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Energetics / Thermochemistry - Measuring energy changes

Grade 12IBChemistry

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

🔑Concepts

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Temperature is a measure of the average kinetic energy of the particles in a substance, whereas heat is the total energy transferred from a warmer body to a cooler body.

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Specific heat capacity (cc) is the amount of heat required to raise the temperature of one unit mass (usually 1 g1\text{ g} or 1 kg1\text{ kg}) 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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Calorimetry is the experimental technique used to measure the heat exchange during a chemical reaction or physical process.

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For aqueous reactions, the heat absorbed or released by the solution is calculated using q=mcΔTq = mc\Delta T, assuming the density of the solution is 1.00 g cm−31.00\text{ g cm}^{-3} and its specific heat capacity is that of water.

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The molar enthalpy change (ΔH\Delta H) is calculated as ΔH=−qn\Delta H = \frac{-q}{n}, where nn is the amount in moles of the limiting reactant. The negative sign indicates that if the temperature of the surroundings increases (qq is positive), the reaction is exothermic (ΔH\Delta H is negative).

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Systematic errors in calorimetry often arise from heat loss to the surroundings and the heat capacity of the calorimeter itself. To compensate for heat loss, a temperature-time graph can be extrapolated back to the time of mixing.

📐Formulae

q=mcΔTq = mc\Delta T

ΔH=−qn\Delta H = \frac{-q}{n}

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

n=mMn = \frac{m}{M}

n=C×Vn = C \times V

💡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 mixed with 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 enthalpy of neutralization (ΔHneut\Delta H_{neut}) in kJ mol−1\text{kJ mol}^{-1}. 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 of solution: m=50.0+50.0=100.0 gm = 50.0 + 50.0 = 100.0\text{ g}.
  2. Calculate temperature change: ΔT=27.5−21.0=6.5 K\Delta T = 27.5 - 21.0 = 6.5\text{ K}.
  3. Calculate heat change (qq): q=mcΔT=100.0×4.18×6.5=2717 J=2.717 kJq = mc\Delta T = 100.0 \times 4.18 \times 6.5 = 2717\text{ J} = 2.717\text{ kJ}.
  4. Calculate moles of H2OH_2O 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=−2.7170.0500=−54.34 kJ mol−1\Delta H = \frac{-2.717}{0.0500} = -54.34\text{ kJ mol}^{-1}.

Explanation:

The reaction is exothermic because the temperature increased, so ΔH\Delta H must be negative. The total volume of the solution is used to find the mass (mm) for the qq calculation, while the moles of the limiting reactant (or products) are used to find the molar enthalpy.

Problem 2:

A 0.522 g0.522\text{ g} sample of methanol (CH3OHCH_3OH, M=32.05 g mol−1M = 32.05\text{ g mol}^{-1}) was burned in a spirit burner to heat 200.0 g200.0\text{ g} of water. The temperature of the water rose by 12.5∘C12.5^\circ\text{C}. Calculate the enthalpy of combustion (ΔHc\Delta H_c) in kJ mol−1\text{kJ mol}^{-1}.

Solution:

  1. Calculate qq for water: q=200.0×4.18×12.5=10450 J=10.45 kJq = 200.0 \times 4.18 \times 12.5 = 10450\text{ J} = 10.45\text{ kJ}.
  2. Calculate moles of methanol: n=0.52232.05=0.016287 moln = \frac{0.522}{32.05} = 0.016287\text{ mol}.
  3. Calculate ΔHc\Delta H_c: ΔHc=−10.450.016287≈−642 kJ mol−1\Delta H_c = \frac{-10.45}{0.016287} \approx -642\text{ kJ mol}^{-1}.

Explanation:

The energy released by the combustion of methanol is absorbed by the water. The calculated ΔHc\Delta H_c is typically lower than the data book value due to heat loss to the environment and incomplete combustion.

Measuring energy changes - Revision Notes & Key Formulas | IB Grade 12 Chemistry