Edexcel International A Level (IAL) Chemistry (YCH11) - Unit 2 - 6.5 Calorimetry calculations-Study Notes - New Syllabus

Edexcel International A Level (IAL) Chemistry (YCH11) -Unit 2 – 6.5 Calorimetry calculations- Study Notes- New syllabus

Edexcel International A Level (IAL) Chemistry (YCH11) -Unit 2 – 6.5 Calorimetry calculations- Study Notes -International A Level (IAL) Chemistry (YCH11) – per latest Syllabus.

Key Concepts:

6.5 be able to use experimental data to calculate:
i energy transferred in a reaction recalling and using the expression:
energy transferred (J) = mass (g) × specific heat capacity (J g⁻¹ °C⁻¹) × temperature change (°C)
ii enthalpy change of the reaction in kJ mol⁻¹
This will be limited to experiments where substances are mixed in an insulated container and combustion experiments using a suitable calorimeter.

Edexcel International A Level (IAL) Chemistry (YCH11) -Concise Summary Notes- All Topics

6.5 (i) Energy Transferred in a Reaction

In calorimetry experiments, the energy transferred during a reaction is calculated using temperature change of a substance (usually water or solution).

Key Formula

energy transferred (\( q \)) = mass × specific heat capacity × temperature change

\( q = mc\Delta T \)

Where:

  • \( q \) = energy transferred (J)
  • \( m \) = mass of solution (g)
  • \( c \) = specific heat capacity (J g\(^{-1}\) °C\(^{-1}\))
  • \( \Delta T \) = temperature change (°C or K)

Important Notes

  • For aqueous solutions, \( c = 4.18 \,\mathrm{J\,g^{-1}\,°C^{-1}} \).
  • \( \Delta T = \text{final temperature} – \text{initial temperature} \).
  • Assume density of solution ≈ \( 1.00 \,\mathrm{g\,cm^{-3}} \).

Sign of Energy Change

  • Temperature increase → exothermic → heat released
  • Temperature decrease → endothermic → heat absorbed

Experimental Context

  • Reactions carried out in insulated containers (e.g. polystyrene cups).
  • Minimises heat loss to surroundings.
  • Assumes all heat transfer affects the solution.

Example 1:

50.0 cm³ of solution is heated from 25.0°C to 31.5°C. Calculate the energy transferred.

▶️ Answer/Explanation

Mass = 50.0 g (density ≈ 1.00 g cm\(^{-3}\))

\( \Delta T = 31.5 – 25.0 = 6.5\,°C \)

\( q = mc\Delta T \)

\( q = 50.0 \times 4.18 \times 6.5 \)

\( q = 1358.5 \,\mathrm{J} \approx 1.36 \times 10^3 \,\mathrm{J} \)

Example 2:

In a calorimetry experiment, the temperature of the solution decreases by 4.0°C. Explain what this indicates about the reaction and the sign of \( q \).

▶️ Answer/Explanation

Temperature decrease means heat is absorbed from surroundings.

Reaction is endothermic.

Therefore, \( q \) for the reaction is positive.

6.5 (ii) Enthalpy Change of Reaction (ΔH) in kJ mol\(^{-1}\)

After calculating the energy transferred (\( q \)) using calorimetry, the enthalpy change of the reaction is determined per mole of reactant in kJ mol\(^{-1}\).

Key Steps

  • Step 1: Calculate energy transferred using \( q = mc\Delta T \) (in J).
  • Step 2: Convert \( q \) to kJ → divide by 1000.
  • Step 3: Calculate number of moles of limiting reagent.
  • Step 4: Divide energy by moles.

Formula

\( \Delta H = \frac{q}{n} \)

(convert \( q \) to kJ first)

Sign Convention

  • Temperature increase → exothermic → \( \Delta H < 0 \)
  • Temperature decrease → endothermic → \( \Delta H > 0 \)

Important Notes

  • Use limiting reagent for moles.
  • Units must be kJ mol\(^{-1}\).
  • Assume no heat loss (ideal condition).

Example 1:

25.0 cm³ of \( \mathrm{1.00\,mol\,dm^{-3}} \) HCl reacts with 25.0 cm³ of \( \mathrm{1.00\,mol\,dm^{-3}} \) NaOH. Temperature rises by 6.0°C. Calculate ΔH for the reaction.

▶️ Answer/Explanation

Total volume = 50.0 cm³ → mass = 50.0 g

\( q = mc\Delta T \)

\( q = 50.0 \times 4.18 \times 6.0 = 1254 \,\mathrm{J} = 1.254 \,\mathrm{kJ} \)

Moles of HCl = \( \mathrm{1.00 \times \frac{25.0}{1000} = 0.0250\,mol} \)

\( \Delta H = \frac{1.254}{0.0250} = 50.2 \,\mathrm{kJ\,mol^{-1}} \)

Temperature increases → exothermic

Therefore, \( \Delta H = -50.2 \,\mathrm{kJ\,mol^{-1}} \)

Example 2:

A fuel releases 2.50 kJ of energy when 0.0100 mol is burned. Calculate ΔH\(_c\).

▶️ Answer/Explanation

\( \Delta H = \frac{q}{n} \)

\( \Delta H = \frac{2.50}{0.0100} = 250 \,\mathrm{kJ\,mol^{-1}} \)

Combustion releases heat → exothermic

Therefore, \( \Delta H_c = -250 \,\mathrm{kJ\,mol^{-1}} \)

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