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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?
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 $b$.
