The enthalpies of elements in their standard states are taken as zero. The enthalpy of formation of a compound

NCERT Exemplar · Q12

Q. The enthalpies of elements in their standard states are taken as zero. The enthalpy of formation of a compound

(i) is always negative

(ii) is always positive

(iii) may be positive or negative

(iv) is never negative

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Concept understanding — Standard Enthalpy of Formation

Standard Enthalpy of Formation: From Intuition to Definition

Imagine you're building a house. You don't start from a finished house — you start from raw materials: bricks, cement, wood, steel. The cost of assembling those raw materials into the final house is a useful number. In chemistry, we do the same thing with compounds.

Every chemical compound is made from elements in their natural, most stable forms. The standard enthalpy of formation ( $\Delta_f H^\circ$ ) is the energy change when you build one mole of a compound from its elements, with everything in their standard states.

The Intuition First

Think of it as the "birth certificate" energy of a compound. It tells you:

How much energy is released or absorbed when the compound is formed from scratch.
Whether the compound is more stable (lower energy) or less stable (higher energy) than the elements it came from.

If $\Delta_f H^\circ$ is negative, the compound is more stable than its elements — energy was released during formation. If positive, the compound is less stable — energy had to be absorbed to force the elements together.

The Precise Definition

\Delta_f H^\circ = \text{enthalpy change when 1 mole of a compound is formed from its constituent elements in their standard states, under standard conditions (1 bar pressure, specified temperature, usually 298 K)}

Key points to lock in:

Exactly 1 mole of the compound is formed — not 2, not 0.5.
Elements in their standard states — this means the most stable physical form of the element at 1 bar and the given temperature. For example:
- Carbon: graphite (not diamond)
- Oxygen: $O_2(g)$ (not $O_3$ )
- Hydrogen: $H_2(g)$
- Bromine: $Br_2(l)$ (liquid at room temperature)
Standard conditions: 1 bar pressure (not 1 atm — slight difference, but in most exams they treat them as equivalent unless specified). Temperature is usually 298 K (25°C), but can be any specified temperature.

The Critical Rule: Elements Have Zero Formation Enthalpy

Important

The standard enthalpy of formation of any element in its standard state is zero by definition.

This is not a measurement — it's a convention. We set the zero point of the energy scale at the most stable form of each element. So:

$\Delta_f H^\circ$ of $O_2(g)$ = 0
$\Delta_f H^\circ$ of $C(graphite)$ = 0
$\Delta_f H^\circ$ of $Br_2(l)$ = 0

But $\Delta_f H^\circ$ of $O_3(g)$ is not zero — ozone is not the standard state of oxygen.

Worked Example: Water

Write the formation reaction for liquid water:

H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(l)

The $\Delta_f H^\circ$ for $H_2O(l)$ is $-285.8 \text{ kJ/mol}$ .

What does this tell you? When 1 mole of water is formed from hydrogen gas and oxygen gas (both in their standard states), 285.8 kJ of heat is released. The water molecule is more stable than the separate elements.

Common Mistake to Avoid …

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The standard enthalpy of formation is defined relative to elements in their standard states (enthalpy zero). It can be positive or negative depending on whether the compound is more or less stable than its elements — so the correct answer is (iii).

The idea of standard enthalpy of formation ( $\Delta_f H^\circ$ ) is one of the most elegant in thermochemistry. It gives us a common reference point: take every element in its most stable form at 1 bar pressure and a specified temperature (usually 298 K), and assign that state an enthalpy of exactly zero. Then, when we form one mole of a compound from those elements, the heat absorbed or released is its $\Delta_f H^\circ$ .

Why zero for elements? Because we can only measure changes in enthalpy, not absolute values. Setting the elements as a baseline lets us compare the "energy content" of compounds on a single scale. If forming a compound releases heat (exothermic), its $\Delta_f H^\circ$ is negative — the compound is more stable (lower energy) than the elements. If it absorbs heat (endothermic), $\Delta_f H^\circ$ is positive — the compound is less stable (higher energy).

Now, is it always one or the other? Let's check real data.

Negative examples — Many compounds form exothermically. For water:

$H_2(g) + \frac12 O_2(g) \rightarrow H_2O(l)$ , $\Delta_f H^\circ = -285.8 \text{ kJ/mol}$ .

For carbon dioxide: $C(s) + O_2(g) \rightarrow CO_2(g)$ , $\Delta_f H^\circ = -393.5 \text{ kJ/mol}$ .

These are stable relative to their elements.
Positive examples — Some compounds require energy input to form. Acetylene:

$2C(s) + H_2(g) \rightarrow C_2H_2(g)$ , $\Delta_f H^\circ = +227 \text{ kJ/mol}$ .

Nitric oxide: $\frac12 N_2(g) + \frac12 O_2(g) \rightarrow NO(g)$ , $\Delta_f H^\circ = +90.3 \text{ kJ/mol}$ .

These are less stable — they tend to decompose or react readily. …

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Q.The enthalpies of elements in their standard states are taken as zero. The enthalpy of formation of a compound

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