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General Chemistry I · Energy and Thermodynamics

Enthalpy

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On this page 8 sections
  1. In 30 seconds
  2. Why this matters
  3. The college version
  4. Eli explains
  5. Worked example
  6. Key takeaway
  7. Quick check
  8. Study tools

In 30 seconds

Enthalpy H is a state function defined as H = E + PV. At constant pressure its change ΔH equals the heat transferred, qp. A reaction with ΔH < 0 is (releases heat); one with ΔH > 0 is (absorbs heat). Thermochemical equations report ΔH beside a balanced equation. Because enthalpy is a state function, ΔH can be found by (summing steps) or from standard enthalpies of formation: ΔH°rxn = ∑n ΔH°f(products) - ∑m ΔH°f(reactants).

Why this matters

Enthalpy data drives the design of heating and cooling materials used in medicine. Reusable hand warmers exploit the exothermic crystallization of a supersaturated salt solution, while instant cold packs rely on the endothermic dissolving of ammonium salts — the temperature drop absorbs heat from an injured joint. Nutrition and metabolism track the enthalpy of combustion of foods, and hospital labs choose reaction conditions from tabulated ΔH values so clinical assays run safely and reproducibly.

The college version

1. Enthalpy and Its Change

Enthalpy is defined as H = E + PV. Its absolute value cannot be measured, but its change can. At constant pressure, ΔH = ΔE + PΔV, and combined with the first law this gives ΔH = qp, the heat transferred at constant pressure. That is why coffee-cup calorimetry (constant pressure) reports ΔH. A negative ΔH means heat leaves the system (exothermic); a positive ΔH means heat enters it (endothermic).

2. Thermochemical Equations

A pairs a balanced equation with its ΔH. The magnitude of ΔH scales with the coefficients, which are read as moles: double the amounts and ΔH doubles; reverse the reaction and the sign of ΔH flips. Physical states must be written because ΔH depends on them — vaporizing water costs energy, so H2O(g) sits at a higher (less negative) enthalpy than H2O(l).

3. Hess's Law and Standard Enthalpies of Formation

Because H is a state function, the enthalpy change of an overall reaction is the same whether it happens in one step or many. Hess's law lets us add the ΔH values of known reactions to get a target reaction's ΔH, reversing equations (flipping signs) or scaling them as needed. The standard enthalpy of formation ΔH°f is the enthalpy change when 1 mole of a compound forms from its elements in their standard states; elements in their standard states are assigned ΔH°f = 0. From tabulated values:

ΔH°rxn = ∑n ΔH°f(products) - ∑m ΔH°f(reactants)

where n and m are stoichiometric coefficients.

How it works

  1. Write the balanced thermochemical equation with correct physical states.
  2. From tables, sum the products' ΔH°f times their coefficients, then subtract the same sum for reactants.
  3. For Hess's law, arrange known equations so they add to the target, reversing or scaling as needed.
  4. Flip the sign of ΔH when you reverse an equation; scale ΔH when you multiply coefficients.
  5. Cancel identical species on both sides and add the ΔH values.
  6. Interpret the sign: negative = exothermic (heat released), positive = endothermic (heat absorbed).

Common confusions

Do not confuseWithDifference
Enthalpy (H)Internal energy (E)H = E + PV; ΔH = ΔE + PΔV at constant P
ExothermicEndothermicReleases heat (ΔH < 0) vs absorbs heat (ΔH > 0)
ΔH°f of an elementΔH°f of a compoundElement in standard state = 0; compound is nonzero
Hess's law stepReversed Hess's law stepReversing a step flips its ΔH sign
Thermochemical equationOrdinary balanced equationThermochemical equation also reports ΔH

Memory aids

"Products minus reactants" — for ΔH°rxn, subtract the sum of reactant ΔH°f values from the sum of product values, and let the sign tell which way the heat flows.

Quick review

Topic Recap

Enthalpy H = E + PV is a state function whose change at constant pressure equals the heat transferred. Exothermic reactions have negative ΔH and release heat; endothermic reactions have positive ΔH and absorb it. Thermochemical equations report ΔH per the mole amounts shown, scaling and reversing with the equation. Because enthalpy is path-independent, Hess's law sums step enthalpies, and ΔH°rxn can be computed directly from standard enthalpies of formation as products minus reactants.

Knowledge Check

  1. For an exothermic reaction, ΔH is: A) positive B) negative C) zero D) undefined
  2. What is ΔH°f of O2(g)? A) −393.5 kJ/mol B) +393.5 kJ/mol C) 0 D) −285.8 kJ/mol
  3. If A → B has ΔH = -40 kJ, then 2B → 2A has: A) −40 kJ B) +40 kJ C) +80 kJ D) −80 kJ
  4. ΔH°rxn equals: A) ∑ΔH°f(reactants) - ∑ΔH°f(products) B) ∑ΔH°f(products) - ∑ΔH°f(reactants) C) ∑ΔH°f(products) + ∑ΔH°f(reactants) D) -∑ΔH°f(products)
  5. For 2H2(g) + O2(g) → 2H2O(l) with ΔH°f[H2O(l)] = -285.8 kJ/mol, ΔH°rxn is: A) −285.8 kJ B) −571.6 kJ C) +571.6 kJ D) +285.8 kJ

Answers and Rationales

  1. B — Exothermic processes release heat to the surroundings, so ΔH < 0.
  2. C — O2(g) is an element in its , so ΔH°f = 0.
  3. C — Reversing flips the sign (+40 kJ) and doubling the amount doubles it (+80 kJ).
  4. B — Products minus reactants, each multiplied by its stoichiometric coefficient.
  5. B — ΔH°rxn = 2(-285.8) - [2(0) + 0] = -571.6 kJ, twice the per-mole value because 2 moles of water form.
Eli, the EliExplains learning guide

Eli explains

The same idea, in plain words

Explain it like I’m 10

Think of enthalpy as a chemical "price tag" for a reaction. A receipt shows the total cost no matter how many coupons you stacked along the way; what matters is the final price. A reaction's ΔH is that bottom line — how much heat the reaction absorbs or gives off overall, regardless of the exact route the atoms took. An exothermic reaction pays you back heat; an endothermic reaction costs you heat.

