Standard Enthalpy and Standard State Concepts Explained

 

❓ Question

Which of the following statements is correct?

1️⃣ $\Delta_f H^\circ_{298}$ is zero for O(g)
2️⃣ The standard state of a pure gas is the pure gas at a pressure of 1 bar and temperature 273 K
3️⃣ The term standard state implies that the temperature is 0 °C
4️⃣ $\Delta_f H^\circ_{500}$ is zero for O₂(g)

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🖼️ Question Image

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✍️ Short Explanation

This problem is based on:

👉 Standard enthalpy of formation
👉 Standard state
👉 Thermodynamics conventions.

Main idea:

Standard enthalpy of formation of an element in its most stable form is:

$$\boxed{ 0 }$$

at any temperature.

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🔷 Step 1 — Check Statement 1 💯

$$\Delta_f H^\circ_{298}=0 \text{ for } O(g)$$

This is incorrect because:

Standard enthalpy of formation is zero only for the most stable form of element.

Stable form of oxygen at $298\text{ K}$:

$$\boxed{ O_2(g) }$$

not atomic oxygen $O(g)$.

Thus Statement 1 is false.

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🔷 Step 2 — Check Statement 2

Standard state of gas:

✔ Pure gas
✔ Pressure = 1 bar

But temperature is not fixed at $273\text{ K}$.

Thus Statement 2 is false.

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🔷 Step 3 — Check Statement 3

“Standard state” does NOT mean:

$$0^\circ C$$

Temperature can be any specified value.

Thus Statement 3 is false.

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🔷 Step 4 — Check Statement 4

$$\Delta_f H^\circ_{500}=0 \text{ for } O_2(g)$$

Since $O_2(g)$ is the most stable form of oxygen:

$$\boxed{ \Delta_f H^\circ=0 }$$

even at $500\text{ K}$.

Thus Statement 4 is correct.

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🔷 Step 5 — JEE Trap Alert 🚨

❌ Standard state ko STP samajh lena

❌ Atomic oxygen ko elemental standard form maan lena

Remember:

Standard enthalpy of formation:

$$\boxed{ 0 \text{ for most stable elemental form} }$$

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✅ Final Answer

$$\boxed{ \Delta_f H^\circ_{500}=0 \text{ for } O_2(g) }$$

(Option 4)

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