Worked example 1
Given 2 H₂(g) + O₂(g) → 2 H₂O(l), ΔH = -572 kJ, find ΔH for H₂O(l) → H₂(g) + 1/2 O₂(g).
Try it first: Decide which operations turn the first equation into the second, and in what order.
0 of 2 steps revealed.
What you'll be able to do: Read a thermochemical equation correctly and scale enthalpy with the amount of substance reacting.
Best after: Energy, Heat and the First Law, Lab: Coffee-Cup Calorimetry and Heat of Reaction
An enthalpy value is meaningless without the equation it belongs to. Doubling the coefficients doubles ΔH, reversing the reaction flips its sign, and every enthalpy calculation later in this unit depends on those two rules.
These are recommended, not required. You can start this lesson at any time.
Enthalpy depends only on the current state of a system, not on how it got there. That is what makes indirect routes legitimate: any path from the same reactants to the same products gives the same ΔH.
The ΔH quoted applies to the molar amounts written in that equation, with those exact physical states. Changing water from liquid to gas changes ΔH, because vaporisation costs energy.
CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(l), ΔH = -890 kJ
Halve every coefficient and ΔH is halved. Reverse the arrow and ΔH keeps its magnitude but changes sign, because the same energy now has to be supplied rather than released. These two operations are the whole toolkit for Hess law in the next lesson.
reverse: ΔH → -ΔH; multiply by n: ΔH → n ΔH
Because ΔH is proportional to the amount reacting, it behaves exactly like a coefficient. Convert grams to moles, then use the ratio of kilojoules to moles from the thermochemical equation.
Products are drawn below reactants for an exothermic reaction and above for an endothermic one. The vertical gap is the magnitude of ΔH. The diagram says nothing about how fast the reaction goes, which is a kinetics question.
Heat exchanged at constant pressure for the reaction as written.
A property determined only by the current state, independent of the path taken.
Multiplying a balanced equation by n multiplies ΔH by n.
Reversing a reaction changes the sign of ΔH but not its magnitude.
Different physical states of the same substance give different enthalpy values.
q = n x ΔH(per mole as written)
Given 2 H₂(g) + O₂(g) → 2 H₂O(l), ΔH = -572 kJ, find ΔH for H₂O(l) → H₂(g) + 1/2 O₂(g).
Try it first: Decide which operations turn the first equation into the second, and in what order.
0 of 2 steps revealed.
For CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(l), ΔH = -890 kJ. How much heat is released by burning 8.00 g of methane (M = 16.04 g/mol)?
0 of 3 steps revealed.
A reaction has ΔH = +65 kJ as written. What is ΔH when the equation is multiplied by 3 and then reversed?
0 of 2 steps revealed.
Why it's wrong: The energy that was released must now be supplied.
Check instead: Reverse means negate, every time.
Why it's wrong: Producing steam instead of liquid water absorbs extra energy.
Check instead: Match the states exactly before using a tabulated value.
Why it's wrong: The value is tied to the coefficients written down.
Check instead: Quote ΔH together with the balanced equation it belongs to.
Why it's wrong: Rate is controlled by activation energy, not by enthalpy.
Check instead: Thermodynamics answers whether, kinetics answers how fast.
No practice questions are available for this topic yet. You can still practice the whole unit.
Enthalpy is a state function: it depends only on the initial and final states, never on the route. A thermochemical equation attaches ΔH to a specific balanced equation with states included. Multiply the equation through and ΔH scales by the same factor; reverse the equation and ΔH changes sign. Because ΔH is proportional to amount, it works as a conversion factor in stoichiometry, letting you convert between grams of a reactant and kilojoules released.
Chemistry 2e, OpenStax, Rice University, licensed CC BY 4.0. License