The variation of equilibrium constant with temperature is given below: | Temperature | Equilibrium Constant | |-------------|---------------------| | $\mathrm{T}_{1}=25^{\circ} \mathrm{C}$ | $\mathrm{K}_{1}=10$ | | $\mathrm{T}_{2}=100^{\circ} \mathrm{C}$ | $\mathrm{K}_{2}=100$ | The value of $\Delta \mathrm{H}^{\circ}, \Delta \mathrm{G}^{\circ}$ at $\mathrm{T}_{1}$ and $\Delta \mathrm{G}^{\circ}$ at $\mathrm{T}_{2}$ (in $\mathrm{kJ} \mathrm{mol}^{-1}$ ) respectively, are close to [Use $\mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ ]

JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Chemical Thermodynamics

The variation of equilibrium constant with temperature is given below:

Temperature	Equilibrium Constant
$T_{1} = 2 5^{\circ} C$	$K_{1} = 10$
$T_{2} = 10 0^{\circ} C$	$K_{2} = 100$

The value of

Δ H^{\circ}, Δ G^{\circ}

T_{1}

and

Δ G^{\circ}

T_{2}

(in

kJ mol^{- 1}

) respectively, are close to
[Use

R = 8.314 JK^{- 1} mol^{- 1}

]

Answer: (C)

The temperature dependence of

K

follows the van't Hoff relation

lo g \frac{K _{2}}{K _{1}} = \frac{Δ H ^{\circ}}{2.303 R} (\frac{1}{T _{1}} - \frac{1}{T _{2}})

. Putting

K_{1} = 10

298 K

and

K_{2} = 100

373 K

, the left side is

lo g 10 = 1

, and

\frac{1}{298} - \frac{1}{373} = 6.75 \times 1 0^{- 4}

, so

Δ H^{\circ} = \frac{2.303 \times 8.314}{6.75 \times 1 0 ^{- 4}} \approx 28.4 kJ mol^{- 1}

. The free energies come from

Δ G^{\circ} = - 2.303 R T lo g K

: at

298 K

Δ G^{\circ} = - 2.303 \times 8.314 \times 298 \times lo g 10 \approx - 5.71 kJ mol^{- 1}

, and at

373 K

Δ G^{\circ} = - 2.303 \times 8.314 \times 373 \times lo g 100 \approx - 14.29 kJ mol^{- 1}

. The three values are about

28.4

- 5.71

and

- 14.29 kJ mol^{- 1}

JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Single correct option