6.6 - Introduction to Enthalpy of Reaction

Overview

Every chemical reaction either releases energy or absorbs it, and the enthalpy of reaction,

\Delta H

, tells you how much. This topic connects

\Delta H

to something you can measure: the heat

q

that flows when a known amount of a substance reacts. On the AP exam you'll move between grams, moles, and kilojoules using a balanced equation and its

\Delta H

, and you'll explain where the energy comes from at the particle level.

What the enthalpy of reaction tells you

The enthalpy change of a reaction,

\Delta H_{\mathrm{rxn}}

, is the heat energy a reaction releases or absorbs at constant pressure. Almost every reaction you'll see runs open to the air (a beaker, a foam cup, a flask), so the pressure stays constant and

\Delta H

is simply the heat of the reaction.

Negative $\Delta H$ : exothermic. The reaction releases heat to its surroundings.

Positive $\Delta H$ : endothermic. The reaction absorbs heat from its surroundings.

As in Topic 6.1, the sign is from the point of view of the system (the reacting chemicals). A negative

\Delta H

means the system lost energy and the surroundings gained it.

You may hear chemists talk about internal energy too. The technical difference between enthalpy and internal energy isn't tested on the AP exam. At constant pressure, treat

\Delta H

as the heat of the reaction.

Thermochemical equations

A thermochemical equation is a balanced equation with its

\Delta H

written next to it:

\mathrm{CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)} \qquad \Delta H^\circ = -890.3\ \mathrm{kJ/mol_{rxn}}

The unit

\mathrm{kJ/mol_{rxn}}

means kilojoules per mole of reaction: the heat for the reaction happening once in mole amounts, exactly as written. Here, 1 mol of

\mathrm{CH_4}

reacting with 2 mol of

\mathrm{O_2}

to make 1 mol of

\mathrm{CO_2}

and 2 mol of

\mathrm{H_2O}

releases 890.3 kJ. So the same equation tells you that 890.3 kJ is released per 1 mol of

\mathrm{CH_4}

burned, per 2 mol of

\mathrm{O_2}

used, and per 2 mol of water made. The coefficients are part of the meaning.

The degree sign in

\Delta H^\circ

means standard conditions (1 atm, 1 M for solutions, usually 25 °C). Topic 6.8 says more about that. Two more things follow from "per mole of reaction as written":

$\Delta H$ belongs to the equation as written. Double the equation and $\Delta H$ doubles. Reverse it and the sign flips. Topic 6.9 uses these rules.

Phases matter. The $-890.3\ \mathrm{kJ/mol_{rxn}}$ above is for liquid water. If the water ends up as a gas, less heat is released (Topic 6.8 shows how much).

Where the energy comes from

Chemical potential energy is stored in the arrangement of atoms and electrons, meaning in bonds and the attractions between particles. During a reaction, bonds in the reactants break and new bonds form in the products (Topic 6.7 counts them). Because the bonds are different, the products have a different chemical potential energy than the reactants.

Energy is conserved, so that difference has to go somewhere. It shows up as a change in the kinetic energy of the particles, which you see as a temperature change:

Exothermic: the products have lower potential energy than the reactants. The lost potential energy becomes kinetic energy, so the particles move faster and the temperature rises.

Endothermic: the products have higher potential energy than the reactants. The energy to raise the potential energy comes from the particles' kinetic energy, so they slow down and the temperature drops.

So a temperature change in a reaction mixture is the visible sign that chemical potential energy changed.

Reaching thermal equilibrium with the surroundings

Right after the reaction, the products are at a different temperature than their surroundings. Just like any two objects at different temperatures (Topic 6.3), they exchange thermal energy until they reach thermal equilibrium, the same temperature:

Exothermic: the warm products transfer thermal energy to the surroundings. The surroundings warm up.

Endothermic: the cool products take in thermal energy from the surroundings. The surroundings cool down.

