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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?

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    $\begingroup$ Don't include tuhe water in your calculations of concentrations. The water is $\pu{ 55.555 mol/L}$ at the beginning. It should not be really different at the end of the reaction. And even it is is $\pu{55.554 mol/L}$ at the end, nobody will be able to check that this a different amount from the initial values. $\endgroup$
    – Maurice
    Apr 17, 2021 at 16:40
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    $\begingroup$ Drawing a table like you have done is no help for answering the question. You better report the logarithm of the concentration $[\ce{Fe^{3+}}]$ vs. time. If the points are alined, the reaction is first order. And the slope of the line is the rate constant of the reaction. Go ! $\endgroup$
    – Maurice
    Apr 17, 2021 at 16:45
  • $\begingroup$ If you have access to a calculator or a graphing application it will be an easier task. I don't have the ability to calculate and write a long answer at this point in time, but whenever there is a hydrolysis reaction, it is a pseudo-first order reaction. In other words, as @Maurice pointed out, don't include water concentration in the rate order and equation. $\endgroup$
    – C_Lycoris
    Apr 17, 2021 at 16:45
  • $\begingroup$ I have tried solving it as a first order reaction now, but I am still unsure if the way I did it is correct. I would truly appreciate if someone could just check if my method is correct, thank you! @C_Lycoris $\endgroup$
    – katara
    Apr 17, 2021 at 17:26
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    $\begingroup$ @confused. Calculating log [$\pu{a/(a-x)}$] is a non sense. No. Try first to calculate the natural log of the concentration, on the $Oy$ axis. The first point is a $\pu{t = 10 s}$, and $\pu{y = ln 238 = 5.47}$ . The second point is at $\pu{t = 20 s}$ and $\pu{y = ln 227 = 5.42}$. Do the same for all points. You will see that the points are well alined, except the value for $80$ s. which is a bit too low. But with all other points you will see that the alinement is good. The slope can be calculated with the first and the last point. It is $ 4.89 · 10^{-3}$ par second. $\endgroup$
    – Maurice
    Apr 17, 2021 at 19:09

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The rate equation is $$\ln[\ce{Fe^{3+}}] = \ln[\ce{Fe^{3+}}]_0 − kt$$ So, $a-x=[A]$. Don't subtract the next value. Put it as it is. That's just for your understanding, however. Don't try to use the integrated rate form to plot the graph. $[\ce{Fe^{3+}}]=238 \text{ where } t=10 \text{ and } [\ce{Fe^{3+}}]=227 \text{ where } t=20$. Using the above mentioned equation: $$\ln[A] = \ln[A]_0 − kt$$ Calculate the value of $\mathrm{\ln[\ce{Fe^{3+}}]_0}$ by eliminating k. Thereafter, you can graph the above mentioned equation $\ln[\ce{Fe^{3+}}] \text{ vs time t}$ with $\ln[\ce{Fe^{3+}}]_0$ as the y-intercept. The slope will yield the value of the rate constant.

[...]that would be the rate of the reaction?

For some reason unbeknownst to me, a lot of people tend to confuse between rate and rate constant. The slope will yield the rate constant. Next, $$\textrm{rate}=k[\textrm A]^{a} \quad \text{}$$ where $a$ is the order of the reaction with respect to reactant $\ce{A}$.

As a side-note, since this is a pseudo-first order reaction, the rate law equation, initially this: $$\text{Rate} = k^{'}[\ce{Fe^{3+}}][\ce{H2O}] $$ now appears as this:$$\text{Rate}=k[\ce{Fe^{3+}}]$$ the variable $'k'$ in the second equation is the slope that you obtained from the graph. Here $k=k^{'}[\ce{H2O}]_o$

For more about pseudo-first order reactions you can check one of my previous answers.

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