[plu]

Hello,

I have been given a set of data and have been told to determine the overall order of the examined reaction without the use of graphs.  The reaction is the hydrolysis of 0.2 M potassium chlorosuccinicate with 0.2 M potassium hydroxide at 25^oC.  The data is as follows:

time (s)  [OH^-] (mol L^-1)
10          0.085
20          0.074
30          0.065
45          0.056
60          0.049
80          0.042
100        0.036

Instinctively, the reaction should be first order in both hydroxide and chlorosuccinicate, second order overall.  However, none of the graphs of [OH^-] vs. t, ln[OH^-] vs. t, or 1/[OH^-] vs. t give linear plots!  Why is this and would there be a simple way that I can solve this problem without the use of graphs?

[FeLiXe]

what you'd normally do is integrate the law

then you can insert two values to calculate k

insert a third value into the formula and see if it works

[plu]

Finding k would be the second part of the problem.  However, I still need to find the order of the reaction before I cna integrate anything.  Any suggestions on how I would do that?

[FeLiXe]

just assume the order to be one, then integrate and then try if k is almost the same in every data-point

if it isn't do the same thing with the order two
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if you want to do something fancy: compare the half-life and the time it takes for another quarter of your educts to be gone and compare those to calculated results.

They would be the same with order 1. With a higher order the second part what take longer.
--
of course you can also look at the differences and divide them by the time. It should be about the same result as the differential quotient.
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if you don't like that either look at the equation:
dx/dt = -k * x^n (where x is the concentration of both educts since you started with the same amount)
do a logarithm:
ln(-dx/dt) = ln(k) + n ln (x)
insert the quotient of the differences for dx/dt, take a value in between for x

then plot the graph or calculate the regression line:
the slope of the straight line will be n and the axis intercept ln(k)

--

however you want to do it
either way works. the problem is accuracy