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Energetics / Thermochemistry - Measuring energy changes

Grade 11IBChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Enthalpy change (ΔH\Delta H) is the heat energy change measured under conditions of constant pressure. It is typically measured in kJ mol−1\text{kJ mol}^{-1}.

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Specific heat capacity (cc) is the amount of heat energy required to raise the temperature of 1 g1\text{ g} of a substance by 1 K1\text{ K} (or 1∘C1^\circ\text{C}). For water, c=4.18 J g−1 K−1c = 4.18\text{ J g}^{-1}\text{ K}^{-1}.

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Heat energy change (qq) is calculated using the mass of the substance being heated, its specific heat capacity, and the temperature change: q=mcΔTq = mc\Delta T.

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In calorimetry, the surroundings (usually water or an aqueous solution) absorb or release heat. If the temperature of the surroundings increases, the reaction is exothermic (ΔH<0\Delta H < 0). If it decreases, the reaction is endothermic (ΔH>0\Delta H > 0).

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Standard enthalpy changes (ΔHθ\Delta H^\theta) are measured under standard conditions: 100 kPa100\text{ kPa} pressure, 298 K298\text{ K} temperature, and 1 mol dm−31\text{ mol dm}^{-3} concentration for solutions.

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Systematic errors in calorimetry include heat loss to the surroundings, heat absorbed by the calorimeter (calorimeter constant), and incomplete combustion when using a spirit burner.

📐Formulae

q=mcΔTq = mc\Delta T

ΔH=−qn\Delta H = \frac{-q}{n}

ΔT=Tfinal−Tinitial\Delta T = T_{final} - T_{initial}

ΔHreaction=∑ΔHfθ(products)−∑ΔHfθ(reactants)\Delta H_{reaction} = \sum \Delta H_{f}^{\theta}(\text{products}) - \sum \Delta H_{f}^{\theta}(\text{reactants})

💡Examples

Problem 1:

When 50.0 cm350.0\text{ cm}^3 of 1.00 mol dm−31.00\text{ mol dm}^{-3} HCl(aq)HCl(aq) is added to 50.0 cm350.0\text{ cm}^3 of 1.00 mol dm−31.00\text{ mol dm}^{-3} NaOH(aq)NaOH(aq), the temperature rises from 21.0∘C21.0^\circ\text{C} to 27.5∘C27.5^\circ\text{C}. Calculate the molar enthalpy of neutralization. Assume the density of the solution is 1.00 g cm−31.00\text{ g cm}^{-3} and c=4.18 J g−1 K−1c = 4.18\text{ J g}^{-1}\text{ K}^{-1}.

Solution:

  1. Calculate mass: m=50.0 g+50.0 g=100.0 gm = 50.0\text{ g} + 50.0\text{ g} = 100.0\text{ g}.
  2. Calculate ΔT\Delta T: ΔT=27.5−21.0=6.5 K\Delta T = 27.5 - 21.0 = 6.5\text{ K}.
  3. Calculate qq: q=100.0×4.18×6.5=2717 Jq = 100.0 \times 4.18 \times 6.5 = 2717\text{ J}.
  4. Calculate moles of water formed: n=c×V=1.00×0.0500=0.0500 moln = c \times V = 1.00 \times 0.0500 = 0.0500\text{ mol}.
  5. Calculate ΔH\Delta H: ΔH=−27170.0500=−54340 J mol−1=−54.3 kJ mol−1\Delta H = \frac{-2717}{0.0500} = -54340\text{ J mol}^{-1} = -54.3\text{ kJ mol}^{-1}.

Explanation:

First, the total mass of the reaction mixture is determined. The heat absorbed by the water (qq) is calculated using the calorimetry equation. Since the temperature rose, the reaction is exothermic, so we apply a negative sign to the enthalpy change. Finally, divide by the number of moles of the limiting reactant to find the molar enthalpy.

Problem 2:

A student burns 0.50 g0.50\text{ g} of ethanol (C2H5OHC_2H_5OH) to heat 150 g150\text{ g} of water in a copper calorimeter. The temperature of the water increases by 20.0∘C20.0^\circ\text{C}. Calculate the enthalpy of combustion of ethanol in kJ mol−1\text{kJ mol}^{-1} (MrM_r of ethanol = 46.0846.08).

Solution:

  1. Calculate qq: q=150×4.18×20.0=12540 J=12.54 kJq = 150 \times 4.18 \times 20.0 = 12540\text{ J} = 12.54\text{ kJ}.
  2. Calculate moles of ethanol: n=0.5046.08=0.01085 moln = \frac{0.50}{46.08} = 0.01085\text{ mol}.
  3. Calculate ΔHc\Delta H_c: ΔHc=−12.540.01085=−1155.76 kJ mol−1\Delta H_c = \frac{-12.54}{0.01085} = -1155.76\text{ kJ mol}^{-1}. Final answer: −1160 kJ mol−1-1160\text{ kJ mol}^{-1} (3 sig figs).

Explanation:

The energy transferred to the water is calculated first. Then, the number of moles of fuel burned is found. The enthalpy of combustion is the energy per mole, and it is negative because combustion is an exothermic process.

Measuring energy changes - Revision Notes & Key Formulas | IB Grade 11 Chemistry