1. Tardigrade
  2. Question
  3. Chemistry
  4. The heat of combustion of benzene at 27° C found by bomb calorimeter i . e ., for the reaction C 6 H 6(l)+(15/2) O 2(g) longrightarrow 6 CO 2( g )+3 H 2 O ( l ) is 780 kcal mol -1. The heat evolved on burning 39 g of benzene in an open vessel will be

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Q. The heat of combustion of benzene at $27^{\circ} C$ found by bomb calorimeter $i . e .,$ for the reaction
$C _{6} H _{6(l)}+\frac{15}{2} O _{2(g)} \longrightarrow 6 CO _{2( g )}+3 H _{2} O _{( l )}$
is $780\, kcal\, mol ^{-1}$. The heat evolved on burning $39 \,g$ of benzene in an open vessel will be

Thermodynamics

Solution:

$\Delta n=6-\frac{15}{2}=-\frac{3}{2}$

$\Delta H=\Delta E+\Delta n_{(g)} R T$

$=-780-\frac{3}{2} \times \frac{2}{1000} \times 300$

$=-780.9 kcal\, mol ^{-1}$

This is the heat evolved from 1 mole of benzene i.e., 78 g.

$\therefore $ From $39 g ,$ heat evolved

$=-\frac{780.9}{2}=-390.45$ kcal