Slope of a Graph help?

Question

So for this lab I am supposed to find the slope of the graph... 

Make a graph of the concentration versus the reaction rate. Draw a best-fit line and calculate the slope of the line.

Based on this data

Concentration
Reaction Rate
0.01 mol/L   1/ 5.4 sec = 0.18518518518 sec = 0.19 sec
0.009 mol/L  1/ 5.9 sec = 0.16949152542 sec = 0.17 sec
0.008 mol/L  1/ 6.8 sec = 0.14705882352 sec = 0.15 sec
0.007 mol/L   1/8.3 sec = 0.12048192771 sec = 0.12 sec 
0.006 mol/L  1/10.6 sec = 0.09433962264 sec = 0.0943 sec 
0.005 mol/L  1/14.0 sec = 0.07142857142 sec = 0.0714 sec 

I was told to use reaction rate = 1/ reaction time (inverse) and I got  a slope of 0.048444 mol/L/ sec  which I rounded off to 0,05 mol/L / sec

pisgahchemist · Accepted Answer

Your data, as entered into Excel
time........1/time .......... conc
5.4..... 0.185185185..... 0.01
5.9..... 0.169491525..... 0.009
6.8..... 0.147058824..... 0.008
8.3..... 0.120481928..... 0.007
10.6... 0.094339623..... 0.006
14...... 0.071428571..... 0.005

Plotting concentration vs time will give you a graph which has this shape
|*
|..*
|.....*
|............*
|...........................*
|________________________
............time(sec)

A plot of concentration vs 1/time looks like this
|..............................*
|.........................*
|...................*
|............*
|......*
|________________________
........... 1 / time(s^-1)

For your data, the slope is 0.0424 mol s / L, as computed with Excel.

We use the graphs of time and concentration graphs to determine the order of a reaction. 
 
If the graph of concentration vs time is linear, then the reaction is zero order in that reactant.

If the graph of ln(concentration) vs time is linear, then the reaction is first order in that reactant.

If the graph of 1/concentration vs time is linear, then the reaction is second order in that reactant.

You appear to be dealing with a second order reaction.

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Slope of a Graph help?

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