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

The moles of potassium manganate(VII) used can be calculated as:

ext{Moles of KMnO}_4 = ext{concentration} imes ext{volume} = 0.0200 imes rac{35.85}{1000} = 0.000717

Using the stoichiometry from the balanced equation, the moles of hydrogen peroxide are: ext{Moles of H}_2O_2 = 0.000717 imes rac{5}{2} = 0.001793

This was from a 25.0 cm³ sample, therefore the concentration in the sample is: ext{Concentration of H}_2O_2 = rac{0.001793}{0.0250} = 0.0717 ext{ mol dm}^{-3}

Since the original solution was 5.00% of that concentration: ext{Concentration in original solution} = 0.0717 imes rac{100}{5} = 1.434 ext{ mol dm}^{-3}

An indicator is not added because potassium manganate(VII) is self-indicating, changing the color of the solution from purple to colorless as it is reduced.

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

The decomposition reaction of hydrogen peroxide is:

$2 ext{H}_2O_2 → 2 ext{H}_2O + ext{O}_2$

First, we convert the volume of oxygen gas to m³: $V = 185 ext{ cm}^3 = 0.185 ext{ dm}^3 = 0.185 imes 10^{-3} ext{ m}^3$

Using the ideal gas equation: $PV = nRT$ Substituting values: n = rac{PV}{RT} = rac{100 imes 0.185 imes 10^{-3}}{8.31 imes 298} ≈ 0.00743 ext{ mol}

From the stoichiometry of the equation, $2 ext{H}_2O_2 → ext{O}_2$ Thus, the moles of hydrogen peroxide needed are: $ext{Moles of } H_2O_2 = 0.00743 imes 2 = 0.01486 ext{ mol}$

Mean bond enthalpy is defined as the average energy required to break one mole of bonds in a gaseous state, leading to the formation of gaseous atoms.

Using the average bond enthalpy data:

The reaction involves breaking 1 O–O bond and forming 4 O–H bonds, so:

$ext{Bond enthalpy} = ext{mean bond enthalpy of products} - ext{mean bond enthalpy of reactants}$

$ext{Enthalpy Change} = 4( ext{Bond enthalpy of O–H}) - 1( ext{Bond enthalpy of O–O})$

Plugging values from Table 3:

$ext{Enthalpy Change} = 4(463) - x$

And since the reaction data states:

$-789 = 4(463) - x$

Solving for x gives: $x = 1852 - 789 = 1063 ext{ kJ mol}^{-1}$ Therefore, the O–O bond enthalpy in hydrogen peroxide is approximately 1063 kJ mol⁻¹.