Answer:
4.16 × 10⁻⁴ M/s
Explanation:
The rate of appearance of a product is the change in its concentration divided by the time elapsed.
[tex]r_{P}=\frac{\Delta [P] }{\Delta t}[/tex]
The rate of disappearance of a reactant is minus the change in its concentration divided by the time elapsed.
[tex]r_{R}=-\frac{\Delta [R] }{\Delta t}[/tex]
The rate of the reaction is equal to the rates of appearance or disappearance divided by its stoichiometric coefficient.
Let's consider the following reaction.
2 S₂O₃²⁻(aq) + I₂(aq) → S₄O₆²⁻(aq) + 2 I⁻(aq)
The rate of dissapearance of thiosulfate is:
[tex]r(S_{2}O_{3}^{2-} )=-\frac{(0-4.56 \times 10^{-3}M )}{11.0 s} =4.15 \times 10^{-4}M/s[/tex]
The rate of the reaction with respect to thiosulfate is:
[tex]r=\frac{r(S_{2}O_{3}^{2-} )}{2} =2.08 \times 10^{-4}[/tex]
The rate of the reaction with respect to the iodide ion is:
[tex]r=\frac{r(I^{-} )}{2} \\r(I^{-}) = 2 r = 4.16 \times 10^{-4} M/s[/tex]
