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What is meant by the term enthalpy of neutralisation? Write an equation for the reaction that occurred. Calculate the enthalpy of neutralisation.

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Question 17

What is meant by the term enthalpy of neutralisation?

The enthalpy of neutralisation is the heat energy released when one mole of water is formed from the reaction of an acid with a base under standard conditions.

i) Write an equation for the reaction that occurred.

The reaction is between sulfuric acid (H2SO4\text{H}_2\text{SO}_4) and sodium hydroxide (NaOH\text{NaOH}).

H2SO4(aq)+2NaOH(aq)⟶Na2SO4(aq)+2H2O(l)\text{H}_2\text{SO}_4(\text{aq}) + 2\text{NaOH}(\text{aq}) \longrightarrow \text{Na}_2\text{SO}_4(\text{aq}) + 2\text{H}_2\text{O}(\text{l})

ii) Calculate the enthalpy of neutralisation of sodium hydroxide (specific heat capacity of solution = 4.2 J g−1 K−14.2 \text{ J g}^{-1} \text{ K}^{-1}, density of water = 1 g cm−31 \text{ g cm}^{-3}).

  1. Calculate the moles of reactants.

    Moles of H2SO4\text{H}_2\text{SO}_4:

    Moles=Concentration×Volume=1 M×501000 L=0.050 mol\text{Moles} = \text{Concentration} \times \text{Volume} = 1 \text{ M} \times \frac{50}{1000} \text{ L} = 0.050 \text{ mol}

    Moles of NaOH\text{NaOH}:

    Moles=Concentration×Volume=2 M×501000 L=0.100 mol\text{Moles} = \text{Concentration} \times \text{Volume} = 2 \text{ M} \times \frac{50}{1000} \text{ L} = 0.100 \text{ mol}

  2. Determine the limiting reactant and moles of water formed.

    From the balanced equation, 1 mol of H2SO41 \text{ mol of H}_2\text{SO}_4 reacts with 2 mol of NaOH2 \text{ mol of NaOH} to produce 2 mol of H2O2 \text{ mol of H}_2\text{O}.
    If 0.050 mol of H2SO40.050 \text{ mol of H}_2\text{SO}_4 reacts, it would require 2×0.050=0.100 mol of NaOH2 \times 0.050 = 0.100 \text{ mol of NaOH}.
    Since we have exactly 0.100 mol of NaOH0.100 \text{ mol of NaOH}, both reactants are completely consumed.
    The moles of water formed will be 2×0.050 mol=0.100 mol2 \times 0.050 \text{ mol} = 0.100 \text{ mol}.

  3. Calculate the total mass of the solution.

    Total volume of solution = 50 cm3(H2SO4)+50 cm3(NaOH)=100 cm350 \text{ cm}^3 (\text{H}_2\text{SO}_4) + 50 \text{ cm}^3 (\text{NaOH}) = 100 \text{ cm}^3.
    Assuming the density of the solution is 1 g cm−31 \text{ g cm}^{-3}:

    Mass of solution=Volume×Density=100 cm3×1 g cm−3=100 g\text{Mass of solution} = \text{Volume} \times \text{Density} = 100 \text{ cm}^3 \times 1 \text{ g cm}^{-3} = 100 \text{ g}

  4. Calculate the heat absorbed by the solution (qq).

    Given ΔT=13.6∘C\Delta T = 13.6^\circ\text{C} (or 13.6 K13.6 \text{ K}) and specific heat capacity (cc) = 4.2 J g−1 K−14.2 \text{ J g}^{-1} \text{ K}^{-1}.

    q=mcΔTq = mc\Delta T q=100 g×4.2 J g−1 K−1×13.6 Kq = 100 \text{ g} \times 4.2 \text{ J g}^{-1} \text{ K}^{-1} \times 13.6 \text{ K} q=5712 Jq = 5712 \text{ J}

    Convert to kilojoules:

    q=57121000 kJ=5.712 kJq = \frac{5712}{1000} \text{ kJ} = 5.712 \text{ kJ}

  5. Calculate the enthalpy of neutralisation (ΔHneut\Delta H_{\text{neut}}).

    Since heat is released (temperature rose), ΔH\Delta H is negative.

    ΔHneut=−Heat releasedMoles of water formed\Delta H_{\text{neut}} = -\frac{\text{Heat released}}{\text{Moles of water formed}} ΔHneut=−5.712 kJ0.100 mol\Delta H_{\text{neut}} = -\frac{5.712 \text{ kJ}}{0.100 \text{ mol}} ΔHneut=−57.12 kJ mol−1\Delta H_{\text{neut}} = \boxed{-57.12 \text{ kJ mol}^{-1}}

    Question 18

    i) What does point S represent?

    Point S represents the equivalence point (or neutralization point) of the reaction, where the acid and base have completely reacted with each other, and the maximum temperature of the solution is reached.

    ii) Why does the temperature rise from R to S?

