Question

consider the balanced chemical equation h₂o₂(aq)+3i⁻(aq) + 2h⁺(aq) → i₃⁻(aq) + 2h₂o(l). in the first 15.0 s of the reaction, the concentration of i⁻ drops from 1.00 m to 0.0843 m. predict the change in the concentration of h₂o₂ (δh₂o₂)/δt. $\frac{1}{2}(\frac{0.0843 - 1.00}{15.0}m)$

Explanation:

Step1: Determine the stoichiometric ratio

The balanced chemical equation is $H_2O_2(aq)+3I^-(aq)+2H^+(aq)
ightarrow I_3^-(aq)+2H_2O(l)$. The stoichiometric ratio between $H_2O_2$ and $I^-$ is $1:3$.

Step2: Calculate the change in $[I^-]$

The change in the concentration of $I^-$, $\Delta[I^-]=0.0843 - 1.00=- 0.9157M$.

Step3: Calculate the change in $[H_2O_2]$

Since the ratio of $\frac{\Delta[H_2O_2]}{\Delta[I^-]}=\frac{1}{3}$, then $\Delta[H_2O_2]=\frac{1}{3}\Delta[I^-]$. In terms of the rate - $\text{Rate}=-\frac{1}{1}\frac{\Delta[H_2O_2]}{\Delta t}=-\frac{1}{3}\frac{\Delta[I^-]}{\Delta t}$. So $\Delta[H_2O_2]=\frac{1}{3}(0.0843 - 1.00)M$. And $\frac{\Delta[H_2O_2]}{\Delta t}=\frac{1}{3}\times\frac{0.0843 - 1.00}{15.0}M/s$.
\[

$$\begin{align*} \frac{\Delta[H_2O_2]}{\Delta t}&=\frac{1}{3}\times\frac{0.0843 - 1.00}{15.0}\\ &=\frac{1}{3}\times\frac{- 0.9157}{15.0}\\ &=\frac{-0.9157}{45.0}\\ &\approx - 0.0204M/s \end{align*}$$

\]

Answer:

The change in the concentration of $H_2O_2$ per unit time $\frac{\Delta[H_2O_2]}{\Delta t}\approx - 0.0204M/s$