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

An indicator is not added because potassium manganate(VII) is self-indicating. It changes color (from purple to colorless) upon reacting and clearly shows the endpoint of the titration.

The oxidation state of oxygen in hydrogen peroxide (H₂O₂) is -1.

The decomposition reaction of hydrogen peroxide is:

To calculate the moles of hydrogen peroxide needed:

1. Use the ideal gas equation: $PV = nRT$ where:

   • P = 100 kPa = 100,000 Pa
   • V = 185 cm³ = 0.185 dm³
   • R = 8.31 J K⁻¹ mol⁻¹
   • T = 298 K

2. Rearranging for n gives:

n = rac{PV}{RT} = rac{(100,000)(0.185)}{(8.31)(298)} = 0.749 ext{ mol}

3. Given that 2 moles of H₂O₂ produce 1 mole of O₂, the amount of H₂O₂ needed is:

$ext{Thereby, } n(H_2O_2) = 2 imes 0.749 = 1.498 ext{ mol}$

Mean bond enthalpy is defined as the average energy required to break one mole of a given type of bond in a gaseous state, averaged over various compounds containing that bond.

To calculate the O–O bond enthalpy:

1. Identify the bonds broken and formed during the reaction for bond enthalpy calculations:

   • Breaking 1 O–O bond and 2 N–H bonds, and 2 H–O bonds are relevant.

2. Using the mean bond enthalpy values from Table 3:

   • Mean bond enthalpy (O–H) = 463 kJ mol⁻¹
   • Mean bond enthalpy (N–H) = 388 kJ mol⁻¹
   • Mean bond enthalpy (O–O) = X kJ mol⁻¹ (unknown)

3. Set up the equation for enthalpy change:

$ΔH = ext{Bonds Broken} - ext{Bonds Formed}$

4. Calculate:

$-789 = [X + 2 imes 388] - [2 imes 463]$

5. Solve for X, leading to:

$X = 146 ext{ kJ mol}^{-1}$

Thus, the O–O bond enthalpy in hydrogen peroxide is approximately 146 kJ mol⁻¹.