Concentration and Half Life of reactants A and B

Question

$3.12$ A reaction $\ce{A + B ->C}$, follows first order kinetics with respect to each reactant, with an overall rate constant $k=\pu{2.0 x 10^{-3} M-1s-1}$, the initial concentrations of A and B are $\pu{0.10 M}$ and $\pu{6.93 M}$. respectively. Given $\ln 2 = 0.693$, choose the correct statement(s).

(A) The concentration of A remaining after $100$ seconds is approximately $\pu{2.5 x 10^{-2} M}$.
(B) The concentration of B remaining after $100$ seconds is approximately $\pu{1.73 M}$.
(C) The half life for A is about $50$ seconds
(D) The half life for B is about $50$ seconds 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?

Answer 1

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 $b$.