JEE MAIN - Chemistry (2020 - 2nd September Evening Slot - No. 5)
The heat of combustion of ethanol into carbon
dioxide and water is – 327 kcal at constant
pressure. The heat evolved (in cal) at constant
volume and 27oC (if all gases behave ideally) is
(R = 2 cal mol–1 K–1) ________.
Answer
326400
Explanation
C2H5OH (l) + 3O2(g) $$ \to $$ 2CO2(g) + 3H2O(l)
$$\Delta $$H = –327 kcal; $$\Delta $$ng = – 1
$$\Delta $$H = $$\Delta $$U + $$\Delta $$ngRT
$$ \Rightarrow $$ $$\Delta $$U = – 327 + 2 × 10–3 × 300
= – 326.4 kcal
= – 326400 cal
$$ \therefore $$ Heat evolved = 326400 cal.
$$\Delta $$H = –327 kcal; $$\Delta $$ng = – 1
$$\Delta $$H = $$\Delta $$U + $$\Delta $$ngRT
$$ \Rightarrow $$ $$\Delta $$U = – 327 + 2 × 10–3 × 300
= – 326.4 kcal
= – 326400 cal
$$ \therefore $$ Heat evolved = 326400 cal.
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