Topic 6 · page 6 of 15

From joules to kJ mol⁻¹

Divide q by 1000, divide by the moles of the limiting reactant, then attach the sign from the direction of the temperature change. Worked for a solution in a cup and for a spirit burner.

14 min read6.56.5(ii)

Read first: Measuring the energy transferred

Five steps 6.56.5(ii)

q is the energy for the amount you actually reacted. ΔH is the energy per mole. The conversion is the same in every experiment.

MethodEnthalpy change from a temperature change

  1. ΔT: final temperature minus initial temperature, in °C.
  2. Energy: q = m × c × ΔT, in joules. The mass is the water or solution.
  3. Convert to kJ: divide by 1000.
  4. Moles: find the amount, in mol, of the limiting reactant (or of the fuel burned, or of the water formed), from the equation.
  5. Divide and sign: ΔH = ± q ÷ n, in kJ mol⁻¹. The sign is negative if the temperature rose and positive if it fell.

The five steps give the size of ΔH. The sign comes last and depends only on whether the surroundings got warmer or colder.

Mixing in a cup 6.56.5(ii)

Worked exampleEnthalpy change of neutralisation

25.0 cm³ of 1.00 mol dm⁻³ HCl is added to 25.0 cm³ of 1.00 mol dm⁻³ NaOH in an insulated cup. The temperature rises by 6.8 °C. Calculate the enthalpy change of neutralisation. (c = 4.18 J g⁻¹ °C⁻¹; 1.00 g cm⁻³)

  1. Energy: m = 25.0 + 25.0 = 50.0 g. q = 50.0 × 4.18 × 6.8 = 1421 J = 1.42 kJ.
  2. Moles: n(HCl) = 0.0250 dm³ × 1.00 mol dm⁻³ = 0.0250 mol. One mole of water forms per mole of HCl, so n(H₂O) = 0.0250 mol.
  3. Per mole: 1.421 ÷ 0.0250 = 56.8 kJ mol⁻¹.
  4. Sign: the temperature rose, so the reaction is exothermic: ΔneutH\Delta_\text{neut}H = −56.8 kJ mol⁻¹.

The result is close to the standard value of about −57 kJ mol⁻¹. The mass is of the solution (50.0 g), but the moles come from the limiting reactant. Here both reactants are in exactly equal amounts.

Worked exampleDisplacement with an excess of metal

Excess zinc powder is added to 50.0 cm³ of 0.500 mol dm⁻³ copper(II) sulfate solution. The temperature rises by 25.5 °C. Calculate the enthalpy change for Zn(s)+CuX2+(aq)→ZnX2+(aq)+Cu(s)\ce{Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s)}. (The data are illustrative.)

  1. q = 50.0 × 4.18 × 25.5 = 5329.5 J = 5.33 kJ
  2. Zinc is in excess, so copper(II) sulfate is limiting: n(Cu²⁺) = 0.0500 × 0.500 = 0.0250 mol
  3. 5.3295 ÷ 0.0250 = 213.18, so the size is 213 kJ mol⁻¹
  4. The temperature rose, so ΔH = −213 kJ mol⁻¹

Worked exampleAn endothermic reaction

50.0 cm³ of solution contains 0.0200 mol of the limiting reactant. The temperature falls from 21.0 °C to 16.8 °C. Calculate ΔH. (Illustrative data.)

  1. ΔT = 16.8 − 21.0 = −4.2 °C, a fall of 4.2 °C.
  2. q = 50.0 × 4.18 × 4.2 = 877.8 J = 0.8778 kJ
  3. 0.8778 ÷ 0.0200 = 43.89
  4. The temperature fell, so the reaction is endothermic: ΔH = +43.9 kJ mol⁻¹

Burning a fuel in a spirit burner 6.56.5(ii)

Here the moles come from the fuel burned. The burner is weighed before and after, and the loss in mass is the mass of fuel that burned.