6.6 - Introduction to Enthalpy of Reaction

The graphs show a reaction mixture in an open, uninsulated cup. The quick change is the reaction itself, turning potential energy into kinetic energy (or the reverse). The slow return to room temperature is heat flowing between the mixture and the room.

\Delta H

counts all the heat that flows by the time everything is back at the starting temperature. That's why, for an exothermic reaction, the heat the surroundings gain equals the energy the reaction released.

Calculating heat from moles

\Delta H^\circ

gives the heat per mole of reaction. To find the heat for any amount of a reactant or product, convert to moles, then use the coefficient of that substance as a ratio:

q = n_\mathrm{A} \times \frac{\Delta H^\circ}{a}

Here

n_\mathrm{A}

is the moles of substance A that react (or form), and

a

is A's coefficient in the balanced equation. Dividing

n_\mathrm{A}

a

gives the moles of reaction,

n_{\mathrm{rxn}}

, so this is the same as

q = n_{\mathrm{rxn}} \times \Delta H^\circ

Sign: $q$ for the reaction has the same sign as $\Delta H$ . The surroundings gain or lose the same amount with the opposite sign: $q_{\mathrm{surr}} = -q_{\mathrm{rxn}}$ .

Wording: if a question asks "how much heat is released," answer with a positive number and the word released (554 kJ released), or write $q = -554$ kJ. Don't write " $-554$ kJ released": that double negative says energy was absorbed.

Limiting reactant: if amounts of two reactants are given, the heat depends on the one that runs out. Use its moles.

Example 1: Heat from burning propane

Propane burns in a gas grill:

\mathrm{C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l)} \qquad \Delta H^\circ = -2219.9\ \mathrm{kJ/mol_{rxn}}

How much heat is released when 11.0 g of propane burns completely?

Step 1: Find moles of propane.

\mathcal{M}_{\mathrm{C_3H_8}} = 3(12.01) + 8(1.008) = 44.09\ \mathrm{g/mol}

n = \frac{11.0\ \mathrm{g}}{44.09\ \mathrm{g/mol}} = 0.2495\ \mathrm{mol\ C_3H_8}

(Keep an extra digit until the end.)

Step 2: Use the ratio from the equation. The coefficient of

\mathrm{C_3H_8}

is 1, so each mole of propane goes with 2219.9 kJ.

q = 0.2495\ \mathrm{mol\ C_3H_8} \times \frac{-2219.9\ \mathrm{kJ}}{1\ \mathrm{mol\ C_3H_8}} = -554\ \mathrm{kJ}

Answer: burning 11.0 g of propane releases 554 kJ of heat (

q = -554

kJ, three significant figures from 11.0 g). The surroundings gain 554 kJ.

Example 2: An iron hand warmer

Inside an air-activated hand warmer, iron powder reacts with oxygen from the air. The reaction can be modeled as:

\mathrm{4Fe(s) + 3O_2(g) \rightarrow 2Fe_2O_3(s)} \qquad \Delta H^\circ = -1648.4\ \mathrm{kJ/mol_{rxn}}

(a) A hand warmer contains 22.0 g of iron. How much heat can it release?

n_{\mathrm{Fe}} = \frac{22.0\ \mathrm{g}}{55.85\ \mathrm{g/mol}} = 0.3939\ \mathrm{mol\ Fe}

The 1648.4 kJ goes with 4 mol of iron, so divide by the coefficient:

q = 0.3939\ \mathrm{mol\ Fe} \times \frac{-1648.4\ \mathrm{kJ}}{4\ \mathrm{mol\ Fe}} = -162\ \mathrm{kJ}

Answer: the iron can release 162 kJ of heat.

(b) What mass of iron must react to release 50.0 kJ? Work the road map backward:

n_{\mathrm{Fe}} = 50.0\ \mathrm{kJ} \times \frac{4\ \mathrm{mol\ Fe}}{1648.4\ \mathrm{kJ}} = 0.1213\ \mathrm{mol\ Fe}

m = 0.1213\ \mathrm{mol} \times 55.85\ \mathrm{g/mol} = 6.78\ \mathrm{g\ Fe}

Answer: 6.78 g of iron. Here we used the magnitude of the heat (50.0 kJ released), so the moles come out positive.

