Formation cycles: going round by the elements

8 minAS Level5.2.15.2.2

AI-drafted, independently checked · not yet teacher-reviewed

By the end of this lesson

By the end you can draw a cycle that goes from reactants to products via the elements, and use standard enthalpy changes of formation to calculate $\Delta H_r$ for any reaction.

The one reaction we can always quote

Every compound has an enthalpy change of formation: the enthalpy change when one mole of it forms from its elements in their standard states.

That gives a universal "meeting point" for a cycle. Whatever the reaction, the reactants can, in principle, be broken back down to their elements — and those elements can then be built up into the products.

Chemists never actually do that. Hess's law lets us pretend, and add the numbers.

Definition

Standard enthalpy change of formation

The enthalpy change when one mole of a compound is formed from its elements in their standard states, under standard conditions (298 K, 100 kPa).

Symbol: $\Delta H_f^{\ominus}$ , in $\text{kJ mol}^{-1}$ .

By definition, the $\Delta H_f^{\ominus}$ of an element in its standard state is zero.

Why the signs come out that way

Start at the reactants and travel down to the elements. Forming a compound from its elements is the reaction we have data for, so going the other way — compound to elements — is the reverse, and its sign flips.

Then travel up from the elements to the products: that is exactly formation, so the signs stay as they are.

So the total round trip is:

$\Delta H_r = \sum \Delta H_f^{\ominus}(\text{products}) - \sum \Delta H_f^{\ominus}(\text{reactants})$

"Products minus reactants" is worth memorising, but if you can draw the arrows you can always rebuild it — and you will not mix up a sign when the question is unusual.

Worked example: Worked example: combustion of methane

Calculate $\Delta H_r^{\ominus}$ for

$\text{CH}_4\text{(g)} + 2\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(l)}$

given $\Delta H_f^{\ominus}$ : CH₄(g) $= -74.8$ , CO₂(g) $= -393.5$ , H₂O(l) $= -285.8\ \text{kJ mol}^{-1}$ .

Have a go first?

Worked example: Worked example: an endothermic decomposition

Calculate $\Delta H_r^{\ominus}$ for the thermal decomposition of calcium carbonate:

$\text{CaCO}_3\text{(s)} \rightarrow \text{CaO(s)} + \text{CO}_2\text{(g)}$

given $\Delta H_f^{\ominus}$ : CaCO₃(s) $= -1207$ , CaO(s) $= -635$ , CO₂(g) $= -393.5\ \text{kJ mol}^{-1}$ .

Have a go first?

Quick check

Multiple choiceQuestion 1 · Recall

CalculateQuestion 2 · Recall

Answer the 3 questions above to continue.