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← General Chemistry I

9. Energy Changes in Chemical Reactions

Where does the heat of a reaction come from?

Do thermochemical calculations: heat, enthalpy, calorimetry and Hess's law.

ExperimentsHeat is energyWeights fall, water warms. What else can we weigh by its warming?
Experiment 1 · Joule, 1845

Work becomes heat

A falling weight does work. A hot body has heat. James Joule asked whether these are two different things or one, and he settled it with a tank of water.

In the 1840s heat was still often pictured as an invisible fluid (the "caloric") that flowed from hot to cold. A brewer's son from Manchester, James Prescott Joule, doubted it. He built a tank of water with a paddle wheel driven by falling weights, and measured the temperature to a fraction of a hundredth of a degree. Drop the weights, watch the water warm, and compare the work with the heat.
102.0 m20.000 °C1.00 kg of water
Weight
Step 1 of 2 · Warm the waterDrop the weight and watch the thermometer. A single fall barely moves it; Joule needed many. Warm the water by at least 1 K.
Falls0
Work done0 J
Warming0.000 K
The maths
Work of one fall w = m g h. Heat gained by the water q = m c ΔT. If work becomes heat, they are equal.
One fall of 10 kg through 2.0 m: 10 × 9.81 × 2.0 = 196.1 J of work.
After 0 falls: work = 0 J. Water: 1000 g × 4.181 J/(g K) × 0.000 K = 0 J.
For scale: 0.047 K per fall of 10 kg. Joule's thermometers read to 1/200 of a degree Fahrenheit.
Questions you might have (3)
Why did Joule need such a sensitive thermometer?

Because a weight falling the height of a room warms a kilogram of water by only about a tenth of a degree. He repeated the fall many times and read thermometers to about a two-hundredth of a degree Fahrenheit. Even the warmth of the room or of his own hand could spoil a result, so he took great care over it.

What is a calorie, and why is a food Calorie so much bigger?

One calorie warms one gram of water by one kelvin, which is 4.184 J. A food "Calorie" (capital C) is 1000 of those, a kilocalorie, which is what a kilogram of water would need to warm by one kelvin. The joule (J) is the SI unit and the one used in chemistry.

What does the sign of q mean?

The sign is from the point of view of the system you are watching. Heat flowing into it is positive and heat flowing out is negative. A reaction that warms the water around it is losing energy, so its q is negative (exothermic). One that cools the water around it, like a cold pack, takes energy in (endothermic).

Run the experiment to see what it showed.
Experiment 2 · Black, 1760s; Dulong and Petit, 1819

Whose metal is it?

The same heat warms a gram of lead far more than a gram of water. What does a material's specific heat say about it, and can you name a metal by it?

Joseph Black, a Scottish chemist, saw around 1760 that equal weights of different substances need different amounts of heat to rise by the same number of degrees, and he called this a body's capacity for heat. Four unlabelled metal samples sit in a boiling-water bath at 100.0 °C. Drop each, 50.0 g, into 100.0 g of water at 22.0 °C in a foam cup and watch the temperature settle. From where it stops, work out which metal it is.
50 g at 100.0 °C22.00 °C100 g water at 22.0 °C
Step 1 of 2 · MeasurePick a sample and drop it in. The water warms as the metal cools, until both are at the same temperature.
Final temperature—
Specific heat found—
Named0 / 4
The maths
Heat lost by the metal = heat gained by the water: m_s c_s (T_s − T_f) = m_w c_w (T_f − T_w) → c_s = m_w c_w (T_f − T_w) ÷ [m_s (T_s − T_f)]
Questions you might have (3)
Why do you stir and use a foam cup?

Stirring brings the whole of the water to one temperature, so the thermometer reads the temperature of everything. Foam is a poor conductor, so little heat leaks to the room; the experiment treats the cup as a perfect wall. In a real measurement the cup itself absorbs a little heat and one calibrates for it (the calorimeter constant).

Why is water so slow to change temperature?

At 4.18 J/(g K) its specific heat is among the highest of common substances, because hydrogen bonds absorb energy as well as the motion of the molecules. Coastal weather is mild for the same reason: a large body of water takes a long time to warm or cool.

Does Dulong and Petit work for every element?

For most solid metals at room temperature, yes, to within about ten per cent. It fails for light, hard elements such as diamond, beryllium and boron, whose atoms are bound so stiffly that they cannot soak up heat that easily at 25 °C. Quantum theory explained the failure only after 1900.

Run the experiment to see what it showed.
TheoryAdding reactionsThe heat of a reaction nobody can run.
Theory 1 · Hess, 1840

Energy adds up like arithmetic

Some reactions cannot be run on their own, and some make a mess of products. Can you know the heat of a reaction nobody can measure?