    The temperature rises from R to S because the reaction between HCl\text{HCl} (acid) and NaOH\text{NaOH} (base) is an exothermic reaction. As more NaOH\text{NaOH} is added, more heat is released, causing the temperature of the solution to increase until the equivalence point (S) is reached.

    iii) Determine the molarity of the HCl\text{HCl} acid.

  6. Identify the volume of NaOH\text{NaOH} at point S.

    From the graph, point S occurs when 15 cm315 \text{ cm}^3 of NaOH\text{NaOH} has been added.

  7. Write the balanced chemical equation.

    The reaction is between HCl\text{HCl} and NaOH\text{NaOH}:

    HCl(aq)+NaOH(aq)⟶NaCl(aq)+H2O(l)\text{HCl}(\text{aq}) + \text{NaOH}(\text{aq}) \longrightarrow \text{NaCl}(\text{aq}) + \text{H}_2\text{O}(\text{l})

    The mole ratio of HCl\text{HCl} to NaOH\text{NaOH} is 1:11:1.

  8. Calculate the moles of NaOH\text{NaOH} at the equivalence point.

    Given concentration of NaOH=0.5 M\text{NaOH} = 0.5 \text{ M} and volume = 15 cm315 \text{ cm}^3.

    Moles of NaOH=Concentration×Volume\text{Moles of NaOH} = \text{Concentration} \times \text{Volume} Moles of NaOH=0.5 M×151000 L=0.0075 mol\text{Moles of NaOH} = 0.5 \text{ M} \times \frac{15}{1000} \text{ L} = 0.0075 \text{ mol}

  9. Calculate the moles of HCl\text{HCl}.

    Since the mole ratio is 1:11:1, at the equivalence point:

    Moles of HCl=Moles of NaOH=0.0075 mol\text{Moles of HCl} = \text{Moles of NaOH} = 0.0075 \text{ mol}

  10. Calculate the molarity of HCl\text{HCl}.

    Given volume of HCl=20 cm3\text{HCl} = 20 \text{ cm}^3.

    Molarity of HCl=Moles of HClVolume of HCl\text{Molarity of HCl} = \frac{\text{Moles of HCl}}{\text{Volume of HCl}} Molarity of HCl=0.0075 mol201000 L=0.0075 mol0.020 L\text{Molarity of HCl} = \frac{0.0075 \text{ mol}}{\frac{20}{1000} \text{ L}} = \frac{0.0075 \text{ mol}}{0.020 \text{ L}} Molarity of HCl=0.375 M\text{Molarity of HCl} = \boxed{0.375 \text{ M}}

    Question 19

    a) Give the name of H2\text{H}_2.

    The name of H2\text{H}_2 is hydrogen gas (or simply hydrogen).

    b) Determine the value of ΔH3\Delta H_3.

  11. Apply Hess's Law to the energy cycle.

    Hess's Law states that the total enthalpy change for a reaction is independent of the pathway taken.
    From the diagram, the direct path from H2(g)+O2(g)\text{H}_2(\text{g}) + \text{O}_2(\text{g}) to H2O2(l)\text{H}_2\text{O}_2(\text{l}) is ΔH1\Delta H_1.
    The indirect path is from H2(g)+O2(g)\text{H}_2(\text{g}) + \text{O}_2(\text{g}) to H2O(l)+12O2(g)\text{H}_2\text{O}(\text{l}) + \frac{1}{2}\text{O}_2(\text{g}) (which is ΔH2\Delta H_2), and then from H2O(l)+12O2(g)\text{H}_2\text{O}(\text{l}) + \frac{1}{2}\text{O}_2(\text{g}) to H2O2(l)\text{H}_2\text{O}_2(\text{l}) (which is ΔH3\Delta H_3).
    Therefore, according to Hess's Law:

    ΔH1=ΔH2+ΔH3\Delta H_1 = \Delta H_2 + \Delta H_3

  12. Substitute the given values and solve for ΔH3\Delta H_3.

    Given ΔH1=−188 kJ mol−1\Delta H_1 = -188 \text{ kJ mol}^{-1} and ΔH2=−286 kJ mol−1\Delta H_2 = -286 \text{ kJ mol}^{-1}.

    −188 kJ mol−1=−286 kJ mol−1+ΔH3-188 \text{ kJ mol}^{-1} = -286 \text{ kJ mol}^{-1} + \Delta H_3 ΔH3=−188 kJ mol−1−(−286 kJ mol−1)\Delta H_3 = -188 \text{ kJ mol}^{-1} - (-286 \text{ kJ mol}^{-1}) ΔH3=−188 kJ mol−1+286 kJ mol−1\Delta H_3 = -188 \text{ kJ mol}^{-1} + 286 \text{ kJ mol}^{-1} ΔH3=98 kJ mol−1\Delta H_3 = \boxed{98 \text{ kJ mol}^{-1}}
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