Where this stops being exact: a price tag has one true value, but ΔH also depends on the state of every substance — whether the water made is liquid or steam, the temperature and pressure, and how many moles react. Change any of those and the price changes. Chemists remove this ambiguity by agreeing on a standard reference state (the ΔH° notation), so everyone quotes the same sticker price.

Simple Example

Burning methane, CH4(g) + 2O2(g) → CO2(g) + 2H2O(l), has ΔH = -890.3 kJ. The negative sign says the reaction gives off 890.3 kJ of heat per mole of methane burned — which is why natural gas heats homes.

Worked example

At constant pressure, ΔH = qp. Standard enthalpies of formation (kJ/mol, 25 °C, standard state): ΔH°f[CH4(g)] = -74.8, ΔH°f[CO2(g)] = -393.5, ΔH°f[H2O(l)] = -285.8, ΔH°f[CO(g)] = -110.5; elements O2(g), H2(g), and graphite C(s) are 0.

Worked Example 1 — Enthalpy from standard enthalpies of formation

Find ΔH°rxn for the combustion of methane:

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)

Step 1. Write products minus reactants:

ΔH°rxn = [ΔH°f(CO2) + 2ΔH°f(H2O)] - [ΔH°f(CH4) + 2ΔH°f(O2)]

Step 2. Insert values:

= [(-393.5) + 2(-285.8)] - [(-74.8) + 2(0)]

Step 3. Evaluate:

= (-393.5 - 571.6) - (-74.8) = -965.1 + 74.8 = -890.3 kJ

The negative sign confirms exothermic combustion, releasing 890.3 kJ per mole of methane.

Worked Example 2 — Hess's law

Find ΔH for C(s) + 12O2(g) → CO(g) using:

C(s) + O2(g) → CO2(g)  ΔH1 = -393.5 kJ

CO(g) + 12O2(g) → CO2(g)  ΔH2 = -283.0 kJ

Step 1. Keep reaction 1 as written (-393.5 kJ); it supplies C(s) + O2(g) on the left.

Step 2. Reverse reaction 2 so CO(g) appears on the product side; reversing flips the sign:

CO2(g) → CO(g) + 12O2(g)  ΔH = +283.0 kJ

Step 3. Add the equations. CO2(g) cancels, and the net oxygen is O2 - 12O2 = 12O2 on the left:

C(s) + 12O2(g) → CO(g)

Step 4. Add the enthalpy changes:

ΔH = (-393.5 kJ) + (+283.0 kJ) = -110.5 kJ

This matches the tabulated ΔH°f of CO, confirming the path-independent answer.

Common setup errors: treating ΔH°f of an element as nonzero (it is 0 only for elements in their standard states, e.g. H2(g), not H2O); using the wrong physical state (liquid vs gas water differ by ~44 kJ/mol); and forgetting to flip the sign when reversing a reaction in Hess's law.

Key takeaways

  • High yield: ΔH = qp at constant pressure; ΔH < 0 is exothermic, ΔH > 0 is endothermic.
  • High yield: ΔH°rxn = ∑n ΔH°f(products) - ∑m ΔH°f(reactants).
  • High yield: ΔH°f = 0 for elements in their standard states.
  • Reversing a reaction flips the sign of ΔH; multiplying coefficients scales ΔH.
  • Enthalpy is a state function, which is why Hess's law works.
  • Physical states matter: H2O(g) and H2O(l) have different ΔH°f values.
  • The magnitude of ΔH is per the mole amounts in the balanced equation.

Quick check

1 question here. Answers stay hidden until you check.

Question 1 of 1

Calculate ΔH° for 2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(s) given: ΔH_f°(Fe2O3) = -824.2 kJ/mol and ΔH_f°(Al2O3) = -1675.7 kJ/mol.

Choose an answer, then check it.

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Study tools & related lessonsYou’ll learn to · Key vocabulary · Related

You’ll learn to

  • Define enthalpy H and explain why ΔH equals the heat at constant pressure, qp.
  • Distinguish exothermic (ΔH < 0) from endothermic (ΔH > 0) processes and read thermochemical equations.
  • Use Hess's law to find a reaction's ΔH from the steps of a pathway.
  • Calculate ΔH°rxn from standard enthalpies of formation ΔH°f.

Key vocabulary

Enthalpy (H)
State function H = E + PV
Δ H
Change in enthalpy of a process
Exothermic
Releases heat, ΔH < 0
Endothermic
Absorbs heat, ΔH > 0
Thermochemical equation
Balanced equation with its ΔH
Hess's law
Overall ΔH is the sum of step ΔH values
Standard enthalpy of formation (Δ H°f)
ΔH to form 1 mol from elements in standard states
Standard state
Most stable form at 1 atm and a stated T (usually 25 °C)

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