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

Question

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 (A) 380 kcal (B) 780.9 kcal (C) 390.45 kcal (D) 780 kcal

Tardigrade Admin · Accepted Answer

Δ n=6-(15/2)=-(3/2) Δ H=Δ E+Δ n(g) R T =-780-(3/2) × (2/1000) × 300 =-780.9 kcal mol -1 This is the heat evolved from 1 mole of benzene i.e., 78 g. ∴ From 39 g , heat evolved =-(780.9/2)=-390.45 kcal

Q. The heat of combustion of benzene at $2 7^{\circ} C$ found by bomb calorimeter $i . e .,$ for the reaction
$C_{6} H_{6 (l)} + \frac{15}{2} O_{2 (g)} ⟶ 6 C O_{2 (g)} + 3 H_{2} O_{(l)}$
is $780 k c a l m o l^{- 1}$ . The heat evolved on burning $39 g$ of benzene in an open vessel will be

$380$ kcal

$780.9$ kcal

$390.45$ kcal

$780$ kcal

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

$Δ H = Δ E + Δ n_{(g)} RT$

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

$= - 780.9 k c a l m o l^{- 1}$

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

$∴$ From $39 g,$ heat evolved

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

Q. The heat of combustion of benzene at 27∘C found by bomb calorimeter i.e., for the reaction C6​H6(l)​+215​O2(g)​⟶6CO2(g)​+3H2​O(l)​ is 780kcalmol−1. The heat evolved on burning 39g of benzene in an open vessel will be

380 kcal

780.9 kcal

390.45 kcal

780 kcal