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