# Determine the order of the reaction, and the value of the rate constant. Predict the concentration of AB at 25s

I've never posted a question about chemistry before so I'm not really sure how this community works but here is the question.

I do have the answer to this question but I have no idea on going about solving it.

I do know:

• that second order is equal to all of the exponents added up
• Rate is equal to $k[A]^{p1} [B]^{p2}$
• Rate is also equal to $([A_2]-[A_1])/([t_2]-[t_1])$

And I believe $M$ stands for molarity.

I have little to no idea how I am to solve this other than those 4 things.

Here are the answers the book gives me:

1. Second Order
2. $k = 2.25\times 10^{-2}$
3. $[AB]$ at $25s = 0.619 M$
• The rate of the reaction depends on the concentration of reactants, not products. – orthocresol Oct 9 '15 at 20:57
• but all they have given me are the products to work with. – Ryan Henry Oct 9 '15 at 21:01
• By the way, this previous question is the same as yours, just the numbers are changed. The answer should be applicable. And I don't quite get what you're talking about. AB is the reactant and the concentration of AB at various times is given to you in that table. – orthocresol Oct 9 '15 at 21:04

If n is the order of this reaction, then the rate equation is given by $$-\frac{d[AB]}{dt}=k[AB]^n$$ If we take the log of both sides of this equation, we obtain: $$\log\left(-\frac{d[AB]}{dt}\right)=n\log([AB])+\log k$$ This equation is of the mathematical form $y=mx+b$ (i.e., a straight line), where $y=\log\left(-\frac{d[AB]}{dt}\right)$, $x=\log([AB])$, m = n, and $b=\log k$. So, if we could plot a graph of $y=\log\left(-\frac{d[AB]}{dt}\right)$ vs $x=\log([AB])$, we should get a straight line with a slope of m and and intercept of logk. So we could determine both the reaction order and the rate constant all in one shot from this single graph.
But how would we reduce the given data to plot such a graph? Well, we could use finite difference approximations for the average reaction rate and the average concentration over each time interval. If $[AB]_t$ represents the molar concentration of reactant AB at time t, we would approximate the average reaction rate over each time interval between two successive values of time $t_1$ and $t_2$ in the table as: $$\left(-\frac{d[AB]}{dt}\right)_{ave}=\frac{[AB]_{t_1}-[AB]_{t_2}}{t_2-t_1}$$ The average concentration over this same time interval would be: $$[AB]_{ave}=\frac{[AB]_{t_1}+[AB]_{t_2}}{2}$$ So we would plot the log of the average reaction rate as a function of the log of the average reactant concentration for all the time intervals in the table. The result should be a straight line with a slope equal to the reaction order and an intercept equal to the log of the reaction rate constant.