JEE MAIN - Chemistry (2020 - 7th January Morning Slot - No. 3)
For the reaction :
A($$l$$) $$ \to $$ 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)
$$\Delta U = 2.1\,kcal,\,\Delta S = 20\,cal\,{K^{ - 1}}$$ at 300 K
Hence $$\Delta $$G in kcal is :
Answer
-2.7
Explanation
A($$l$$) $$ \to $$ 2B(g)
We know, $$\Delta $$H = $$\Delta $$U + $$\Delta $$ngRT
and $$\Delta $$G = $$\Delta $$H - T$$\Delta $$S
$$ \therefore $$ $$\Delta $$G = $$\Delta $$U + $$\Delta $$ngRT - 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
We know, $$\Delta $$H = $$\Delta $$U + $$\Delta $$ngRT
and $$\Delta $$G = $$\Delta $$H - T$$\Delta $$S
$$ \therefore $$ $$\Delta $$G = $$\Delta $$U + $$\Delta $$ngRT - 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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