# Second order reaction with different reactants

In a laser photolysis reaction $$\ce{C6H5NH2}$$ $$(C_0 = \pu{60 µM})$$ is oxidized to its radical cation. The disappearance of the cation can follow three pathways: (1) Dimerzing of the cation, (2) Reaction between the cation of the initial substance, (3) Reaction between the cation and the solvent. Decide which pathway is most likely to occur.

$$\begin{array}{l|cccccccc} \hline t/\pu{µs}~(\text{efter puls}) & 0 & 5 & 10 & 15 & 20 & 25 & 30 & 35 \\ \hline [\text{Radikalkatjon}]/\pu{µM} & 20 & 11.12 & 6.86 & 4.47 & 3.00 & 2.06 & 1.43 & 1 \\ \hline \end{array}$$

I first checked for the dimerizing of the cation by assuming it to be a second order reaction with the same reactants and plotting $$1/[\mathrm{cation}]$$ against $$t,$$ which didn't give a good fit. So I wanted to plot it as a second order reaction with different reactants $$(\ce{A + B -> products})$$ but I don't understand how that is done.

I should plot $$\displaystyle\ln\frac{a-x}{b-x}$$ against $$t,$$ but what exactly is $$(a-x)$$ and $$(b-x)?$$

I know that I should set up a reaction like:

$$\begin{array}{l|ccc} \hline & \text{A} & \text{B} & \text{product} \\ \hline t=0 & a & b & 0 \\ t=t & a-x & b-x & x \\ \hline \end{array}$$

And I would assume that my $$a = \pu{20 µM}$$ and $$b = \pu{60 µM}.$$ But where do I go from there? All help is appreciated!

The general solution, if the reaction is $$aA+bB \to product$$, is to allow an amount $$x$$ to react at time $$t$$ then there will be concentrations $$(A_0-ax)$$ and $$(B_0-bx)$$ of species A and B respectively at time $$t$$ if $$A_0,B_0$$ are the concentrations at time zero. The rate equation is $$\displaystyle -\frac{1}{a}\frac{d(A_0-ax)}{dt}=\frac{dx}{dt}=k(A_0-ax)(B_0-bx)$$ which can be integrated ($$0\to x, 0\to t)$$ to give

$$\ln\left(\frac{A_0-ax}{B_0-bx} \right)=k(A_0b-B_0a)t +\ln\left(\frac{A_0}{B_0} \right)$$

If the reaction follows this scheme plot the left hand side vs $$t$$ and this should produce a straight line of slope $$k(A_0b-B_0a)$$.

Should it happen that concentrations are initially $$A_0/B_0 = a/b$$ so that species are present in stoichiometric amounts then as $$A_0b-B_0a=0$$ the solution above cannot be used. In this case the rate equation becomes $$\displaystyle \frac{dA}{dt}=-bkA^2$$ which gives $$\displaystyle \frac{1}{A}-\frac{1}{A_0}=bkt$$.

Before jump into rate equation, I'd like to approach this in different direction.

A laser photolysis reaction of aniline, $$\ce{C6H5-NH2}$$ ($$\ce{A}$$), has created an $$\ce{A^{.+}}$$ radical cation, which has decayed with time to (unknown) products. No other additional data were given except for aniline concentration $$(\ce{[A]} = \pu{60 \mu M})$$. When plotted experimental values of $$\ce{[A^{.+}]}$$ against time $$t$$, you obtained a hyperbola ($$xy = a$$ type). Thus, I plotted $$\ln \ce{[A^{.+}]}$$ against time $$t$$ to see what happens. I got a very straightline with $$R^2 = 0.993$$ agreement:

Thus, you can conclude that the decay process follows first order kinetics:

$$\text{disappearance rate of } \ce{[A^{.+}]} = k\ce{[A^{.+}]}$$

That means:

$$-\frac{d\ce{[A^{.+}]}}{dt} = k\ce{[A^{.+}]}$$ $$\frac{d\ce{[A^{.+}]}}{\ce{[A^{.+}]}} = -kdt$$

Integrate both sides:

$$\int^\ce{A}_\ce{A_\circ}\frac{d\ce{[A^{.+}]}}{\ce{[A^{.+}]}} = -\int^t_0kdt$$ $$[\ln\ce{[A^{.+}]}]^\ce{A}_\ce{A_\circ} = -k[t]^t_0$$ $$\ln\ce{[A^{.+}]} = -kt + \ce{[A_\circ]} \tag1$$

The equation $$(1)$$ is a straight line. The equation of best trend line of the plot is:

$$y = -0.0839 x + 2.8439 \tag2$$

Hence, $$k = \pu{0.0839 \mu s-1}$$ and $$\ln\ce{[A_\circ]} = 2.8439$$ or $$\ce{[A_\circ]} = e^{2.8439} = \pu{17.18 \mu M}$$ (actual $$\ce{[A_\circ]} = \pu{20 \mu M}$$).

Based on this fact you may able to determine which path it takes to decay. Keep in mind that decay process only depends on the concentration of radical cation concentration.

• @ Mathew Mahindaratne. Well. The alignment of the points ln A vs. t is not ideal. The points at the beginning (t = 0) and at the end (t = 35) are above the line. The points in the middle of the graph are under the line. It looks as if the order is higher than 1, but smaller than 2, as I state in my previous comment. It may be due to two simultaneous reactions, one 1st order and another one 2nd order. – Maurice May 25 '20 at 20:37

A kinetics may be proven to be second-order when the initial concentrations of A and B are different, by using Frost and Pearson's treatment. In page $$16$$ of their manual, they say that the kinetics law is the following : $$\ce{\frac{1}{B_o - A_o}ln\frac{A_oB}{B_oA} = kt}$$ In order to simply prove that the reaction is second order, they recommend to simply plot log(B/A) against t.

Ref.: Arthur A. Frost, Ralph G. Pearson, Kinetics and Mechanism, A Study of Homogeneous Chemical Reactions, John Wiley & Sons, Inc. New York, 1965

The trouble with the numerical values is that the only known values are those of the radical cation $$\ce{A}$$. The corresponding values of $$\ce{B (= C_6H_6NH_2)}$$ are not known. They are $$60$$ µM at the beginning. But they remain unknown at t > $$0$$. I have made different hypothesis about these variations, with $$\ce{\Delta A}$$ = a $$\ce{\Delta B}$$, with $$\ce{a}$$ being $$0, 1$$ or $$2$$. The obtained values are never alined, as if the reaction is not of $$2$$nd order.

But the reaction should be "about" $$2$$nd order, since the half-life is growing. Interpolation of the given data shows that it is first $$6$$ µs, then $$7$$µs., then $$9$$ µs, then $$11$$ µs. This is typical for an order greater than $$1$$, but smaller than to $$2$$.