Question

consider the balanced chemical equation. h₂o₂(aq) + 3i⁻(aq) + 2h⁺(aq) → i₃⁻(aq) + 2h₂o(l) in the first 10.0 s of the reaction, the concentration of i⁻ drops from 1.000 m to 0.767 m. part a predict the rate of change in the concentration of h₂o₂ (δh₂o₂/δt). express the rate to three significant figures and include the appropriate units. δh₂o₂/δt = value units

Explanation:

Step1: Calculate the rate of change of $I^-$

The rate of change of $I^-$ is given by $\frac{\Delta[I^-]}{\Delta t}$. The initial concentration of $I^-$ is $[I^-]_0 = 1.000\ M$ and the final concentration is $[I^-]_t=0.767\ M$ in a time - interval $\Delta t = 10.0\ s$. So, $\frac{\Delta[I^-]}{\Delta t}=\frac{[I^-]_t - [I^-]_0}{\Delta t}=\frac{0.767\ M-1.000\ M}{10.0\ s}=- 0.0233\ M/s$.

Step2: Use the stoichiometric relationship

From the balanced chemical equation $H_2O_2(aq)+3I^-(aq)+2H^+(aq)\to I_3^-(aq)+2H_2O(l)$, the stoichiometric ratio between $H_2O_2$ and $I^-$ is $\frac{\Delta[H_2O_2]}{\Delta[I^-]}=\frac{1}{3}$.
We can express the rate of change of $H_2O_2$ in terms of the rate of change of $I^-$ as $\frac{\Delta[H_2O_2]}{\Delta t}=\frac{1}{3}\times\frac{\Delta[I^-]}{\Delta t}$.
Substitute $\frac{\Delta[I^-]}{\Delta t}=-0.0233\ M/s$ into the above equation: $\frac{\Delta[H_2O_2]}{\Delta t}=\frac{1}{3}\times(- 0.0233\ M/s)\approx - 0.00777\ M/s$.

Answer:

• 0.00777, M/s