Question

Consider the reaction of peroxydisulfate ion $\left(\mathrm{S}_{2} \mathrm{O}_{8}{ }^{2-}\right)$withiodideion(I$^{-}$ ) in aqueous solution: $${\mathrm{S}}_2{\mathrm{O}}_8{}^{2-}(aq)+3{\mathrm{I}}^{-}(aq)⟶2{\mathrm{SO}}_4{}^{2-}(aq)+{\mathrm{I}}_3^{-}(aq)$$ At a particular temperature the rate of disappearance of $\mathrm{S}_{2} \mathrm{O}_{8}{ }^{2-}$ varies with reactant concentrations in the following manner: $$\text{Unknown environment 'tabular'}$$ (a) Determine the rate law for the reaction. (b) What is the average value of the rate constant for the disappearance of $\mathrm{S}_{2} \mathrm{O}_{8}{ }^{2-}$ based on the four sets of data? (c) How is the rate of disappearance of ${\mathrm{S}}_2{\mathrm{O}}_8{}^{2-}$ related to the rate of disappearance of $I^{-}?(\mathrm{d})$ What is the rate of disappearance of I when $\left[{\mathrm{S}}_2{\mathrm{O}}_8{}^{2-}\right]=0.025\mathrm{M}$ and $\left[{\mathrm{I}}^{-}\right]=0.050\mathrm{M}?$

Short answer

The rate law for the reaction is $rate=k[{\mathrm{S}}_2{\mathrm{O}}_8{}^{2-}][{\mathrm{I}}^{-}{]}^2$ with an average rate constant of $k_{avg}=6.09\times {10}^3{M}^{-2}{s}^{-1}$. The rate of disappearance of the iodide ion is three times the rate of disappearance of the peroxydisulfate ion ($Rat{e}_{{\mathrm{I}}^{-}}=3\cdot Rat{e}_{{\mathrm{S}}_2{\mathrm{O}}_8{}^{2-}}$). For given initial concentrations ($[{\mathrm{S}}_2{\mathrm{O}}_8{}^{2-}]=0.025M$ and $[{\mathrm{I}}^{-}]=0.050M$), the rate of disappearance of iodide is $11.4\times {10}^{-6}M/s$.

Step-by-step solution

1. Determine the rate law equation

To determine the rate law equation, we need to find the relationship between the rate of the reaction (the rate of disappearance of peroxydisulfate) and the initial concentrations of the reactants. The general rate law equation is: $rate=k[{\mathrm{S}}_2{\mathrm{O}}_8{}^{2-}{]}^m[{\mathrm{I}}^{-}{]}^n$ where $k$ is the rate constant, $m$ and $n$ are the reaction orders with respect to the peroxydisulfate and iodide ions, respectively. To find the reaction orders (m and n), we can use the initial rate data from the table. Let's start with experiments 1 and 2 since they both have the same initial concentration of iodide ions: Experiments 1 and 2: ${\displaystyle \frac{rat{e}_2}{rat{e}_1}}=\frac{3.9\times {10}^{-6}}{2.6\times {10}^{-6}}=1.5$ ${\displaystyle \frac{[{\mathrm{S}}_2{\mathrm{O}}_8{}^{2-}{]}_2^m[{\mathrm{I}}^{-}{]}_2^n}{[{\mathrm{S}}_2{\mathrm{O}}_8{}^{2-}{]}_1^m[{\mathrm{I}}^{-}{]}_1^n}}$ Since the concentration of iodide ions is the same, it will cancel out: ${\displaystyle \frac{[{\mathrm{S}}_2{\mathrm{O}}_8{}^{2-}{]}_2^m}{[{\mathrm{S}}_2{\mathrm{O}}_8{}^{2-}{]}_1^m}}=1.5\Rightarrow {\displaystyle \frac{(0.027{)}^m}{(0.018{)}^m}}=1.5$ After isolating m and solving, we find that $m\approx 1$. Next, let's use the data from experiments 1 and 3 since they both have the same initial concentration of peroxydisulfate ions: Experiments 1 and 3: ${\displaystyle \frac{rat{e}_3}{rat{e}_1}}=\frac{7.8\times {10}^{-6}}{2.6\times {10}^{-6}}=3$ ${\displaystyle \frac{[{\mathrm{S}}_2{\mathrm{O}}_8{}^{2-}{]}_3^m[{\mathrm{I}}^{-}{]}_3^n}{[{\mathrm{S}}_2{\mathrm{O}}_8{}^{2-}{]}_1^m[{\mathrm{I}}^{-}{]}_1^n}}$ Since the concentration of peroxydisulfate ions is the same, it will cancel out: ${\displaystyle \frac{[{\mathrm{I}}^{-}{]}_3^n}{[{\mathrm{I}}^{-}{]}_1^n}}=3\Rightarrow {\displaystyle \frac{(0.054{)}^n}{(0.036{)}^n}}=3$ After isolating n and solving, we find that $n\approx 2$. Thus, the rate law equation is: $rate=k[{\mathrm{S}}_2{\mathrm{O}}_8{}^{2-}][{\mathrm{I}}^{-}{]}^2$

