Questions 32-34 refer to the following. 5 H2O2(aq) + 2 MnO4 −(aq) + 6 H+(aq) → 2 Mn2+(aq) + 8 H2O(l) — Kinetics Chemistry Question

STEPS:

1. Analyze the balanced chemical equation: The reaction stoichiometry is given as:
$$5\text{ H}_2\text{O}_2(aq) + 2\text{ MnO}_4^-(aq) + 6\text{ H}^+(aq) \rightarrow 2\text{ Mn}^{2+}(aq) + 8\text{ H}_2\text{O}(l) + 5\text{ O}_2(g)$$
From the coefficients, we can see that for every 2 moles of $\text{MnO}_4^-$ that are consumed, 5 moles of $\text{O}_2$ are produced.
2. Relate the rates of chemical species to overall reaction rate: The rate of a reaction can be expressed using the change in concentration of any reactant or product over time, divided by its stoichiometric coefficient (with reactants being negative to represent disappearance and products being positive to represent appearance):
$$\text{Reaction Rate} = -\frac{1}{2}\frac{\Delta[\text{MnO}_4^-]}{\Delta t} = +\frac{1}{5}\frac{\Delta[\text{O}_2]}{\Delta t}$$
3. Set up the relative rate relationship: To find the rate of disappearance of permanganate ($-\frac{\Delta[\text{MnO}_4^-]}{\Delta t}$) relative to the rate of appearance of oxygen ($+\frac{\Delta[\text{O}_2]}{\Delta t}$), isolate the permanganate term by multiplying both sides of the rate equality by $2$:
$$\text{Rate of disappearance of }\text{MnO}_4^- = \frac{\mathbf{2}}{\mathbf{5}} \times \left(\text{Rate of appearance of }\text{O}_2\right)$$
4. Substitute the given values: Plug in the rate of appearance of $\text{O}_2(g)$, which is $1.0 \times 10^{-3}\text{ mol/(L}\cdot\text{s)}$:
$$\text{Rate of disappearance of }\text{MnO}_4^- = \frac{2}{5} \times \left(1.0 \times 10^{-3}\text{ mol/(L}\cdot\text{s)}\right)$$
5. Perform the final calculation:
$$\text{Rate of disappearance of }\text{MnO}_4^- = 0.40 \times \left(1.0 \times 10^{-3}\text{ mol/(L}\cdot\text{s)}\right) = \mathbf{4.0 \times 10^{-4}\text{ mol/(L}\cdot\text{s)}}$$
This matches Option D.

*

WHY_OTHERS_WRONG:

• Option A is incorrect: This value ($6.0 \times 10^{-3}$) is obtained if a student incorrectly multiplies the rate of oxygen appearance by $6$ (the coefficient of $\text{H}^+$) instead of using the proper 2:5 mole ratio between $\text{MnO}_4^-$ and $\text{O}_2$.
• Option B is incorrect: This value ($4.0 \times 10^{-3}$) represents a math error where the exponent is off by a factor of 10. A student might incorrectly simplify $0.40 \times 10^{-3}$ as $4.0 \times 10^{-3}$ instead of shifting the decimal point to get $4.0 \times 10^{-4}$.
• Option C is incorrect: This value ($6.0 \times 10^{-4}$) results from an incorrect mole ratio. A student might use a 3:5 ratio (such as using $\frac{1}{2}$ of the $\text{H}^+$ coefficient) rather than the correct 2:5 ratio between permanganate and oxygen.