This question is about hydrogen peroxide, H₂O₂ - AQA - A-Level Chemistry - Question 3 - 2022 - Paper 1

To calculate the concentration of hydrogen peroxide in the original solution, use the titration data:

1. Calculate moles of ext{KMnO}_4 used:

   • Volume of ext{KMnO}_4 = 35.85 cm³ = 0.03585 dm³
   • Concentration of ext{KMnO}_4 = 0.0200 mol dm⁻³
   • Moles of ext{KMnO}_4 = concentration × volume = $0.0200 imes 0.03585 = 7.17 imes 10^{-3} ext{ mol}$

2. The stoichiometry from the ionic equation gives:

   • $5 ext{H}_2 ext{O}_2 : 2 ext{MnO}_4^ ext{-}$ implies that $n( ext{H}_2 ext{O}_2) = 5/2 imes n( ext{KMnO}_4)$
   • Thus, $n( ext{H}_2 ext{O}_2) = 5/2 imes 7.17 imes 10^{-3} = 1.79 imes 10^{-2} ext{ mol}$

3. The original concentration (before dilution) is calculated as:

   • Original concentration of solution was 5 ext{ %} of normal, therefore ext{C}_{ ext{original}} = rac{0.0179}{0.025} imes 100 = 1.43 ext{ mol dm}^{-3}.

No indicator is needed because potassium manganate(VII) is self-indicating; it changes from purple to colorless as it reacts.

In hydrogen peroxide (H₂O₂), the oxidation state of oxygen is -1.

The decomposition of hydrogen peroxide can be represented by the equation:

To find the moles of O₂ produced:

1. Use the ideal gas equation:

   • $PV = nRT$ where $R = 8.31 ext{ J K}^{-1} ext{ mol}^{-1}$
   • Converting volume from cm³ to dm³ gives $185 ext{ cm}^3 = 0.185 ext{ dm}^3$

2. Rearranging gives:

   • n = rac{PV}{RT} = rac{(100 imes 10^3 ext{ Pa}) imes 0.185 ext{ dm}^3}{8.31 ext{ J K}^{-1} ext{ mol}^{-1} imes 298 ext{ K}}
   • Therefore, $n ext{ (O}_2 ext{)} = 0.00747 ext{ mol}.$

3. From the decomposition equation, $2 ext{H}_2 ext{O}_2 ightarrow ext{O}_2$, the molar ratio is:

   • $2 ext{ moles of } ext{H}_2 ext{O}_2 : 1 ext{ mole of } ext{O}_2$, thus:
   • $n( ext{H}_2 ext{O}_2) = 2 imes 0.00747 = 0.01494 ext{ mol}$.

Mean bond enthalpy is the average energy required to break one mole of a particular bond in a gaseous state across a range of molecules.

To find the O–O bond enthalpy from the reaction:

1. Calculate the total bond enthalpy of bonds broken:

   • Bonds broken: 2 O–H bonds and 1 O–O bond.
   • Using Table 3:
     • Mean O–H bond enthalpy = 463 kJ mol⁻¹,
     • Mean O–O bond enthalpy = x (to be calculated).
   • Total energy = $(2 imes 463) + x = 926 + x$ kJ mol⁻¹.

2. Calculate total bond enthalpy of bonds formed:

   • Bonds formed: 4 O–H bonds.
   • Total bond enthalpy = $4 imes 463 = 1852$ kJ mol⁻¹.

3. The reaction enthalpy ( ΔH) for the reaction gives:

   • $926 + x - 1852 = -789$ kJ mol⁻¹.
   • This simplifies to $x = 37 kJ mol⁻¹$ for the O–O bond enthalpy.