First or second order reaction 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} &amp; \ce{[Fe(CN)_{6}^4-]}/\pu{mM} \\ \hline
10\ &amp; 20.3\\
20 &amp; 33.7\\ 
30\ &amp; 43.3\\ 
40\ &amp; 50.4 \\ 
50\ &amp; 56\\ 
60\ &amp; 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!
EDIT
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:
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. $B=A_0(1-e^{-kt}$)