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State and explain whether the reaction is exothermic or endothermic. Calculate the enthalpy change when 10 grams of NaOH is dissolved in water, given that the molar mass of sodium hydroxide is 40 gram

Chemistry

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Question 14

i) State and explain whether the reaction is exothermic or endothermic.

  1. Analyze the diagram and given ΔH\Delta H value.

    The diagram shows that the energy of the products (NaOH(aq)) is lower than the energy of the reactants (NaOH(s) + H2_2O(l)). The given enthalpy change, ΔH=−42.8 kJ mol−1\Delta H = -42.8 \text{ kJ mol}^{-1}, is negative.

  2. Conclude and explain.

    The reaction is exothermic. This is because the enthalpy change (ΔH\Delta H) is negative, indicating that heat is released from the system to the surroundings. The products have lower energy than the reactants.

    ii) Given that the molar mass of sodium hydroxide is 40 grams. Calculate the enthalpy change when 10 grams of NaOH is dissolved in water.

  3. Calculate the number of moles of NaOH.

    Given molar mass of NaOH = 40 g mol−140 \text{ g mol}^{-1} and mass of NaOH = 10 g10 \text{ g}.

    Moles of NaOH=MassMolar mass=10 g40 g mol−1=0.25 mol\text{Moles of NaOH} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{10 \text{ g}}{40 \text{ g mol}^{-1}} = 0.25 \text{ mol}
  4. Calculate the total enthalpy change.

    The molar enthalpy change for dissolution is ΔH=−42.8 kJ mol−1\Delta H = -42.8 \text{ kJ mol}^{-1}.

    Enthalpy change=Moles of NaOH×ΔH\text{Enthalpy change} = \text{Moles of NaOH} \times \Delta H Enthalpy change=0.25 mol×(−42.8 kJ mol−1)\text{Enthalpy change} = 0.25 \text{ mol} \times (-42.8 \text{ kJ mol}^{-1}) Enthalpy change=−10.7 kJ\text{Enthalpy change} = \boxed{-10.7 \text{ kJ}}

    Question 15

    i) Define molar heat of combustion of a substance.

    The molar heat of combustion is the heat energy released when one mole of a substance undergoes complete combustion in excess oxygen under standard conditions.

    ii) 1.26 grams of element Q was completely burned in air. The heat evolved was used to heat 250 cm³ of water, rising the temperature from 32 to 50°C. Calculate the relative atomic mass of element Q given that molar enthalpy of combustion is -360 kJ mol⁻¹ (density of water is 1 g cm⁻³ and specific heat capacity of water = 4.2 J g⁻¹ K⁻¹).

  5. Calculate the mass of water.

    Given volume of water = 250 cm3250 \text{ cm}^3 and density of water = 1 g cm−31 \text{ g cm}^{-3}.

    Mass of water=Volume×Density=250 cm3×1 g cm−3=250 g\text{Mass of water} = \text{Volume} \times \text{Density} = 250 \text{ cm}^3 \times 1 \text{ g cm}^{-3} = 250 \text{ g}
  6. Calculate the temperature change of water.

    Initial temperature = 32∘C32^\circ\text{C}, Final temperature = 50∘C50^\circ\text{C}.

    ΔT=Tfinal−Tinitial=50∘C−32∘C=18∘C\Delta T = T_{\text{final}} - T_{\text{initial}} = 50^\circ\text{C} - 32^\circ\text{C} = 18^\circ\text{C}

    (Note: A change of 18∘C18^\circ\text{C} is equivalent to a change of 18 K18 \text{ K}).

  7. Calculate the heat absorbed by water (qq).

    Given specific heat capacity of water (cc) = 4.2 J g−1 K−14.2 \text{ J g}^{-1} \text{ K}^{-1}.

    q=mcΔTq = mc\Delta T q=250 g×4.2 J g−1 K−1×18 Kq = 250 \text{ g} \times 4.2 \text{ J g}^{-1} \text{ K}^{-1} \times 18 \text{ K} q=18900 Jq = 18900 \text{ J}

    Convert Joules to kilojoules:

    q=189001000 kJ=18.9 kJq = \frac{18900}{1000} \text{ kJ} = 18.9 \text{ kJ}
  8. Calculate the number of moles of Q that reacted.

    The heat evolved from the combustion of Q is equal to the heat absorbed by water.
    Molar enthalpy of combustion of Q = −360 kJ mol−1-360 \text{ kJ mol}^{-1}.

    Moles of Q=Heat absorbed by waterMagnitude of molar enthalpy of combustion\text{Moles of Q} = \frac{\text{Heat absorbed by water}}{\text{Magnitude of molar enthalpy of combustion}} Moles of Q=18.9 kJ360 kJ mol−1=0.0525 mol\text{Moles of Q} = \frac{18.9 \text{ kJ}}{360 \text{ kJ mol}^{-1}} = 0.0525 \text{ mol}
  9. Calculate the relative atomic mass of Q.

    Given mass of Q = 1.26 g1.26 \text{ g}.

    Relative atomic mass of Q=Mass of QMoles of Q\text{Relative atomic mass of Q} = \frac{\text{Mass of Q}}{\text{Moles of Q}} Relative atomic mass of Q=1.26 g0.0525 mol=24 g mol−1\text{Relative atomic mass of Q} = \frac{1.26 \text{ g}}{0.0525 \text{ mol}} = \boxed{24 \text{ g mol}^{-1}}

    Question 16

    a) Explain why a plastic container is used instead of a metallic one.

    A plastic container is used because plastic is a poor conductor of heat (an insulator). This minimizes heat loss to the surroundings, ensuring that the measured heat change for the reaction is as accurate as possible. A metallic container, being a good conductor of heat, would allow significant heat loss, leading to inaccurate results.

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