Calculate rate constant of second order reaction and pseudo first order reaction


> 
> The reaction A + B -> Prod follows the second order kinetics. In an experiment, you initially have ten times as much B
> as A. The concentration of A varies with time according to the following data:
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> [![enter image description here](https://i.stack.imgur.com/dZIVe.png)](https://i.stack.imgur.com/dZIVe.png)
> 
> 
> How big will the error in the determined rate constant be if you choose to analyze the reaction as a pseudo-1st?
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> 


I began by plotting the first order reaction for this problem as ln[A] against t. From this I got that the slope = -$9.06 \* 10^{-4}$ s$^{-1}$ and so the rate constant is k$\_1$ = $9.06 \* 10^{-4}$ s$^{-1}$.


However, I am having trouble understanding how to plot for the second order reaction. I know that I should plot ln(a-x)/(b-x) but I don't understand how this is gotten from the given data.


How can I determine (a-x) and (b-x)?


I thought that x = [A]$\_0$ - [A]$\_t$ = $0.01$ - $1.11\*10^{-6}$ = $0.0099$. But then do I subtract that from all the given data points? I did this and it gives me negative values and then I can't do the natural log in order to plot the points. Also if x = $0.0099$ and b = $10$\*[A)$\_0$ = $0.1$ then I get that (b-x) = $0.0901$ for all the points?




The general rate law for a second order kinetics if $\pu{A\_o < B\_o}$, is $$\pu{\frac{1}{B\_o - A\_o} \ln\frac{A\_o B}{A B\_o} = kt}$$ If you only want to check that it is second oder, you can forget about the constants $\pu{A\_o}$ and $\pu{B\_o}$, and plot the logarithm of $\pu{\frac{B}{A}}$ versus the time $t$. The points should be well alined if it is second order.


Ref. Arthur A. Frost, Ralph G. Pearson, Kinetics and Mechanism, J. Wiley, NY, 1965, p. 16.


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