2. Calculate the average rate constant

Now, we can use any experiment's data along with the determined rate law to calculate the rate constant for that experiment. Then, we can do this for all experiments and calculate the average rate constant. For experiment 1: $rat{e}_1=k[{\mathrm{S}}_2{\mathrm{O}}_8{}_1^{2-}][{\mathrm{I}}_1^{-}{]}^2\Rightarrow k={\displaystyle \frac{rat{e}_1}{[{\mathrm{S}}_2{\mathrm{O}}_8{}_1^{2-}][{\mathrm{I}}_1^{-}{]}^2}}=7.14\times {10}^3{M}^{-2}{s}^{-1}$ Similarly, for experiments 2, 3, and 4: For experiment 2: $k=5.96\times {10}^3{M}^{-2}{s}^{-1}$ For experiment 3: $k=5.71\times {10}^3{M}^{-2}{s}^{-1}$ For experiment 4: $k=5.56\times {10}^3{M}^{-2}{s}^{-1}$ Now, we can take the average of these values: $k_{avg}={\displaystyle \frac{(7.14+5.96+5.71+5.56)\times {10}^3}{4}}{M}^{-2}{s}^{-1}=6.09\times {10}^3{M}^{-2}{s}^{-1}$

3. Relate the rate of disappearance of peroxydisulfate to iodide

Using stoichiometry from the balanced equation, for every molecule of peroxydisulfate (${\mathrm{S}}_2{\mathrm{O}}_8{}^{2-}$) that disappears, there are three molecules of iodide ion (${\mathrm{I}}^{-}$) that also disappear. Thus, the rate of disappearance of the iodide ion (Rate${}_{{\mathrm{I}}^{-}}$) is three times the rate of disappearance of the peroxydisulfate ion (Rate${}_{{\mathrm{S}}_2{\mathrm{O}}_8{}^{2-}}$): $Rat{e}_{{\mathrm{I}}^{-}}=3\cdot Rat{e}_{{\mathrm{S}}_2{\mathrm{O}}_8{}^{2-}}$

4. Calculate the rate of disappearance of iodide for given initial concentrations

We are given the initial concentrations of peroxydisulfate and iodide ions: $[{\mathrm{S}}_2{\mathrm{O}}_8{}^{2-}]=0.025M$ and $[{\mathrm{I}}^{-}]=0.050M$. We can calculate the rate of peroxydisulfate disappearance using the rate law equation and average rate constant: $Rat{e}_{{\mathrm{S}}_2{\mathrm{O}}_8{}^{2-}}=k_{avg}[{\mathrm{S}}_2{\mathrm{O}}_8{}^{2-}][{\mathrm{I}}^{-}{]}^2=6.09\times {10}^3\cdot 0.025\cdot (0.050{)}^2=3.8\times {10}^{-6}M/s$ Now, we can calculate the rate of iodide ion disappearance using the relationship found in step 3: $Rat{e}_{{\mathrm{I}}^{-}}=3\cdot Rat{e}_{{\mathrm{S}}_2{\mathrm{O}}_8{}^{2-}}=3\cdot (3.8\times {10}^{-6})=11.4\times {10}^{-6}M/s$