JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Chemical Thermodynamics

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Choose the reaction(s) from the following options, for which the standard enthalpy of reaction is equal to the standard enthalpy of formation.

(A) $\frac{3}{2} O_{2} (g) \to O_{3} (g)$ ✓ Correct
(B) $\frac{1}{8} S_{8} (s) + O_{2} (g) \to SO_{2} (g)$ ✓ Correct
(C) $2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (l)$
(D) $2 C (g) + 3 H_{2} (g) \to C_{2} H_{6} (g)$

Answer: (A), (B)

Solution

Standard enthalpy of formation (

Δ H_{f}^{\circ}

) is defined for the formation of 1 mole of a compound from its constituent elements in their standard states (most stable form at 1 bar, 298 K).

(1)

\frac{3}{2} O_{2} (g) \to O_{3} (g)

: Forms 1 mole

O_{3}

from

O_{2}

(standard state of oxygen). This is

Δ H_{f}^{\circ} (O_{3})

. TRUE.

(2)

\frac{1}{8} S_{8} (s) + O_{2} (g) \to SO_{2} (g)

: Forms 1 mole

SO_{2}

from

S_{8} (s)

(standard state of sulfur) and

O_{2} (g)

. This is

Δ H_{f}^{\circ} (SO_{2})

. TRUE.

(3)

2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (l)

: Forms 2 moles of

H_{2} O

, not 1. This is

2 \times Δ H_{f}^{\circ} (H_{2} O)

. FALSE.

(4)

2 C (g) + 3 H_{2} (g) \to C_{2} H_{6} (g)

: Carbon is in gaseous state, not its standard state (graphite, solid). FALSE.

Correct answers: (1) and (2).

JEE Advanced · 2019 · 27 May 2019 · Paper 1 · One or more correct options

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JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Chemical Thermodynamics

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