Energy Changes and Energy Profile Diagrams for Chemical Reactions

Lessons

Senior 4 · Chemistry

Energy Changes and Energy Profile Diagrams for Chemical Reactions

4 lessons

Lesson 2 of 4

Internal Energy and the First Law of Thermodynamics

30 min

Warming water without a flame

In the 1840s James Prescott Joule let falling weights turn a paddle wheel inside a closed can of water. The churning water got slightly warmer, although no flame was used. Doing work on the water had the same effect as heating it: both raised its temperature. Joule's experiments showed that heat and work are two ways of transferring energy, and that a fixed amount of work always produces the same rise in temperature. This lesson names the energy stored inside a system, its internal energy, and states the law that connects it to heat and work.

Joule's apparatus of 1845 in the Science Museum, London. Falling weights (not shown) turned the brass paddle wheel (left) inside the water-filled container (right) by means of the cord wound on the drum; the work done on the water raised its temperature. Photo: Gaius Cornelius, public domain, via Wikimedia Commons

Definition

Internal energy, U: the total energy of all the particles in a system: the kinetic energy of their motion (moving, rotating, vibrating) and the potential energy stored in chemical bonds and in the forces between particles. Only changes, ΔU, can be measured.

First law of thermodynamics: energy cannot be created or destroyed; the change in internal energy of a system equals the heat supplied to it plus the work done on it: $\Delta U = q + w$ .

Enthalpy, H: $H = U + p V$ . For a change at constant pressure, the enthalpy change equals the heat transferred: $\Delta H = q_p$ .

Figure 17.2: Heat flowing into a gas raises its internal energy; work done by the gas as it expands takes energy out. The first law adds the two.

Signs

Chemists (following IUPAC) count everything from the point of view of the system:

Quantity	Positive (+) when	Negative (−) when
q (heat)	heat flows into the system	heat flows out of the system
w (work)	work is done on the system (it is compressed)	the system does work on the surroundings (it expands)
ΔU	the system gains energy	the system loses energy

For an isolated system q = 0 and w = 0, so $\Delta U = 0$ : the energy of the universe is constant. A machine that produces work forever without an energy supply is therefore impossible.

Work done by a gas

When a reaction produces gas against a constant outside pressure p, the gas pushes back the atmosphere and the system does work: $w = -p\,\Delta V$ with p in Pa, ΔV in m³ and w in J. If the volume cannot change (a sealed steel "bomb" calorimeter), w = 0 and $\Delta U = q_V$ the heat measured at constant volume.

From internal energy to enthalpy

Most reactions are carried out in open vessels at the constant pressure of the atmosphere. Then the heat transferred is not ΔU but the enthalpy change: $\Delta H = \Delta U + p\Delta V$ For a reaction involving gases, $p\Delta V = \Delta n_{\text{gas}} RT$ , where $\Delta n_{\text{gas}}$ is the change in the number of moles of gas: $\Delta H = \Delta U + \Delta n_{\text{gas}} RT$ When no gas is formed or used up, ΔH and ΔU are almost equal.

Worked example 1: heat in, work out

A gas in a cylinder absorbs 500 J of heat and, as it expands, does 200 J of work pushing back a piston. Calculate ΔU.

Solution. Heat enters the system: $q = +500\ \mathrm{J}$ . The system does work on the surroundings: $w = -200\ \mathrm{J}$ . $\Delta U = q + w = 500 + (-200) = +300\ \mathrm{J}$ The internal energy of the gas rises by 300 J; the other 200 J left as work.

Worked example 2: work done by a reaction that makes gas

0.100 mol of magnesium reacts with excess dilute hydrochloric acid in an open flask at 25 °C (298.15 K) and 101 325 Pa: $\mathrm{Mg(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + H_2(g)}$ . Calculate the work done by the system. ( $R = 8.314\ \mathrm{J\,K^{-1}\,mol^{-1}}$ )

Solution. 0.100 mol of $\mathrm{H_2}$ is formed. Its volume: $V = \frac{nRT}{p} = \frac{0.100 \times 8.314 \times 298.15}{101\,325} = 2.45 \times 10^{-3}\ \mathrm{m^3}$ $w = -p\Delta V = -101\,325 \times 2.45 \times 10^{-3} = -248\ \mathrm{J}$ (equal to $-\Delta n RT$ ). The hydrogen pushes back the atmosphere, so 248 J of the reaction's energy leaves as work rather than as heat.

Worked example 3: ΔH and ΔU for burning methane

Use standard enthalpies of formation ( $\mathrm{CH_4(g)}$ −74.8, $\mathrm{CO_2(g)}$ −393.5, $\mathrm{H_2O(l)}$ −285.8 kJ mol⁻¹) to find ΔH for $\mathrm{CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)}$ at 298.15 K, then find ΔU.

Solution. $\Delta H = [-393.5 + 2(-285.8)] - [-74.8 + 0] = -890.3\ \mathrm{kJ\,mol^{-1}}$ Moles of gas: 1 ( $\mathrm{CO_2}$ ) after, 3 ( $\mathrm{CH_4} + 2\mathrm{O_2}$ ) before, so $\Delta n_{\text{gas}} = -2$ . $\Delta U = \Delta H - \Delta n_{\text{gas}} RT = -890.3 - (-2)(8.314 \times 10^{-3})(298.15) = -890.3 + 4.96 = -885.3\ \mathrm{kJ\,mol^{-1}}$ At constant pressure the gas volume shrinks, the atmosphere does about 5 kJ of work on the system, and that energy also leaves as heat, so slightly more heat (890.3 kJ) is released than at constant volume (885.3 kJ).

