The heat of combustion of benzene at 270C found by a bomb calorimeter i.e. for the reaction C6H6(l)+712O2(g)→6CO2(g)+3H2O(l) is 780kCal mol-1 . The heat evolved on burning 39g of benzene in an open vessel will be

ΔΗ=ΔΕ+ΔngRT =−780+(−32)×2×3001000 =-780.9 Kcal/mode ⇒for 39 g i.e.,n=3918=12 ⇒ heat evolved =780.92=390.45Kcal

The heat of combustion of benzene at 270C found by a bomb calorimeter i.e. for the reaction C6H6(l)+712O2(g)→6CO2(g)+3H2O(l) is 780kCal mol-1 . The heat evolved on burning 39g of benzene in an open vessel will be

ΔΗ=ΔΕ+ΔngRT =−780+(−32)×2×3001000 =-780.9 Kcal/mode ⇒for 39 g i.e.,n=3918=12 ⇒ heat evolved =780.92=390.45Kcal

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The heat of combustion of benzene at $27^{0} C$ found by a bomb calorimeter i.e. for the reaction $C_{6} H_{6 (l)} + 7 \frac{1}{2} O_{2 (g)} \to 6 C O_{2 (g)} + 3 H_{2} O_{(l)}$ 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

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$390.45 k C a l$

$360 k C a l$

$780.9 k C a l$

$780 k C a l$

answer is C.

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$Δ Η = Δ Ε + Δ ngRT = - 780 + (\frac{- 3}{2}) \times \frac{2 \times 300}{1000} = - 780.9 Kcal / mode \Rightarrow for 39 g i . e ., n = \frac{39}{18} = \frac{1}{2} \Rightarrow heat evolved = \frac{780.9}{2} = 390.45 Kcal$

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The heat of combustion of benzene at 270C found by a bomb calorimeter i.e. for the reaction C6H6(l)+712O2(g)→6CO2(g)+3H2O(l) is 780kCal mol-1 . The heat evolved on burning 39g of benzene in an open vessel will be

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The heat of combustion of benzene at $27^{0} C$ found by a bomb calorimeter i.e. for the reaction $C_{6} H_{6 (l)} + 7 \frac{1}{2} O_{2 (g)} \to 6 C O_{2 (g)} + 3 H_{2} O_{(l)}$ 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