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

For the reaction:

X_{2} O_{4} (l) \to 2 X O_{2} (g)

∆ U = 2 .1 k cal, ∆ S = 20 cal K^{- 1}

at

300 K

Hence,

∆ G

is:

$2.7 k c a l$
$- 2.7 k c a l$
$9.3 k c a l$
$- 9.3 k c a l$

Solution

$∆ H = ∆ U + ∆ n_{g} RT$

$= 2 .1 + \frac{2 \times 2 \times 300}{1000} = 3.3 k cal$

$∆ G = ∆ H - T ∆ S$

$= 3.3 - 300 \times \frac{20}{1000} = 3.3 - 6 = - 2.7 k c a l$

Asked in: NEET 2014

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