Lesson 1 of 5Energy Changes and Rates of Reaction

Unit 3 · Lesson 1 of 5

Enthalpy Changes and Calorimetry

System versus surroundings, exothermic and endothermic changes, thermochemical equations, and measuring heat with $q = m c Δ T$ in a calorimeter to find molar enthalpies.

5 min read10 practice questionsAsk AI about this →

Key things to remember

1 $Δ H = H_{products} - H_{reactants} :$ negative means exothermic (surroundings warm), positive means endothermic (surroundings cool).
2 $q = m c Δ T$ , with $c = 4.18\text{ J/{(g}\cdot{}^\circ\text{C)}}$ for water and dilute aqueous solutions.
3In a calorimeter, $q_{reaction} = - q_{surroundings} .$
4Molar enthalpy: $Δ H_{molar} = \frac{q _{reaction}}{n}$ , and $Δ H = n Δ H_{molar}$ .
5Reversing a thermochemical equation flips the sign of $Δ H$ ; multiplying the coefficients multiplies $Δ H$ by the same factor.

A hand warmer heats up when you snap it, an instant cold pack turns icy when you squeeze it, and a barbecue runs on the heat from burning propane. Every one of these is a chemical or physical change that moves energy into or out of its surroundings. In Grade 11 you balanced equations and counted moles; now you will attach an energy value to those equations, so you can predict exactly how many kilojoules a given amount of fuel will release, or how cold a cold pack will get. The key quantity is enthalpy change, written $Δ H$ .

To track energy, chemists split the universe into two parts. The system is the set of chemicals you are studying — the reactants turning into products. The surroundings are everything else: the water they are dissolved in, the beaker, the air, your hand. Energy is conserved, so any energy the system loses is gained by the surroundings, and vice versa. That is why a thermometer placed in the surroundings can tell you what is happening inside the system.

An open system can exchange both matter and energy with the surroundings (an open beaker).
A closed system exchanges energy but not matter (a sealed flask).
An isolated system exchanges neither — the ideal a calorimeter tries to approach.

Exothermic and Endothermic Changes

Enthalpy ( $H$ ) is the total heat content of a system at constant pressure. You can never measure $H$ itself, but you can measure how much it changes: $Δ H = H_{products} - H_{reactants} .$ In an exothermic change the products have less enthalpy than the reactants, so energy leaves the system, the surroundings warm up, and $Δ H$ is negative. In an endothermic change the products have more enthalpy, energy flows in from the surroundings, the surroundings cool down, and $Δ H$ is positive.

Fig 1Enthalpy diagrams: in an exothermic change the products sit lower than the reactants; in an endothermic change they sit higher.

AExothermic

Products lower in enthalpy than reactants
$Δ H < 0$
Surroundings get warmer
Examples: combustion, neutralization, freezing, condensation

BEndothermic

Products higher in enthalpy than reactants
$Δ H > 0$
Surroundings get cooler
Examples: dissolving ammonium nitrate, photosynthesis, melting, boiling

Thermochemical Equations

A thermochemical equation is a balanced equation, with states of matter, that also reports the energy change. There are two equivalent ways to write it. You can write $Δ H$ beside the equation, or you can write the energy as a term inside the equation — on the product side for an exothermic reaction (energy is released, like a product) and on the reactant side for an endothermic reaction (energy is absorbed, like a reactant). The $Δ H$ value belongs to the equation exactly as written, in the mole amounts given by the coefficients.

CH_{4} (g) + 2 O_{2} (g) \to CO_{2} (g) + 2 H_{2} O (l) Δ H = - 890.3 kJ CH_{4} (g) + 2 O_{2} (g) \to CO_{2} (g) + 2 H_{2} O (l) + 890.3 kJ

Two ways to write the same fact: burning 1 mol of methane releases 890.3 kJ, shown as Delta H beside the equation or as an energy term on the product side.

Enthalpy scales with the equation

Double every coefficient and

Δ H

doubles. Reverse the equation and

Δ H

changes sign. The molar enthalpy,

Δ H_{molar}

in kJ/mol, is the enthalpy change per mole of one named substance, so

Δ H = n Δ H_{molar}

Measuring Heat with a Calorimeter

When heat $q$ flows into a substance, its temperature rises by an amount that depends on three things: how much of it there is (mass $m$ ), how hard it is to heat (its specific heat capacity $c$ , the energy needed to warm 1 g by 1 °C), and the temperature change $Δ T = T_{final} - T_{initial}$ . So $q = m c Δ T$ . Water has a large specific heat capacity, $c = 4.18\text{ J/{(g}\cdot{}^\circ\text{C)}}$ , which is why it is the standard liquid in a calorimeter: it absorbs a lot of energy with a modest, easily measured temperature change. A simple coffee-cup calorimeter is two nested polystyrene cups with a lid and a thermometer. Polystyrene is a good insulator, so very little heat escapes, and the reaction happens right inside the water, which acts as the surroundings. You measure the temperature change of the water, calculate $q$ for the water, and then use conservation of energy: whatever the water gained, the reaction lost. For dilute aqueous solutions, assume the solution has the density ( $1.00 g/mL$ ) and specific heat capacity of pure water.

q_{reaction} = - q_{surroundings} = - m c Δ T Δ H_{molar} = \frac{q _{reaction}}{n}

The sign flip connects what the thermometer sees to what the reaction did.

