Topic 6 · page 12 of 15

Core Practical 2: results, cycle and evaluation

Turn two temperature changes into two enthalpy changes, close the cycle with the formation of water, and judge the answer against the data-book value and the uncertainties.

15 min read6.76.66.6(ii)6.8

Read first: Core Practical 2: method

From temperatures to ΔH₁ and ΔH₂ 6.7

The results below are illustrative: they are typical of a good class set, not measured data. The temperature changes are the corrected ΔT from cooling-line graphs, as on the earlier page.

	Reaction 1: Mg + HCl	Reaction 2: MgO + HCl
Mass of solid	0.120 g	0.400 g
Corrected ΔT	11.0 °C	6.9 °C
q = 50.0 × 4.18 × ΔT	2299 J	1442 J
Moles of solid	0.120 ÷ 24.3 = 0.00494 mol	0.400 ÷ 40.3 = 0.00993 mol
ΔH = −q ÷ n	−465.5 kJ mol⁻¹	−145.3 kJ mol⁻¹

Worked exampleThe two enthalpy changes

Reaction 1: q = 50.0 × 4.18 × 11.0 = 2299 J = 2.299 kJ. n(Mg) = 0.120 ÷ 24.3 = 0.004938 mol. ΔH₁ = −2.299 ÷ 0.004938 = −465.5 kJ mol⁻¹.

Reaction 2: q = 50.0 × 4.18 × 6.9 = 1442 J = 1.442 kJ. n(MgO) = 0.400 ÷ 40.3 = 0.009926 mol. ΔH₂ = −1.442 ÷ 0.009926 = −145.3 kJ mol⁻¹.

Both temperatures rose, so both signs are negative.

The enthalpy cycle 6.76.66.6(ii)

A square cycle. The top arrow is the reaction you want; the other three sides are known.

The top arrow is the formation of MgO. The route from Mg to MgO round the bottom goes down the left arrow, along the bottom, and back up the right arrow. That last step is against ΔH₂.

\Delta_\text{f}H(\text{MgO}) = \Delta H_1 + \Delta_\text{f}H(\text{H}_2\text{O}) - \Delta H_2

Worked exampleEnthalpy change of formation of magnesium oxide

Down the left, with the arrow: ΔH₁ = −465.5.
Along the bottom, with the arrow: $\Delta_\text{f}H$ (H₂O(l)) = −285.8 (from a data book).
Up the right, against the arrow: −ΔH₂ = −(−145.3) = +145.3.
$\Delta_\text{f}H$ (MgO) = −465.5 + (−285.8) + 145.3 = −606.0 kJ mol⁻¹, which is −606 kJ mol⁻¹ to 3 significant figures.

You can get the same answer by adding equations. Keep reaction 1, reverse reaction 2 (so ΔH₂ changes sign) and add the formation of water:

Equation	ΔH / kJ mol⁻¹
$\ce{Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g)}$	−465.5
$\ce{MgCl2(aq) + H2O(l) -> MgO(s) + 2HCl(aq)}$	+145.3
$\ce{H2(g) + 1/2 O2(g) -> H2O(l)}$	−285.8
Sum: $\ce{Mg(s) + 1/2 O2(g) -> MgO(s)}$	−606.0

On adding, MgCl₂(aq), 2HCl(aq), H₂(g) and H₂O(l) each appear once on each side and cancel, leaving the formation of MgO.

Evaluating the result 6.8

The data-book value of $\Delta_\text{f}H$ (MgO) is about −602 kJ mol⁻¹. The difference is 606 − 602 = 4 kJ mol⁻¹, a percentage difference of 4 ÷ 602 × 100 = 0.7%, under 1%. That is a very good result for a cup and a thermometer.