In 1840 Germain Hess measured the heat given off when sulfuric acid takes up water and then is neutralized, by different stages, and noticed that the total always came out the same. Some reactions cannot be measured at all: burn carbon in too little air and you get a mixture of CO and CO₂. Hess's idea lets you get round that. Each problem gives you reactions whose heats are known and a target; reverse them (the heat changes sign) and scale them (the heat scales too) until they add up to the target.
Carbon to carbon monoxide
TargetC(s) + ½O₂(g) → CO(g)
C(s) + O₂(g) → CO₂(g) ΔH = -393.5 kJ
×
CO(g) + ½O₂(g) → CO₂(g) ΔH = -283.0 kJ
×
Your steps added up
—
ΔH of the sum: —
Thermite
Target2Al(s) + Fe₂O₃(s) → Al₂O₃(s) + 2Fe(s)
2Al(s) + 3⁄2O₂(g) → Al₂O₃(s) ΔH = -1675.7 kJ
×
2Fe(s) + 3⁄2O₂(g) → Fe₂O₃(s) ΔH = -824.2 kJ
×
Your steps added up
—
ΔH of the sum: —
Burning limestone
TargetCaCO₃(s) → CaO(s) + CO₂(g)
Ca(s) + C(s) + 3⁄2O₂(g) → CaCO₃(s) ΔH = -1207.0 kJ
×
Ca(s) + ½O₂(g) → CaO(s) ΔH = -634.9 kJ
×
C(s) + O₂(g) → CO₂(g) ΔH = -393.5 kJ
×
Your steps added up
—
ΔH of the sum: —
Step 1 of 2 · Add the stepsStart with carbon to CO. Set how many of each step to take (× ½, 1 or 2) and whether to reverse it, so that everything except the target cancels.
Targets reached0 / 3
Carbon → CO—
Limestone—
The maths
Hess's law: ΔH(sum of reactions) = sum of the ΔH values. Reverse a reaction → change the sign of ΔH. Multiply it → multiply ΔH by the same number.
Carbon to CO: -393.5 + (−1)(-283.0) = -110.5 kJ
Thermite: -1675.7 − (-824.2) = -851.5 kJ
Limestone: −(-1207.0) + (-634.9) + (-393.5) = +178.6 kJ
Questions you might have (3)
Why does the order of the steps not matter?

Because the energy of a substance is a fixed property of its state (what it is, how much, how hot, at what pressure), not of its history. Two paths between the same start and the same finish change the energy by the same amount. A quantity like this is called a state function. Heat and work separately are not: how much of each you get depends on the path, but their sum does not.

Why does reversing a reaction change the sign?

If making CO₂ from carbon and oxygen releases 393.5 kJ, then splitting CO₂ back into carbon and oxygen must take back exactly 393.5 kJ, or one could cycle round and create energy from nothing.

Is thermite really that hot?

Yes. About 850 kJ is released for every two moles of iron produced, and nothing carries the heat away from the small amount of material, so the iron comes out molten, above 2000 °C. It has been used to weld railway track since the 1890s.

Work through it to see what it shows.
Theory 2 · Berthelot and Thomsen, 1860s–1880s

One number for every compound

Thousands of reactions, thousands of heats to measure. Hess's law says you only need one number per compound. Which?

Marcellin Berthelot in Paris and Julius Thomsen in Copenhagen spent the 1860s to the 1880s burning every substance they could get into a calorimeter and recording the heat. Their tables ran to thousands of reactions. Hess's law shows how to compress them: give each compound the heat of forming it from its elements, and every reaction among them follows by subtraction. Pick a reaction, look up the formation enthalpies, and compute the heat of the reaction.
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)
SpeciesΔHf° (kJ/mol)Source
reactant · CH₄(g)-74.87NIST WebBook
reactant · O₂(g)0element in its standard state: 0 by definition
product · CO₂(g)-393.51NIST WebBook
product · H₂O(l)-285.83NIST WebBook
What is ΔH° for the reaction as written, in kJ? (Multiply by the coefficients.)
kJ
Step 1 of 2 · Compute oneChoose a reaction, read the formation enthalpies above, and compute ΔH° for the equation as written.
Reactions solved0 / 4
ΔH° (this one)—
Best fuel per gram—
The maths
ΔH°(reaction) = Σ n ΔHf°(products) − Σ n ΔHf°(reactants). ΔHf° of an element in its standard state = 0.
Products: 1 × (-393.51) + 2 × (-285.83) = -965.2 kJ
Reactants: 1 × (-74.87) + 2 × (0) = -74.9 kJ
ΔH° = -965.2 − (-74.9) = -890.3 kJ
Questions you might have (3)
Why is the formation enthalpy of liquid water different from gaseous water?

Condensing water vapour to liquid releases about 44 kJ per mole (−285.8 compared with −241.8). That is why the heat of burning a fuel depends on whether the water ends up as vapour or as liquid. Tables that give the lower value assume the water leaves as steam.

Why is the enthalpy of formation of oxygen gas zero?

It is defined that way: the standard state of an element is its most stable form at 1 bar and 298 K, and the formation of an element from itself involves no change at all. Zero is a zero of the scale, like sea level. Only differences matter, so any consistent zero will do.

Is a reaction that releases heat sure to happen?

No. Limestone breaking down takes in 178 kJ per mole, yet a lime kiln does it every day. Heat alone does not decide which way a reaction goes; the freedom of the particles to spread out matters as well. That is for a later course. For now, the enthalpy tells you how much heat is given out or taken in, not whether it will happen.

Work through it to see what it shows.
PracticeProblemsNumbers from this unit, in the style of a college problem set.
Practice

Practice

Numbers from the experiments, in the style of a college problem set. Every problem has a hint and a worked solution.

q = mcΔT0 right of 0 · streak 0

How much heat, in joules, is needed to warm 119 g of iron (specific heat 0.449 J/(g·K)) by 12 K?

J
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