Where calorimetry goes wrong

Real calorimeters lose some heat to the air, the cup and the thermometer, and some reactions are not complete. A burner heating a can of water loses a large share of its heat, so the measured enthalpy is much smaller in size than the accepted value. More sealed designs, such as a bomb calorimeter (a sealed steel vessel immersed in water, used for combustion), use

q = C Δ T

with the measured heat capacity

C

of the whole apparatus and give far more accurate results.

Type of change	Example	Typical size of energy change
Physical (phase change)	Melting ice: $Δ H = + 6.01 kJ/mol$ ; boiling water at 100 °C: $+ 40.7 kJ/mol$	Tens of kJ per mole — only intermolecular forces change
Chemical	Burning methane: $- 890.3 kJ/mol$	Hundreds to thousands of kJ per mole — chemical bonds break and form
Nuclear	Fission of uranium-235	Roughly $1 0^{10} kJ/mol$ — millions of times larger, because nuclei change and a little mass becomes energy

Energy released or absorbed grows enormously as you go from physical to chemical to nuclear change.

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

How much heat, in kilojoules, is needed to warm

250.0 g

of water from

18.5^{\circ} C

64.0^{\circ} C

? Use

c = 4.18\text{ J/{(g}\cdot{}^\circ\text{C)}}

Example 2

In a coffee-cup calorimeter,

50.0 mL

2.00 mol/L

HCl (a q)

is mixed with

50.0 mL

2.00 mol/L

NaOH (a q)

. The temperature rises from

21.0^{\circ} C

34.5^{\circ} C

. Find the molar enthalpy of neutralization per mole of water formed. Assume the solution has a density of

1.00 g/mL

and

c = 4.18\text{ J/{(g}\cdot{}^\circ\text{C)}}

Example 3

An ethanol burner heats

200.0 g

of water in a metal can from

20.0^{\circ} C

38.5^{\circ} C

. The burner loses

0.925 g

of ethanol,

C_{2} H_{5} OH

(

M = 46.08 g/mol

). Find the molar enthalpy of combustion in J/mol, and compare it with the accepted value of

- 1367 kJ/mol

Example 4

Using

CH_{4} (g) + 2 O_{2} (g) \to CO_{2} (g) + 2 H_{2} O (l),

Δ H = - 890.3 kJ

, how much heat is released when

40.0 g

of methane (

M = 16.05 g/mol

) burns completely?

Where marks get lost

Using $Δ T = T_{initial} - T_{final}$ , which flips every sign that follows. Always take final minus initial.
Reporting the heat gained by the water as the enthalpy of the reaction without changing the sign — a temperature rise means the reaction is exothermic, so $Δ H$ must be negative.
Using only the mass of one solution in a mixing experiment instead of the total mass of everything the heat went into (both solutions, or the water plus the dissolved solid).
Mixing units: $q = m c Δ T$ with $c$ in J/(g·°C) gives joules, but molar enthalpies are usually reported in kJ/mol — convert before comparing.
Forgetting that $Δ H$ in a thermochemical equation refers to the coefficients as written, so burning 2 mol of fuel releases twice the listed energy.

Flashcards 8

System vs. surroundings

Sign of Delta H for an exothermic change

Sign of Delta H for an endothermic change

Heat equation

+ 4 more

Practice

A solid dissolves in water and the temperature of the solution drops from

22.0^{\circ} C

17.5^{\circ} C

. Which statement is correct?

How much heat, in kJ, does

150.0 g

of water absorb when its temperature rises by

12.0^{\circ} C

? Give 3 significant figures. Use

c = 4.18\text{ J/{(g}\cdot{}^\circ\text{C)}}

8.00 g

of ammonium nitrate,

NH_{4} NO_{3}

(

M = 80.06 g/mol

), dissolves in

100.0 g

of water in a calorimeter. The temperature falls from

22.00^{\circ} C

16.30^{\circ} C

. Taking the solution's mass as

108.0 g

and

c = 4.18\text{ J/{(g}\cdot{}^\circ\text{C)}}

, find the molar enthalpy of solution in kJ/mol (include the sign; 3 significant figures).

kJ/mol

Dissolving sodium hydroxide has

Δ H = - 44.5 kJ/mol

. If

5.00 g

NaOH

(

M = 40.00 g/mol

) dissolves in water to give

105.0 g

of solution, by how many degrees Celsius does the temperature rise? Assume no heat is lost and

c = 4.18\text{ J/{(g}\cdot{}^\circ\text{C)}}

. Give 3 significant figures.

°C

Using

CH_{4} (g) + 2 O_{2} (g) \to CO_{2} (g) + 2 H_{2} O (l),

Δ H = - 890.3 kJ

, how many kilojoules of heat are released when

2.50 mol

of methane burns? Round to the nearest 10 kJ.

Which thermochemical equation correctly shows an endothermic reaction?

What is the name for the energy needed to raise the temperature of 1 g of a substance by 1 °C?

Put the steps for finding a molar enthalpy from coffee-cup calorimetry data in order.

1Calculate the heat gained by the solution with $q = m c Δ T$
2Divide the heat of the reaction by the moles of the limiting or named substance
3Change the sign to get the heat of the reaction
4Calculate $Δ T$ as final minus initial temperature

Match each change to its type of energy change.

Water boiling

Propane burning in a barbecue

Uranium-235 splitting in a reactor

Water vapour condensing

A student's measured molar enthalpy of combustion of a fuel is

- 720 kJ/mol

, but the accepted value is

- 1367 kJ/mol

. Which is the most likely cause?

Spot a mistake on this page? Let us know