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

First, calculate the amount of potassium manganate(VII) used:

$ext{Volume of } KMnO_4 = 35.85 ext{ cm}^3 = 0.03585 ext{ dm}^3$

Using the formula:

$n = C imes V$

The moles of potassium manganate(VII) used are:

$n_{KMnO_4} = 0.0200 ext{ mol dm}^{-3} imes 0.03585 ext{ dm}^3 = 7.17 imes 10^{-3} ext{ mol}$

Using the stoichiometric ratio from the equation, where 2 moles of manganate react with 5 moles of hydrogen peroxide:

rac{5}{2} imes n_{KMnO_4} = n_{H_2O_2}

So,

n_{H_2O_2} = rac{5}{2} imes 7.17 imes 10^{-3} = 1.79 imes 10^{-2} ext{ mol}

Because the original hair bleach solution is 5.00% of the diluted solution, we find the concentration in the original solution:

For 25.0 cm³ diluted, the total amount is:

C_{original} = rac{1.79 imes 10^{-2}}{0.025 ext{ dm}^3} imes rac{1}{0.05} = 1.43 ext{ mol dm}^{-3}

No indicator is needed because potassium manganate(VII) is self-indicating. It changes color at the endpoint of the titration, providing a clear visual cue without the need for additional indicators.

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

The equation for the decomposition of hydrogen peroxide is:

Using the ideal gas equation:

$PV = nRT$

Rearranging gives:

n = rac{PV}{RT}

Convert the volume to cubic meters:

$185 ext{ cm}^3 = 0.185 ext{ dm}^3$

Now substituting in:

$P = 100,000 ext{ Pa}$ $R = 8.31 ext{ J K}^{-1} ext{ mol}^{-1}$ $T = 298 ext{ K}$

Then,

Mean bond enthalpy is defined as the average energy required to break one mole of a particular type of bond in a molecule in the gas phase.

Using the data from Table 3: Mean bond enthalpy values:

• Mean bond enthalpy for O–H: 463 kJ mol⁻¹
• Mean bond enthalpy for N–N: 163 kJ mol⁻¹
• Mean bond enthalpy for O–O: X (unknown)

Considering the reaction and applying Hess's law, we can calculate it:

Using the equation:

$ext{Bond enthalpy change} = ext{Total bonds broken} - ext{Total bonds formed}$

We can set up the equation to find the O–O bond enthalpy:

$ext{ΔH} = 2 imes ext{O–H} + ext{N–N} - 2 imes ext{O–O}$

Substituting values given:

$-789 = 2(463) + 163 - 2X$ X = rac{926 + 163 + 789}{2}

Thus, we find: $X = 420.5 ext{ kJ mol}^{-1}$