Calorimetry Calculations (College Board AP® Chemistry): Study Guide

The energy from 0.0100 mol of 1-propanol was used to heat 250 g of water. The temperature of the water rose from 298K to 310K (the specific heat capacity of water is 4.18 J g^-1 K^-1).

Calculate the enthalpy of combustion.

Answer:

• Step 1: q = m × c × ΔT

  • m (of water) = 250 g

  • c (of water) = 4.18 J g^-1 K^-1

  • ΔT (of water) = 310 − 298 K = 12 K

• Step 2: q = 250 × 4.18 × 12

  • q = 12 540 J

• Step 3:  This is the energy released by 0.01 mol of 1-propanol

  • The water gained 12 540 J, so the combustion released it:

  • q_reaction = −q_water = −12 540 J

  • ΔH = $\frac{q_{\mathrm{reaction}}}{n}$ = $\frac{-12540\mathrm{J}}{0.0100\mathrm{mol}}$ = −1 254 000 J/mol = −1254 kJ/mol

  • ΔH = −1.3 × 10³ kJ/mol (2 significant figures, from ΔT = 12 K)

• There's no need to convert the temperature units in calorimetry as the change in temperature in °C is equal to the change in temperature in K

• If the solution warms up, the reaction is exothermic and ΔH is negative

  • Show the sign change: q_reaction = −q_solution

• Use the mass of whatever changed temperature:

  • The whole solution in a coffee-cup calorimeter, the water alone when a fuel heats a can of water

• ΔH as written is already per mole of reaction as the equation is balanced: check the coefficients before scaling anything

  • Divide q_reaction by the moles of reaction, found from the limiting reactant and its coefficient

• Heat lost to the surroundings makes the measured ΔT smaller, so the calculated q and the magnitude of ΔH are smaller too

• Give your answer to the number of significant figures in the data, with kJ/mol_rxn as the unit of ΔH

Excess iron powder was added to 100.0 mL of 0.200 M copper(II) sulfate solution in a calorimeter. The reaction equation was as follows

Fe (s) + CuSO₄ (aq) → FeSO₄ (aq) + Cu (s)

The maximum temperature rise was 7.5 °C. Determine the enthalpy of reaction, in kJ/mol_rxn.

Answer:

• Step 1: Calculate q:

  • q = m × c × ΔT

    • q = 100 g x 4.18 J g^-1 K^-1 × 7.5 K = 3135 J (m is the mass of the solution, taking its density as 1.00 g/mL)

  • The solution gained this heat, so the reaction released it:

    • q_reaction = −q_solution = −3135 J

• Step 2: Calculate the amount of CuSO₄ (aq)

  • Iron is in excess, so CuSO₄ is the limiting reactant:

    • moles CuSO₄ = volume in L × molarity = 0.1000 L × 0.200 mol/L = 0.0200 mol

  • The equation has a 1:1 ratio, so this is 0.0200 mol of reaction

• Step 3: Calculate ΔH

  • ΔH = $\frac{q_{\mathrm{reaction}}}{n}$ = $\frac{-3135\mathrm{J}}{0.0200\mathrm{mol}}$ = −156 750 J/mol_rxn = −156.75 kJ/mol_rxn

  • ΔH = −1.6 × 10² kJ/mol_rxn (2 significant figures, from ΔT = 7.5 °C)

1.023 g of 1-propanol (M = 60.11 g mol^-1) was burned in a spirit burner and used to heat 200 g of water in a copper calorimeter. The temperature of the water rose by 30 °C.

Calculate the enthalpy of combustion of 1-propanol using this data.

Answer:

• Step 1: Calculate q:

  • q = m × c × ΔT

  • q_water = 200 g × 4.18 J g^-1 °C^-1 × 30 °C = 25 080 J (m is the mass of the water alone: the burning 1-propanol does not change temperature with it)

  • q_reaction = −q_water = −25 080 J

• Step 2: Calculate the amount of 1-propanol burned:

  • moles = $\frac{\mathrm{mass}}{\mathrm{molar}\mathrm{mass}}$ = $\frac{1.023\mathrm{g}}{60.11\mathrm{g}{\mathrm{mol}}^{-1}}$ = 0.01702 mol

• Step 3: Calculate ΔH:

  • ΔH = $\frac{q_{\mathrm{reaction}}}{n}$ = $\frac{-25080\mathrm{J}}{0.01702\mathrm{mol}}$ = −1 474 000 J/mol = −1474 kJ/mol

  • ΔH = −1.5 × 10³ kJ/mol