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This question is about hydrogen peroxide, H₂O₂ - AQA - A-Level Chemistry - Question 3 - 2022 - Paper 1
Question 3
This question is about hydrogen peroxide, H₂O₂. The half-equation for the oxidation of hydrogen peroxide is H₂O₂ → O₂ + 2H⁺ + 2e⁻ Hair bleach solution contains hydr... show full transcript
Worked Solution & Example Answer:This question is about hydrogen peroxide, H₂O₂ - AQA - A-Level Chemistry - Question 3 - 2022 - Paper 1
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2. Calculate the concentration, in mol dm⁻³, of hydrogen peroxide in the original hair bleach solution.
Answer
First, calculate the moles of potassium manganate(VII) used:
Molarity (M) = number of moles / volume in dm³.
Volume used = 35.85 cm³ = 0.03585 dm³.
Moles of KMnO₄ = 0.0200 mol dm⁻³ × 0.03585 dm³ = 0.000717 mol.
According to the stoichiometry of the reaction, the ratio of KMnO₄ to H₂O₂ is 2:5. Therefore:
Moles of H₂O₂ = (5/2) × 0.000717 mol = 0.001793 mol.
Since the diluted solution is 5% of the original:
Concentration of H₂O₂ in the diluted solution = 0.001793 mol / 0.025 dm³ = 0.0717 mol dm⁻³.
Concentration in original solution = 0.0717 mol dm⁻³ / 0.05 = 1.434 mol dm⁻³.
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Step 5
5. Give an equation for the decomposition of hydrogen peroxide.
Answer
The equation for the decomposition of hydrogen peroxide is:
ightarrow 2 ext{H}_2 ext{O} + ext{O}_2 $$ To find the amount of hydrogen peroxide needed to produce 185 cm³ of oxygen gas at 100 kPa and 298 K, use the ideal gas equation: $$ PV = nRT $$ Where: - P = 100 kPa = 100,000 Pa - V = 185 cm³ = 0.185 dm³ - R = 8.314 J K⁻¹ mol⁻¹ - T = 298 K Rearranging the equation: $$ n = rac{PV}{RT} $$ Calculating: $$ n = rac{100,000 imes 0.185}{8.314 imes 298} ≈ 0.748 ext{ mol O}_2 $$ From the equation, 2 moles of H₂O₂ produce 1 mole of O₂, so: $$ ext{Moles of H₂O₂} = 2 imes 0.748 ≈ 1.496 ext{ mol} $$Step 6
6. Define the term mean bond enthalpy.
Answer
Mean bond enthalpy is the average energy required to break one mole of a given type of bond in a molecule in the gas phase.
To calculate the O–O bond enthalpy in hydrogen peroxide, we can use the mean bond enthalpies provided in Table 3:
In the reaction 2 H₂O₂ → 2 H₂O + O₂, we can express the change in enthalpy: Using the bond enthalpies:
- 2 O–H bonds in H₂O₂ (for 2 H₂O) = 2 × 463 kJ/mol = 926 kJ/mol.
- 1 O–O bond formed = O–O enthalpy value.
Thus, rearranging: