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Difficulty: HardRate of Reaction and Collision Theory
During the catalytic decomposition of hydrogen peroxide according to the equation:
2H2O2(aq)→2H2O(l)+O2(g)2\text{H}_2\text{O}_2(aq) \rightarrow 2\text{H}_2\text{O}(l) + \text{O}_2(g)
a student reacts 100 cm3100\text{ cm}^3 of a 0.40 mol dm−30.40\text{ mol dm}^{-3} solution of H2O2\text{H}_2\text{O}_2. If the peroxide decomposes completely in 80 s80\text{ s}, what is the average rate of formation of oxygen gas in cm3 s−1\text{cm}^3\text{ s}^{-1} at STP? (Molar volume of gas at STP = 22.4 dm3 mol−122.4\text{ dm}^3\text{ mol}^{-1})

Answer: 5.6 cm^3 s^-1

Answer

The average rate of formation of oxygen gas at STP is 5.6 cm3 s−15.6\text{ cm}^3\text{ s}^{-1}.
First, the amount of hydrogen peroxide in moles is calculated as 0.40 mol dm−3×0.100 dm3=0.040 mol0.40\text{ mol dm}^{-3} \times 0.100\text{ dm}^3 = 0.040\text{ mol}. From the reaction stoichiometry, 2 moles of H2O22\text{ moles of H}_2\text{O}_2 produce 1 mole of O21\text{ mole of O}_2, meaning 0.020 mol of O20.020\text{ mol of O}_2 is evolved. At STP, 0.020 mol×22400 cm3 mol−1=448 cm30.020\text{ mol} \times 22400\text{ cm}^3\text{ mol}^{-1} = 448\text{ cm}^3 of oxygen. Dividing by the total time of 80 s80\text{ s} yields 5.6 cm3 s−15.6\text{ cm}^3\text{ s}^{-1}.

Step-by-Step Solution

1
Calculate the total number of moles of H₂O₂ present in the reaction solution
Moles of H2O2=0.40 mol dm−3×0.100 dm3=0.040 mol\text{H}_2\text{O}_2 = 0.40\text{ mol dm}^{-3} \times 0.100\text{ dm}^3 = 0.040\text{ mol}
Concentration and volume determine the total amount of reactant available.
2
Determine the total moles of O₂ gas formed using stoichiometric ratios
Moles of O2=0.040 mol2=0.020 mol\text{O}_2 = \frac{0.040\text{ mol}}{2} = 0.020\text{ mol}
The balanced chemical equation shows a 2:1 mole ratio between H2O2\text{H}_2\text{O}_2 and O2\text{O}_2.
3
Convert moles of O₂ into volume in cm³ at STP
Volume of O2=0.020 mol×22400 cm3 mol−1=448 cm3\text{O}_2 = 0.020\text{ mol} \times 22400\text{ cm}^3\text{ mol}^{-1} = 448\text{ cm}^3
1 mole of any ideal gas occupies 22.4 dm322.4\text{ dm}^3 (22400 cm322400\text{ cm}^3) at STP.
4
Divide total volume of O₂ produced by the elapsed reaction time
Rate of O2\text{O}_2 formation =448 cm380 s=5.6 cm3 s−1= \frac{448\text{ cm}^3}{80\text{ s}} = 5.6\text{ cm}^3\text{ s}^{-1}
Reaction rate with respect to gas product evolution is change in volume per unit time.

Key Concept

Stoichiometric rate of reaction and gas volume calculations
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Topics

Rate of Reaction and Collision Theory