Warning

Common mistakes. (1) Mixing sign conventions: some physics books write $\Delta U = Q - W$ , with W the work done by the system. In chemistry, with w the work done on the system, $\Delta U = q + w$ . (2) Forgetting the minus sign in $w = -p\Delta V$ : an expanding gas does work, so w is negative. (3) Using dm³ with Pa: convert to m³ (1 dm³ = 10⁻³ m³), or use kPa with dm³. (4) Counting liquids and solids in $\Delta n_{\text{gas}}$ : only gases count.

Safety. Never heat a sealed container: as the internal energy of the gas inside rises, so does its pressure, and the container can explode. Reactions that make gas are done in open or vented apparatus.

Tip

ΔU = q + w, "gains are positive". Heat in and work on the system both add to U. At constant volume ΔU = q; at constant pressure ΔH = q. They differ by $\Delta n_{\text{gas}} RT$ , about 2.5 kJ per mole of gas at 25 °C.

The Organic Chemistry Tutor: First Law of Thermodynamics, Basic Introduction (11 min). Watch for the signs of q and w, and the worked problems on internal energy, heat and work.

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Practice

State the first law of thermodynamics in words and as an equation, defining each symbol.
A system loses 250 J of heat and has 400 J of work done on it. Calculate ΔU.
A gas expands from 2.00 dm³ to 5.00 dm³ against a constant pressure of 100 kPa. Calculate the work done, with its sign.
Explain why ΔU = q in a sealed bomb calorimeter.
For $\mathrm{N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)}$ , ΔH = −92.2 kJ mol⁻¹ at 298 K. Calculate ΔU (R = 8.314 J K⁻¹ mol⁻¹).
For which of these reactions are ΔH and ΔU almost equal? (a) $\mathrm{H_2(g) + Cl_2(g) \rightarrow 2HCl(g)}$ ; (b) $\mathrm{CaCO_3(s) \rightarrow CaO(s) + CO_2(g)}$ . Explain.
The decomposition $\mathrm{CaCO_3(s) \rightarrow CaO(s) + CO_2(g)}$ has ΔH = +178.3 kJ mol⁻¹. Calculate ΔU at 1173 K, the temperature of a lime kiln.
Explain, using the first law, why a bicycle pump gets warm when you pump up a tyre quickly.

Show answersHide answers

Energy is conserved: the change in internal energy of a system equals the heat supplied to it plus the work done on it, $\Delta U = q + w$ .
$q = -250\ \mathrm{J}$ , $w = +400\ \mathrm{J}$ ; $\Delta U = +150\ \mathrm{J}$ .
$\Delta V = 3.00\ \mathrm{dm^3} = 3.00 \times 10^{-3}\ \mathrm{m^3}$ ; $w = -p\Delta V = -1.00 \times 10^{5} \times 3.00 \times 10^{-3} = -300\ \mathrm{J}$ .
The volume cannot change, so no expansion work is done (w = 0) and $\Delta U = q + 0 = q_V$ .
$\Delta n_{\text{gas}} = 2 - 4 = -2$ ; $\Delta U = \Delta H - \Delta nRT = -92.2 - (-2)(8.314 \times 10^{-3})(298) = -92.2 + 4.96 = -87.2\ \mathrm{kJ\,mol^{-1}}$ .
(a): two moles of gas on each side, so $\Delta n_{\text{gas}} = 0$ and ΔH = ΔU. In (b) one mole of gas forms, so they differ by RT.
$\Delta n_{\text{gas}} = +1$ ; $\Delta U = 178.3 - (1)(8.314 \times 10^{-3})(1173) = 178.3 - 9.75 = 168.5\ \mathrm{kJ\,mol^{-1}}$ .
Pushing the handle does work on the air (w > 0). The compression is fast, so little heat escapes (q ≈ 0); $\Delta U = q + w$ is positive, the internal energy of the air rises and its temperature rises.

Quick check

Key points

Internal energy U is the total kinetic and potential energy of the particles of a system; only ΔU can be measured.
First law: $\Delta U = q + w$ (q: heat to the system; w: work done on the system). An isolated system has ΔU = 0.
Expansion work: $w = -p\Delta V$ . At constant volume, $\Delta U = q_V$ ; at constant pressure, $\Delta H = q_p$ .
$\Delta H = \Delta U + \Delta n_{\text{gas}} RT$ ; for methane combustion, ΔH = −890.3 and ΔU = −885.3 kJ mol⁻¹.

Sources and credits

Image: Joule's heat apparatus.JPG by Gaius Cornelius, via Wikimedia Commons, public domain.
Enthalpies of formation from Standard enthalpy of formation, Wikipedia contributors, licensed under CC BY-SA 4.0.
Video: First Law of Thermodynamics, Basic Introduction - Internal Energy, Heat and Work - Chemistry, The Organic Chemistry Tutor, embedded from YouTube.
The text, diagram, table, worked examples and Practice items were written and drawn by Ganzaa.
This lesson uses CC BY-SA material and is shared under CC BY-SA 4.0.

1Warming water without a flame

2Signs

3Work done by a gas

4From internal energy to enthalpy

Warming water without a flame

Signs

Work done by a gas

From internal energy to enthalpy