The heat of combustion of sucrose, $C_{12} H_{22} O_{11} (s)$ at constant volume is $- 1348.9 k c a l m o l^{- 1}$ at $25^{0}$ C, then the heat of reaction at constant pressure, when steam is produced is:

- 1342.344 k c a l

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+ 1342.344 k c a l

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+ 1250 k c a l

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none of the above

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Solution

The correct option is C $- 1342.344 k c a l$
Given, $△ E = - 1348.9 K C a l / m o l$ (Internal Energy Change)
$T = 25^{o} C = 298 K$
$R = 2 c a l / K$ (Gas Constant)

By using the equation,
$△ H = △ E + △ n_{g} R T$

$C_{12} H_{22} O_{11} (s) + 12 O_{2} \to 12 C O_{2} (g) + 11 H_{2} O (g)$

$△ n_{g} = 11 + 12 - 12 = 11 m o l e$

$Δ H = - 1348.9 \times 10^{3} c a l + 11 \times 2 \times 298 c a l$

$△ H = - 1342.34 K c a l$

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The heat of combustion of sucrose, C12H22O11(s) at constant volume is −1348.9 kcal mol−1 at 250C, then the heat of reaction at constant pressure, when steam is produced is:

The correct option is C −1342.344 kcalGiven, △E=−1348.9 KCal/mol (Internal Energy Change)T=25oC=298 KR=2 cal/K (Gas Constant)By using the equation, △H=△E+△ngRTC12H22O11(s)+12O2→12CO2(g)+11H2O(g)△ng=11+12−12=11 moleΔH=−1348.9×103cal+11×2×298cal△H=−1342.34 Kcal

The heat of combustion of sucrose, $C_{12} H_{22} O_{11} (s)$ at constant volume is $- 1348.9 k c a l m o l^{- 1}$ at $25^{0}$ C, then the heat of reaction at constant pressure, when steam is produced is: