JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Chemical Thermodynamics

For the reaction,

N_{2} O_{4} ⇌ 2 NO_{2}

, graph is plotted as shown below. Identify correct statements.

A. Standard free energy change for the reaction is

- 5.40 kJ mol^{- 1}

.
B. As

Δ G^{⊖}

in graph is positive,

N_{2} O_{4}

will not dissociate into

NO_{2}

at all.
C. Reverse reaction will go to completion.
D. When 1 mole of

N_{2} O_{4}

changes into equilibrium mixture, value of

Δ G^{⊖} = - 0.84 kJ mol^{- 1}

E. When 2 mole of

NO_{2}

, changes into equilibrium mixture,

Δ G^{⊖}

for equilibrium mixture is

- 6.24 kJ mol^{- 1}

.

Choose the correct answer from the options given below:

(A)D and E only✓ Correct
(B)C and E only
(C)A and D only
(D)B and C only

Answer: (A)

Solution

On the free-energy plot the pure-reactant point A lies 5.40 kJ mol

^{- 1}

below the pure-product point B, so the standard reaction free energy is

Δ_{r} G^{\circ} = G_{B} - G_{A} = + 5.40 kJ mo l^{- 1}

, a positive value; hence the statement of

- 5.40

is wrong, and a positive

Δ_{r} G^{\circ}

neither prevents partial dissociation nor drives the reverse reaction to completion, because the system settles at the minimum E. Going from A down to the equilibrium minimum E gives

Δ G = G_{E} - G_{A} = - 0.84 kJ mo l^{- 1}

, and going from B down to E gives

Δ G = G_{E} - G_{B} = - (5.40 + 0.84) = - 6.24 kJ mo l^{- 1}

. The correct statements are D and E only.

JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Single correct option

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JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Chemical Thermodynamics

Solution

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