For the reaction: X _ 2 O _ 4 (l) → 2X O _ 2 (g) ∆ U = 2 .1 kcal, ∆ S = 20cal K ^ - 1 at 300K Hence, ∆ G is:

Correct answer: B. - 2.7kcal ∆ H = ∆ U + ∆ n _ g RT = 2 .1 + (2 × 2 × 300)/(1000) = 3 . 3kcal ∆ G = ∆ H - T ∆ S = 3.3 - 300 × (20)/(1000) = 3.3 - 6 = - 2.7kcal

NEET 2014

Exam: NEET
Year: 2014
Subject: Chemistry
Chapter: Thermodynamics (C)

NEET 2014 · Chemistry · Thermodynamics (C)

For the reaction: X _ 2 O _ 4 (l) → 2X O _ 2 (g) ∆ U = 2 .1 kcal, ∆...

For the reaction:

{X}_{2}{O}_{4}\left(l\right) \to 2X{O}_{2}\left(g\right)

∆ U = 2 .1 kcal, ∆ S = 20cal{K}^{- 1}

300K

Hence,

∆ G

is:

Options

A $2.7kcal$
B $- 2.7kcal$
C $9.3kcal$
D $- 9.3kcal$

Correct answer

B. $- 2.7kcal$

Step-by-step solution

$∆ H = ∆ U + ∆ {n}_{g}RT$

$= 2 .1 + \dfrac{2 \times 2 \times 300}{1000} = 3 . 3kcal$

$∆ G = ∆ H - T ∆ S$

$= 3.3 - 300 \times \dfrac{20}{1000} = 3.3 - 6 = - 2.7kcal$

Practice Thermodynamics (C) on Quantrex Academy →

More from Thermodynamics (C)