The enthalpies of all elements in their standard s – Class 11 Chemistry

Exercises · 5.3

Q. The enthalpies of all elements in their standard states are:

(i) unity

(ii) zero

(iii) < 0

(iv) different for each element

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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

Watch out

Do NOT write the formation reaction with coefficients other than those that produce exactly 1 mole of product. For example, writing $2H_2 + O_2 \rightarrow 2H_2O$ gives the enthalpy change for 2 moles of water — that's not the standard enthalpy of formation. You must divide by 2.

Also, the product must be in its standard state. For water, that's liquid at 298 K — not steam ( $H_2O(g)$ ). The formation enthalpy of steam is different ( $-241.8 \text{ kJ/mol}$ ).

Why This Concept Matters

Standard enthalpies of formation are the building blocks of thermochemistry. Once you have a table of $\Delta_f H^\circ$ values for common compounds, you can calculate the enthalpy change for any reaction using Hess's law:

\Delta_r H^\circ = \sum \Delta_f H^\circ(\text{products}) - \sum \Delta_f H^\circ(\text{reactants})

This is the single most powerful tool in thermochemistry — and it all rests on the definition you just learned.

If you've searched "Standard Enthalpy of Formation class 11 chemistry notes" or "Standard Enthalpy of Formation NCERT solutions", this page covers exactly that ground — the concept is a standard part of the Class 11 Chemistry NCERT/CBSE syllabus. It also carries real weight in JEE Main, NEET and state CET Chemistry papers, where questions on standard enthalpy of formation test both conceptual understanding and calculation speed.

The standard enthalpy of formation of any element in its standard state is defined as zero. This is a convention that sets a reference point for all enthalpy calculations. Therefore, the correct answer is (ii) zero.

The question asks about the enthalpies of elements in their standard states. This is a fundamental concept in thermochemistry, and the answer hinges on understanding what "standard state" means and how we define enthalpy changes.

The Concept: Why Zero?

Enthalpy ( $H$ ) is a state function, but we can never measure its absolute value. We can only measure changes in enthalpy ( $\Delta H$ ). To make these changes meaningful and comparable, we need a common reference point.

The Standard Enthalpy of Formation ( $\Delta H_f^\circ$ ) of a compound is defined as the enthalpy change when one mole of the compound is formed from its constituent elements in their standard states under standard conditions (1 bar pressure, usually 298 K).

For this definition to work, we must assign a value to the enthalpy of the elements themselves. By international convention, the standard enthalpy of formation of an element in its most stable allotropic form at 1 bar and the specified temperature is taken as zero.

Important

This is a convention, not a discovery. It's like setting sea level as zero for measuring altitude. It doesn't mean the element has no internal energy; it means we've chosen it as the baseline.

Step-by-Step Reasoning

Identify the core principle. The question is about the "enthalpies of all elements in their standard states." This directly refers to the standard enthalpy of formation ( $\Delta H_f^\circ$ ) of the elements themselves.
Recall the definition. The standard enthalpy of formation of a substance is the enthalpy change when 1 mole of that substance is formed from its elements in their standard states. For an element in its standard state, "forming it from itself" involves no chemical change.
Apply the convention. Since there is no chemical reaction involved in "forming" an element from itself, the enthalpy change is zero. By definition, we set $\Delta H_f^\circ = 0$ for all elements in their standard states.
Consider the exceptions (the nuance). The phrase "most stable allotropic form" is crucial. For example:
- Carbon in the form of graphite has $\Delta H_f^\circ = 0$ .
- Carbon in the form of diamond has $\Delta H_f^\circ = +1.9 \text{ kJ/mol}$ (because it is not the most stable form at standard conditions).
- Oxygen gas ( $O_2$ ) has $\Delta H_f^\circ = 0$ .
- Ozone gas ( $O_3$ ) has $\Delta H_f^\circ = +142.7 \text{ kJ/mol}$ .

Watch out

A common mistake is to think that all forms of an element have zero enthalpy. Only the most stable form at standard conditions has $\Delta H_f^\circ = 0$ . Other allotropes have non-zero values.

Evaluate the options.
- (i) unity: Incorrect. The value is not 1.
- (ii) zero: Correct. This is the standard convention.
- (iii) < 0: Incorrect. The value is exactly zero, not negative.
- (iv) different for each element: Incorrect. While different elements have different absolute enthalpies, the convention sets them all to zero for their standard states.

Tip

Think of it like a bank account. You can't know the total money in the world, but you can track deposits and withdrawals. Setting the "balance" of elements to zero is like opening a new account with a zero balance. All transactions (reactions) are then measured relative to that starting point.

✓Final answer

The correct option is (ii) zero .

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Q.The enthalpies of all elements in their standard states are:

The Concept: Why Zero?

Step-by-Step Reasoning

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