JEE Main 2026 — Jan 21, Shift 1 — Chemistry solution

JEE Main2026Jan 21, Shift 1Chemistry

For the reaction, $N_{2} O_{4} ⇌ 2 N O_{2}$ , 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 $N O_{2}$ 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 $N O_{2}$ , changes into equilibrium mixture, $Δ G^{⊖}$ for equilibrium mixture is $- 6.24$ kJ mol $^{- 1}$ .

Choose the correct answer from the options given below:

Solution

$Δ_{r} G^{\circ} = G_{B}^{\circ} - G_{A}^{\circ} = + 5.40$ kJ/mol (positive). A is wrong.

$N_{2} O_{4}$ partially dissociates — equilibrium establishes at E. B and C wrong.

For 1 mole $N_{2} O_{4} \to$ equilibrium mixture, $Δ G = - 0.84$ kJ/mol. D correct.

For 2 mole $N O_{2} \to$ equilibrium mixture, $Δ G = - (5.40 + 0.84) = - 6.24$ kJ/mol. E correct.

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