathrm d[A]}{\mathrm dt} = Rate$ for a reaction $\ce{aA +bB -> cC + dD}$ here a=1. $\endgroup$ – Safdar Aug 22 '20 at 13:33
  • $\begingroup$ Yes, but am asking why are you multiplying (a-x)(b-x). How do you realise that that is the rate law. Shouldn't it be the product of the initial concetration. Even if you demonstrate it in the form of a rate law by using derivative. $\endgroup$ – Aditya Roychowdhury Aug 22 '20 at 13:59
  • $\begingroup$ @Aditya This would be the rate at some time t.. after x moles of A and B have reacted.. since rate here would be $k[A_t][B_t]$ $\endgroup$ – Safdar Aug 22 '20 at 14:03

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