Consider the following reaction approaching equilibrium at 27 ° C and 1 atm pressure A + B K _ t = 0 ^ 3 ⇌ _ K _ r = 10 ^ 2 C + D The standard Gibb's energy change ( Δ _ r G °) at 27 ° C is (-) _____ kJmol ^ - 1

Correct answer: 0 We know, K _ eq = K _ f K _ b ∴ K _ eq = 10 ^ 3 10 ^ 2 = 10 Using gibbs free energy equation, ∴ Δ G ° = - RTlnK _ eq ⇒ Δ G = - RTln10 ⇒ - (8 . 3 × 300 × 2 . 3) = - 5 . 7kJ mol ^ - 1 6kJ mole ^ - 1 (nearest integer) Answer = 6 .

JEE Main 202329 JanuaryMorning Shift

Exam: JEE Main
Year: 2023
Date: 29 January
Shift: Morning Shift
Subject: Chemistry
Chapter: Thermodynamics (C)

JEE Main 2023 · 29 January · Morning Shift · Chemistry · Thermodynamics (C)

Consider the following reaction approaching equilibrium at 27 ° C...

Consider the following reaction approaching equilibrium at $27 ° C and 1$ atm pressure

$A + B\mathop{\overset{{K}_{t} = {0}^{3}}{⇌}}_{{K}_{r} = {10}^{2}}C + D$

The standard Gibb's energy change \left({\Delta}_{r}G °\right) $at$ 27 ° C $is \left(-\right)$ _____ ${kJmol}^{- 1}$ (Nearest integer).

(Given : $R = 8 . 3J{K}^{- 1}{mol}^{- 1} and \ln 10 = 2 . 3$ )

Correct answer

Step-by-step solution

We know,

${K}_{eq} = \dfrac{{K}_{f}}{{K}_{b}}$

$∴ {K}_{eq} = \dfrac{{10}^{3}}{{10}^{2}} = 10$
Using gibbs free energy equation,

$∴ \Delta G ° = - {RTlnK}_{eq} ⇒ \Delta G = - RTln10$

$⇒ - \left(8 . 3 \times 300 \times 2 . 3\right) = - 5 . 7kJ{mol}^{- 1} \approx 6kJ{mole}^{- 1}$ (nearest integer)

Answer $= 6$ .

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