Rate of reaction, rate constants, and Arrhenius' Equation

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
1 reply · 10K views
physics120
Messages
34
Reaction score
0

Homework Statement



The effect of temperature on the rate of a reaction was studied and the following data obtained:

k (s-1) T (°C)
3.06×10-4 10
4.84×10-4 16
6.50×10-4 20
1.40×10-3 31
2.87×10-3 42
4.16×10-3 48
5.94×10-3 54
7.92×10-3 59


It is known that the variation of the rate constant k with the absolute temperature T is described by the Arrhenius equation:


k = A exp^[( -Ea )/(RT)]


where Ea is the activation energy, R is the universal gas constant and A is the pre-exponential factor (units of the rate constant). Taking the natural logarithm of both sides affords:


ln k = ln A - Ea/RT


a) For a plot of y = ln k versus x = 1/T, calculate the slope of the best straight line using linear regression.

b) Calculate the activation energy Ea.


Homework Equations



Relevant equations listed in part 1

The Attempt at a Solution



Do I need to plot this data? Is there any way to do this without using excel or a graphing calculator? If there isn't, I tried putting these in a spreadsheet, then plot ln k in the y-axis and 1/T on the x-axis and then use Excel's "trendline" to get the slope. (The slope is Ea/R. So Ea is R x slope for part 2). However, I got -38.619 for the slope.. although, I don't know if this is right or wrong as I seem to keep getting the units wrong, are the not the units for (ln k)/(1/T), which is s^(-1)/degC^(-1), or degC/s? (where degC = degrees celsius)

If there is another way to figure this out, and if you know why I am getting the wrong units, help would be greatly appreciated! Once I know the first part, part b is a cinch.

Thank-you!
 
Physics news on Phys.org
Strictly speaking, logarithms are unitless (formally, the rate constants are multiplied by a time unit, giving a unitless quantity for the logarithm/exponential to act on). So I wouldn’t worry too much about that. But other than that, yes, plot the data and take the slope to get the activation energy (pay attention to the negative sign).