For the reaction : A(l) -> 2B(g) \ Delta U = 2.1 kcal, Delta S = 20 cal K^ - 1 at 300 K Hence Delta G in kcal is :

For the reaction Al 2Bg Delta U 21kcalDelta S 20calK 1 ... | (Chemistry) | DivineJEE

Exam

Chemistry

For the reaction :

\\

l

)

\to

2B(g)

\\ \Delta U = 2.1\,kcal,\,\Delta S = 20\,cal\,{K^{ - 1}}

at 300 K

\\

Hence

\Delta

G in kcal is :

A( $l$ ) $\to$ 2B(g)

We know, $\Delta$ H = $\Delta$ U + $\Delta$ n_gRT

and $\Delta$ G = $\Delta$ H - T $\Delta$ S

$\therefore$ $\Delta$ G = $\Delta$ U + $\Delta$ n_gRT - T $\Delta$ S

= 2.1 + ${{2 \times 2 \times 300} \over {1000}}$ - ${{300 \times 20} \over {1000}}$

= 2.1 + 1.2 - 6 = – 2.70 Kcal/mol

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