[XxslbabesxX]

I have to find the integrated rate law and rate law. I am given the following info:

Time (s)     {C₆H₆] (mol/L)
195                  1.6x10^-2
604                  1.5x10^-2
1246                1.3x10^-2
2180                1.1x10^-2
6210                0.68x10^-2

How would I find out if this is a first order reaction or a second order reaction with the given information? I tried graphing the data on my calculator, but that didn't work out. Any help would be very greatly appreciated.

[Mitch]

Doesn't these rate law problems involve a chemical equation of some sort?

[XxslbabesxX]

Whoops, I'm sorry lol There was more info given:

2C₄H₆---->C₈H₁₂


Assume that Rate=Delta C₄H₆ over Delta T (time)

[mike]

Can you try plotting:

ln[] versus time   -----> first order will give a straight line graph with negative slope

and


1/[] versus time ------> second order will give a straight line graph with a positive slope.

[XxslbabesxX]

I guess I'm most likely doing something wrong, but when I plot it with ln[] versus time I get a straight line with a neg. slope and when I plot it with 1/[] versus time I get a straight line w/ a pos. slope.

According to the back of my textbook the naswer is a a second order reaction, but I just can't seem to find out why.

[mike]

I graphed the same data and although they both look straight the second order graph fits the data much better than the first. I did it in excel if that helps.

[XxslbabesxX]

Thanks....I plotted it again on my calculator, but both still look equally equal.

If anyone knows any other way to determine the order, I would love to know:)

[Donaldson Tan]

your calculator has a small display area. it won't be able to show how well the first order and 2nd order models fit the experimental data. do it on excel or other graphing software on the computer.