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JEE Main · 2020 · 07 Jan 2020 · Shift 1 · Chemical Thermodynamics

For the reaction

A (l) ⟶ 2 B (g)

Δ U = 2.1 Kcal, Δ S = 20 cal K^{- 1}

at 300 K
Hence

Δ G

in Kcal is

Answer: -2.7

Solution

For

A (l) \to 2 B (g)

the gas moles change by

Δ n_{g} = 2

, so with

R = 2 cal K^{- 1} mol^{- 1}

,

Δ H = Δ U + Δ n_{g} R T = 2100 + 2 \times 2 \times 300 = 3300 cal

. Then

Δ G = Δ H - T Δ S = 3300 - 300 \times 20 = - 2700 cal = - 2.7 kcal

.

JEE Main · 2020 · 07 Jan 2020 · Shift 1 · Numerical answer

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