Worked exampleEnthalpy change of combustion of ethanol

A spirit burner containing ethanol heats 100 g of water from 20.0 °C to 45.0 °C. The mass of the burner falls by 0.80 g. Calculate the enthalpy change of combustion of ethanol. (MrM_\text{r} of C₂H₅OH = 46.0)

  1. Energy: q = 100 × 4.18 × 25.0 = 10 450 J = 10.45 kJ. The mass is that of the water.
  2. Moles: n(ethanol) = 0.80 ÷ 46.0 = 0.01739 mol
  3. Per mole: 10.45 ÷ 0.01739 = 601 kJ mol⁻¹
  4. Sign: the water got hotter, so ΔcH\Delta_\text{c}H = −601 kJ mol⁻¹

The data-book value is −1367 kJ mol⁻¹. This experiment found only (601 ÷ 1367) × 100 = 44% of it, because most of the heat from the flame missed the water. That is typical of a spirit burner, and the next two pages explain why and how to judge it.

Check yourself

  1. 50.0 cm³ of 2.00 mol dm⁻³ HCl is mixed with 50.0 cm³ of 2.00 mol dm⁻³ NaOH and the temperature rises by 13.6 °C. Calculate ΔneutH\Delta_\text{neut}H.
  2. 1.20 g of propan-1-ol, C₃H₇OH (MrM_\text{r} = 60.0), is burned and heats 200 g of water by 21.0 °C. Calculate ΔcH\Delta_\text{c}H of propan-1-ol.
  3. 50.0 cm³ of 1.00 mol dm⁻³ HCl is mixed with 50.0 cm³ of 1.20 mol dm⁻³ NaOH, and the temperature rises by 6.7 °C. Identify the limiting reactant and calculate the enthalpy change per mole of water formed.
  4. A student calculates the first worked example using m = 25.0 g. State the value she gets for ΔH and explain her error.
  5. The temperature of 50.0 g of a solution falls from 22.5 °C to 18.0 °C when 0.0100 mol of a reactant is used up. Calculate ΔH.
  6. What mass of ethanol would raise the temperature of 250 g of water by 30.0 °C if all the energy from burning reached the water? (ΔcH\Delta_\text{c}H = −1367 kJ mol⁻¹, MrM_\text{r} = 46.0)

Answers cover these until you have tried the questions

  1. m = 100 g. q = 100 × 4.18 × 13.6 = 5684.8 J = 5.685 kJ. n(HCl) = 0.0500 × 2.00 = 0.100 mol = n(H₂O). 5.685 ÷ 0.100 = 56.8. Temperature rose, so −56.8 kJ mol⁻¹.
  2. q = 200 × 4.18 × 21.0 = 17 556 J = 17.556 kJ. n = 1.20 ÷ 60.0 = 0.0200 mol. 17.556 ÷ 0.0200 = 877.8. Exothermic: −878 kJ mol⁻¹.
  3. n(HCl) = 0.0500 mol; n(NaOH) = 0.0500 × 1.20 = 0.0600 mol. They react 1 : 1, so HCl is limiting and 0.0500 mol of water forms. m = 100 g. q = 100 × 4.18 × 6.7 = 2800.6 J = 2.8006 kJ. 2.8006 ÷ 0.0500 = 56.0. ΔH = −56.0 kJ mol⁻¹.
  4. With m = 25.0 g, q = 25.0 × 4.18 × 6.8 = 710.6 J = 0.711 kJ. ΔH = −0.711 ÷ 0.0250 = −28.4 kJ mol⁻¹. Her error is using the mass of one solution. The two solutions together, 50.0 g, are heated.
  5. ΔT = −4.5 °C. q = 50.0 × 4.18 × 4.5 = 940.5 J = 0.9405 kJ. 0.9405 ÷ 0.0100 = 94.05. The temperature fell, so +94.1 kJ mol⁻¹.
  6. q = 250 × 4.18 × 30.0 = 31 350 J = 31.35 kJ. n = 31.35 ÷ 1367 = 0.02293 mol. m = 0.02293 × 46.0 = 1.05 g. (In a real burner far more is needed, because of heat loss.)