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 1.000 m to 0.843 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 = - 0.0204 m/s incorrect; try again; 9 attempts remaining the final concentrations of all reactants will decrease during the reaction, which signifies that δh₂o₂/δt will be a negative value. use the change in concentration of i⁻ and the stoichiometric relationship between i⁻ and h₂o₂ to determine the change in concentration of h₂o₂ over the amount of time.

Explanation:

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

The rate of change of $I^-$ is $\frac{\Delta[I^-]}{\Delta t}=\frac{0.843\ M - 1.000\ M}{15.0\ s}=\frac{- 0.157\ M}{15.0\ s}\approx - 0.0105\ M/s$.

Step2: Use stoichiometry

From the balanced equation $H_2O_2(aq)+3I^-(aq) + 2H^+(aq)\to I_3^-(aq)+2H_2O(l)$, the mole - ratio of $H_2O_2$ to $I^-$ is $1:3$. So $\frac{\Delta[H_2O_2]}{\Delta t}=\frac{1}{3}\times\frac{\Delta[I^-]}{\Delta t}$.

Step3: Calculate $\frac{\Delta[H_2O_2]}{\Delta t}$

Substitute $\frac{\Delta[I^-]}{\Delta t}\approx - 0.0105\ M/s$ into the above formula: $\frac{\Delta[H_2O_2]}{\Delta t}=\frac{1}{3}\times(- 0.0105\ M/s)\approx - 0.00350\ M/s$.

Answer:

$- 0.00350\ \frac{M}{s}$