Example 3: Finding the enthalpy of reaction from a temperature change (AP-style)

A student adds excess zinc powder to 50.0 mL of 0.200 M

\mathrm{CuSO_4}(aq)

in a coffee-cup calorimeter. The reaction is:

\mathrm{Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)}

The temperature of the solution rises from 21.6 °C to 32.0 °C. Assume the solution has a density of 1.00 g/mL and a specific heat of 4.18 J/(g·°C), and that no heat is lost to the cup or the air.

(a) Calculate the heat absorbed by the solution.

m = 50.0\ \mathrm{mL} \times 1.00\ \mathrm{g/mL} = 50.0\ \mathrm{g} \qquad \Delta T = 32.0\ ^\circ\mathrm{C} - 21.6\ ^\circ\mathrm{C} = 10.4\ ^\circ\mathrm{C}

q_{\text{solution}} = mc\Delta T = (50.0\ \mathrm{g})(4.18\ \mathrm{J/(g\cdot{}^{\circ}C)})(10.4\ ^\circ\mathrm{C}) = 2.17 \times 10^{3}\ \mathrm{J} = 2.17\ \mathrm{kJ}

(b) Calculate $\Delta H_{\mathrm{rxn}}$ in $\mathrm{kJ/mol_{rxn}}$ . The solution gained 2.17 kJ, so the reaction released it:

q_{\mathrm{rxn}} = -2.17

kJ. Zinc is in excess, so

\mathrm{Cu^{2+}}

is the limiting reactant:

n_{\mathrm{Cu^{2+}}} = (0.200\ \mathrm{mol/L})(0.0500\ \mathrm{L}) = 0.0100\ \mathrm{mol}

One mole of reaction uses 1 mol of

\mathrm{Cu^{2+}}

, so this is 0.0100 mol of reaction:

\Delta H_{\mathrm{rxn}} = \frac{q_{\mathrm{rxn}}}{n_{\mathrm{rxn}}} = \frac{-2.17\ \mathrm{kJ}}{0.0100\ \mathrm{mol_{rxn}}} = -217\ \mathrm{kJ/mol_{rxn}}

(c) The student repeats the experiment with 100.0 mL of the same $\mathrm{CuSO_4}$ solution and excess zinc. Predict $\Delta T$ and justify. About 10.4 °C again. Twice as much

\mathrm{Cu^{2+}}

reacts, so twice as much heat is released, but that heat warms twice as much solution. Since

\Delta T = \frac{q}{mc}

and both

q

and

m

double,

\Delta T

doesn't change. (

\Delta H

per mole of reaction doesn't change either.)

(d) Explain, in terms of energy, why the temperature rose. The products,

\mathrm{Zn^{2+}}(aq)

and

\mathrm{Cu}(s)

, have lower chemical potential energy than the reactants. As the reaction happened, that potential energy became kinetic energy of the particles, so the solution's temperature went up.

Common mistakes

Forgetting to divide by the coefficient. $\Delta H$ is per mole of reaction as written, not per mole of every substance. In Example 2, 1648.4 kJ goes with 4 mol of iron.

Mixing up system and surroundings. If the water around a reaction warms up, the surroundings gained heat, so $q_{\mathrm{rxn}}$ and $\Delta H$ are negative.

Double negatives. " $-554$ kJ released" means energy was absorbed. Write "554 kJ released" or " $q = -554$ kJ."

Using the excess reactant. The heat depends on how much of the limiting reactant reacts.

Mixing J and kJ. $mc\Delta T$ gives joules, but $\Delta H$ values are in kilojoules. Convert before you divide.

Using the wrong mass in calorimetry. $m$ in $q = mc\Delta T$ is the mass of the solution that changes temperature, not the mass of the reactant.

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