I have that $100$ mM of K$_3$Fe(CN)$_6$ is dissolved in equimolar of the organic ion (which I assume is HCN). In the solution Fe(CN)$_6$$^{3-}$ reacts to form Fe(CN)$_6$$^{4-}$.

The formation of Fe(CN)$_6$$^{4-}$ as a function of time is measured in a Na$_2$SO$_4$ (which does not take part in the reaction) solution of different concentrations.

One of the concentrations of Na$_2$SO$_4$ = $0$ and I want to determine the rate constant from this and:

$$\begin{array}{c|c} t/\pu{min} & \ce{[Fe(CN)_{6}^4-]}/\pu{mM} \\ \hline 10\ & 20.3\\ 20 & 33.7\\ 30\ & 43.3\\ 40\ & 50.4 \\ 50\ & 56\\ 60\ & 60.4 \\ \hline \end{array}$$

I assumed that the reaction would be second order since the reactants are equimolar and so I plotted using the equation:

$$ \frac{1}{[A]}= kt + \frac{1}{[A_0]}$$

When I plotted using this I get that the linear equation is: $−5.78×10^{−4}*x + 0.0463$

and R$^2$ = $0.771$.

So I tried using a first order equation and got that the linear equation is: $0.0204*x + 3.01$

and R$^2$ = $0.878$.

Doesn't this mean that my reaction should actually be first order? I don't see how it could be that and would appreciate it if someone could tell me if it is first or second or what I am doing wrong. Thank you!


I did plot the curves and now when I look at them I do think that the first order line is a bit curved while the second order is more straight. It is really hard to see and my professor usually looks at the R-value which is why I thought that would help me determine the order here. But does that mean the reaction is second order?

Here are my graphs:

enter image description here enter image description here

  • $\begingroup$ It is always a good idea to plot data before any curve fitting is done. Perhaps plot your data and see what it may reveal. $\endgroup$
    – Ed V
    Apr 30, 2021 at 13:12
  • $\begingroup$ You could also try doing a similar analysis to check if your points correspond to a first order reaction. Also, what makes you think that your organic ion is HCN? Thinking about the process would get you closer to understanding what's happening here. $\endgroup$
    – user23638
    Apr 30, 2021 at 13:50
  • $\begingroup$ Using $R^2$ is a very, very poor discriminator of the goodness of a fit, basically never use it. Look at fit and see it is curved vs the data, then plot residuals, (calc data-real data)/real data for example and see if they are random, look at size of error in gradient and error in intercept. $\endgroup$
    – porphyrin
    Apr 30, 2021 at 18:25
  • $\begingroup$ @EdV I have added my plots in my question now but I am still a bit unsure of the order.. $\endgroup$
    – katara
    May 3, 2021 at 9:25
  • 1
    $\begingroup$ I'd be really surprised if either HCl or HCN could be called an organic ion, and I hope you agree with me on this one. The wording of the exercise is a bit weird for sure, but to me it looks like you might want to work assuming pseudo first order conditions. $\endgroup$
    – user23638
    May 3, 2021 at 9:34

1 Answer 1


Use the 2nd order scheme $2A \to B$ with initial amount of $A=A_0$ so $B=A_0-A$ and B is the product you observe. The rate of change is $\displaystyle \frac{1}{2}\frac{dA}{dt}=-k_2A^2$ and I'm not going to use [ ] for concentrations just $A,\,B,\,A_0$. Also the 2 in the equation is incorporated into the rate constant just for convenience.

Solving the equation gives $\displaystyle A=\frac{A_0}{A_0k_2t+1}$ and so the product follows $\displaystyle B=\frac{A_0^2k_2t}{A_0k_2t+1}$. This is shown as the red line in the figure below. I just guessed the rate constant a few time to fit and as you can see the fit is good. You can also linearise this equation by isolating $A_0k_2t$.

The first order plot is also shown (grey line) and this is poor. $B=A_0(1-e^{-kt}$)

2nd order plot


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