I have carried out an experiment on the $\ce{MnO2}$ catalyzed decomposition of hydrogen peroxide. I would have expected the reduction in concentration to be exponential but when I fit an exponential model to the data the residuals are not random. My data curve is too curved. So in other words, the decomposition is faster at the beginning of the experiment and slower at the end of the experiment than I would expect if the process was exponential. It's highly unlikely that it's measurement error - if it was the residuals would be closer to random.

Am I correct in believing the process should follow a negative exponential? I did wonder if the discrepancy might be due to catalyst contamination which would explain why the exponential model is less curved than the data.

EDIT: I have also fitted the data to a reciprocal to model a second order reaction and the reciprocal of concentration squared - just for fun. The residuals retain their smile under all models so I think that M.Farooq is correct and I didn't control the temperature as well as I should have. However, the second order fit is not as good as the exponential one on the whole (SE of residuals is 0.025 for the fist order model and 0.1673 for the second order but it's marginal )so I think it is first order. I've run out of manganese dioxide so I'll have to try it again when I get hold of some. This Paper indicates that the decomposition of hydrogen peroxide in the presence of a solid catalyst is indeed first order.

enter image description here

dat<-read.csv("H:\\code_library\\R\\joe_chem.csv", header=TRUE)
model<-lm(log(Conc) ~ reading_time, data = dat)
plot(dat, main="Conc vs Time",xlab="Time(s)",ylab="Concentration")
legend(90,1,legend="model:  y")
lines(dat$reading_time, y(dat$reading_time, coef(model)[2], coef(model)[1]),col="blue")
plot(dat$reading_time, residuals(model), main="Residuals",xlab="Time(s)", ylab="Residuals")

lm(formula = log(Conc) ~ reading_time, data = dat)

      Min        1Q    Median        3Q       Max 
-0.030231 -0.021338 -0.005329  0.014697  0.079083 

               Estimate Std. Error t value Pr(>|t|)    
(Intercept)  -1.871e+00  6.533e-03  -286.4   <2e-16 ***
reading_time -1.784e-03  2.048e-05   -87.1   <2e-16 ***
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.02477 on 54 degrees of freedom
Multiple R-squared:  0.9929,    Adjusted R-squared:  0.9928 
F-statistic:  7586 on 1 and 54 DF,  p-value: < 2.2e-16
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    $\begingroup$ Please consider checking out FAQ: How can I format math/chemistry expressions on Chemistry Stack Exchange?. (Apart from the use of the $\ce{...}$ macro, which is covered there, you shouldn't type a zero 0 when you mean an oxygen O.) And in general, we try not to use MathJax in titles, unless it's absolutely necessary. $\endgroup$ Commented Jul 9, 2021 at 14:23
  • $\begingroup$ Thanks for the pointer - I'll take a look. $\endgroup$ Commented Jul 9, 2021 at 14:24
  • $\begingroup$ If you posses access to the raw data, and because the dots in the plot of $c = f(t)$ don't overlap each other, may you replot the diagram with solid dots (e.g., black). This could ease accessing the data from the illustration, e.g., by WebPlotDigitizer. What program did you use to plot and fit the data? And while working on the figure, may you add the equation describing for your fit blue curve and some quality indicators (e.g., correlation $r^2$, residual errors)? $\endgroup$
    – Buttonwood
    Commented Jul 9, 2021 at 14:36
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    $\begingroup$ I can imagine several reasons for the curve not to follow a single exponential decay: The simplest one is that as H2O2 begins to decompose, it generates a lot of heat, which further accelerates the rate of reaction. If you did not control the temperature very strictly...then fitting the curve is of no use. $\endgroup$
    – ACR
    Commented Jul 9, 2021 at 14:42
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    $\begingroup$ Clearly the model you used is incorrect as the residuals are not evenly distributed. You should fit with $ae^{-kt}+b$ and vary $a,k,b$ to fit, in which case you would need to have some estimate of the background $b$. Only when the residuals have no systematic deviation can you sensibly apply statistical tests such as std dev of fit or $\chi^2$. Never, never, use $R^2$ it is far, far too insensitive for chemical data but is ok when there is just a general trend. $\endgroup$
    – porphyrin
    Commented Jul 9, 2021 at 15:02

1 Answer 1


Who says it's exponential? That the reaction is first-order?

If the rate is "too high" early in the reaction when concentration is high, but "too low" later on when concentration is low, that indicates you have a higher concentration dependence than the first-order you assumed when you expected an exponential curve.

If we were to assume second-order then we would have


Solving by standard methods leads to the result


So, if $1/c$ is a linear function of time (and it looks that way when I try to pick points off the graph) you have a second-order reaction rate law, not first-order.

PS -- next time show units on the graph. You will need them to extract the rate constant if, indeed, the reaction is other than first-order.

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    $\begingroup$ Again to the point to emphasize for the OP is that the temperature must be held constant in a kinetics experiment. "MerelyUseful" did you control the temperature? If not, it has to be repeated. $\endgroup$
    – ACR
    Commented Jul 9, 2021 at 16:27
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    $\begingroup$ I would still try the second -order model. I assume the concentration was low enough so that the water solvent sank the heat and kept temperature reasonably within range, although it would have been nice to show units onthe graph. $\endgroup$ Commented Jul 9, 2021 at 16:43
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    $\begingroup$ This is a typical mistake in student graphs. People forget to put the units because they are so comfortable with their own data, the units are "understood". $\endgroup$
    – ACR
    Commented Jul 9, 2021 at 16:47
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    $\begingroup$ To my shame I can't remember the units - I think mol/dm-3. When I've got a few spare minutes I'll try it using the second order model. $\endgroup$ Commented Jul 10, 2021 at 13:30
  • $\begingroup$ I am thinking that while not perfect, it will be much more accurate. Since multiple H2O2 molecules, or fragments thereof, would have to combine to get a single molecule of O2, a reaction order greater than 1 should come as no surprise. $\endgroup$ Commented Jul 10, 2021 at 13:41

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