Source of error or uncertainty	Effect	Improvement
Heat lost while the reaction runs (faster for Mg, slower for MgO)	ΔT too small, so both ΔH values are too small in size	lid and stirring; cooling-curve correction for both reactions
Mg ribbon is coated with a layer of oxide	less Mg reacts than weighed, so the moles are overstated and ΔH₁ is too small in size	clean the ribbon or use magnesium powder
MgO has absorbed water and carbon dioxide from the air	the mass of MgO overstates the moles, so ΔH₂ is too small in size	use fresh MgO and keep the bottle closed
Mass of Mg: balance ±0.001 g, two readings	±0.002 g on 0.120 g is 1.7%	use a larger mass of Mg with excess acid and a bigger ΔT
Thermometer ±0.05 °C, two readings	±0.1 °C on 11.0 °C is 0.9%; on 6.9 °C is 1.4%	use a thermometer with finer divisions

The temperature change for MgO is smaller, so the thermometer uncertainty is a bigger fraction of it.

Worked exampleIs the difference from the data book significant?

Combine the uncertainties of the two measurements that matter most.

ΔH₁: Mg mass 1.7% + ΔT 0.9% = 2.6%. 2.6% of 465.5 = ±12 kJ mol⁻¹.
ΔH₂: MgO mass 0.5% + ΔT 1.4% = 1.9%. 1.9% of 145.3 = ±2.8 kJ mol⁻¹.
The cycle adds ΔH₁ and subtracts ΔH₂, so the absolute uncertainties add: ±(12 + 2.8) = about ±15 kJ mol⁻¹.
The result is −606 ± 15. The data-book value, −602, is inside this range, so the difference of 4 kJ mol⁻¹ is smaller than the uncertainty in the experiment.

Check yourself

In a repeat, 0.150 g of magnesium raises the temperature of 50.0 cm³ of acid by 13.8 °C. Calculate ΔH₁.
In the same repeat, 0.500 g of magnesium oxide raises the temperature of 50.0 cm³ of acid by 8.6 °C. Calculate ΔH₂.
Use your answers to questions 1 and 2, and $\Delta_\text{f}H$ (H₂O(l)) = −285.8 kJ mol⁻¹, to calculate $\Delta_\text{f}H$ (MgO).
Which of the two temperature changes (11.0 °C or 6.9 °C) has the greater percentage uncertainty from a thermometer reading to ±0.05 °C, and why?
The student's result is −606 kJ mol⁻¹ and the data-book value is −602 kJ mol⁻¹. Calculate the percentage difference and explain whether the experiment is likely to have a systematic error.
The magnesium ribbon used in an experiment is coated with a thin layer of magnesium oxide. State and explain the effect on the calculated $\Delta_\text{f}H$ (MgO).

Answers cover these until you have tried the questions

q = 50.0 × 4.18 × 13.8 = 2884.2 J = 2.8842 kJ. n(Mg) = 0.150 ÷ 24.3 = 0.006173 mol. ΔH₁ = −2.8842 ÷ 0.006173 = −467 kJ mol⁻¹.
q = 50.0 × 4.18 × 8.6 = 1797.4 J = 1.7974 kJ. n(MgO) = 0.500 ÷ 40.3 = 0.01241 mol. ΔH₂ = −1.7974 ÷ 0.01241 = −145 kJ mol⁻¹.
$\Delta_\text{f}H$ (MgO) = ΔH₁ + $\Delta_\text{f}H$ (H₂O) − ΔH₂ = −467 + (−285.8) − (−145) = −608 kJ mol⁻¹.
The 6.9 °C change. The uncertainty is ±0.1 °C for both, but 0.1 ÷ 6.9 × 100 = 1.4% is larger than 0.1 ÷ 11.0 × 100 = 0.9%. A smaller temperature change makes the same uncertainty a larger fraction.
(606 − 602) ÷ 602 × 100 = 0.7%. This is smaller than the combined uncertainty (about ±15 kJ mol⁻¹ or 2.5%), so there is no evidence of a systematic error beyond the limits of the apparatus.
Some of the "magnesium" that was weighed is oxide, which gives out less heat per gram than magnesium does. The moles of magnesium are therefore overstated, so ΔH₁ comes out too small in size (less negative). Since ΔH₁ is added in the cycle, the calculated $\Delta_\text{f}H$ (MgO) is less negative than the true value.