In a solution of $\ce{Fe^3+}$ the concentration was measured to be: $$\begin{array}{c|c} t/\pu{min} & \ce{[Fe^3+]}/\pu{\mu M} \\ \hline 10\ & 238\\ 20 & 227\\ 40\ & 206\\ 60\ & 187 \\ 80\ & 169\\ 100\ & 154 \\ \hline \end{array}$$ The decrease in $\ce{Fe^3+}$ concentration is due to the hydrolysis of the ion and precipitation of hydroxide. Calculate the reaction order and reaction rate.
I am having trouble understanding how to tackle this problem. I began by find the reaction formula:
$$ \ce{Fe^{3+} + H_2O <=> Fe(OH)^{2+} + H^+} $$
So this is a second order reaction. I then did:
$$\begin{array}{c|c|c} \text{time} & \ce{[Fe^3+]} & \text{H2O} & \ce{[Fe(OH)^2+]} & \ce{[H+]}\\ \hline t=0 & a\ & a & - & -\\ t=t & a-x & b-x & x & x\\ t = \infty & - & b-a & a & a \\ \hline \end{array}$$
But I am unsure if I should include water and the hydrogen ion, and also how I should continue if this is correct. Can someone please push me in the right direction here. Thank you!
EDIT
First order equation:
$$ \ln \frac{a}{a-x} = kt$$
Is a then equal to $238$ and $a-x$ is the initial value ($238$) minus the next value? e.g., $238-227 = 11$.
If I plot:
$$\begin{array}{c|c|c} t & \ln \left (\frac{a}{a-x}\right) \\ \hline 10 & 0\\ 20 & 3.07\\ 40 & 2.43\\ 60 & 2.53\\ 80 & 2.58 \\ 100 & 2.76\\ \hline \end{array}$$
I get that the slope is $0.0168$ and that would be the rate of the reaction?