# Determine reaction order and reaction rate

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?

• 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. – Maurice Apr 17 at 16:40
• 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 ! – Maurice Apr 17 at 16:45
• 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. – C_Lycoris Apr 17 at 16:45
• 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 – confused Apr 17 at 17:26
• @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. – Maurice Apr 17 at 19:09

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.
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 $$k=k^{'}[\ce{H2O}]_o$$