# Determine reaction rate from reaction time?

These are the results of my experiment (solution A and B were reacted and the reaction time was measured).

Data Table Edit: How would I determine the rate of reaction (so I can find the rate law) from the reaction time?

Solution A: $\pu{4.3 g}$ of potassium iodate per $\pu{1 L}$ of water

Solution B: $\pu{2 g}$ of sodium bisulphite, $\rm5~mL~1.0~M~\ce{H2SO4}$, 4 g soluble starch per $\pu{1 L}$ of water

I found the concentrations of $\ce{[KIO3]}$ by dividing mass of $\ce{KIO3}$ by molar mass of $\ce{KIO3} (\rm214.001\frac{~g}{mol})$ and then dividing that by volume of solution A. I also divided the mass of $\ce{NaHSO3}$ by molar mass of $\ce{NaHSO3} (\rm104.061\frac{g}{mol}$) and then dividing that by volume of solution B to find the concentrations of $\ce{[NaHSO3]}$.

I then tried to find the reaction rate by $$\frac{1}{3}\times\frac{\ce{[NaHSO3]}}{t}$$ and $$\frac{\ce{[KIO3]}}{t}$$ as the reaction is $$\ce{IO3- + 3 HSO3- -> I- + 3 HSO4-}$$

Not sure if I'm on the right track or not.

Source: Reaction Rate was found by $[\ce{NaHSO3}]$/Reaction Time and $[\ce{KIO3}]$/Reaction Time.

Rate order was found by log(Reaction Rate)/log([$\ce{NaHSO3}$]) and log(Reaction Rate)/log([$\ce{KIO3}$]).

Would appreciate it if anyone could tell me if I did this correctly.

• This is an example of an iodine clock reaction. If you really want to solve it, here is a hint: 1. Start with your stock solutions A and B and calculate the molar concentrations. 2. Calculate the initial concentrations of iodate and bisulfite in each of your experiments. 3. Group your data. 4. Plot your data. Mar 14, 2016 at 15:22
• Sorry for the late reply. I didn't have access to Internet. The reaction is IO3− + 3 HSO3− → I− + 3 HSO4−, so I tried to find the rate of reaction by [NaHSO3]/t*1/3 and [KIO3]/t. Mar 15, 2016 at 7:05

The reactant orders in iodine clock reactions generally turn out to be first order. Your calculated orders are suggestive of first order dependences, but there is a systematic error evident in the data that possibly stems from the experimental design. In addition, I don't believe that the manner in which you calculated the orders is correct. The approach in this experiment is to use the method of initial rates. Briefly, a differential rate law (which gives instantaneous rate as a function of concentrations) can be written as rate $=k[\ce{A}]^n [\ce{B}]^m$ for a reaction $\ce{A + B -> products}$. To find the order of reactant A, for example, one varies the initial concentration of A and keeps the concentration of B constant. This generates a pseudo-rate law in which the concentration for B is treated as a constant(k'): rate = $kk'[\ce{A}]^n$. This relation can be linearized as follows:
\begin{align} \log \text{rate} &= \log kk'[\ce{A}]^n\\ \log \text{rate} &= \log [\ce{A}]^n + \log kk'\\ \log \text{rate} &= n \log[\ce{A}] + \log kk' \tag{I} \end{align}
Trials are conducted in which [A] is varied and the initial rate is determined. A plot of log rate vs. log[A] should be linear and its slope should equal n, the order of reactant A. If, on the other hand, we solve EQ I for n we get \begin{align} n &= \frac{\log \text{rate} - \log kk'}{\log[\ce{A}]} \\ n &= \frac{\log \frac{\text{rate}}{kk'}}{\log[\ce{A}]} \end{align}
Thus calculating n as $\log \text{rate}/\log[\ce{A}]$ (your approach) is not fully correct because the pseudo-rate constant $kk'$ is not accounted for. I plotted your $[\ce{HSO3-}]$ data in the form of $\log \text{rate}$ vs. $\log [\ce{HSO3-}]$ and obtained a value for n (the slope) of 0.84 with $r = 0.997$ (note: the rates must be treated as negative values because reactant is being consumed). I advise you to attempt such a plot for the data where $\ce{[IO3-]}$ was varied. Regarding the systematic error, you should think about the fact that you determined average rates experimentally, meaning reactant concentration was actually decreasing during the timed intervals and, therefore, the rates you calculated reflected an average reactant concentration, not an initial concentration.