Heat of combustion of glucose at constant pressure at $17^{o} C$ was found to be $- 651, 000 c a l$ . Calculate the heat of combustion of glucose at constant volume considering water to be in the gaseous state.

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Solution

Equation for combustion of glucose
$C_{6} H_{12} O_{6} (s) + 6 O_{2} (g) \to 6 C O_{2} (g) + 6 H_{2} O (g)$
We know, $Q_{p} = Q_{v} + Δ n R T$
where, $Q_{p} = heat of combustion at constant pressure = - 651, 000 c a l$
$Q_{v} = heat of combustion at constant volume$
$Δ n = number of gaseous moles of product - number of gaseous moles of reactant$
$Δ n = 12 - 2 = 6$
$R = Rydberg's constant = 2 c a l K^{- 1} m o l^{- 1}$
$T = temperature in K = 273 + 17 = 290 K$
$- 651, 000 = Q_{v} + (6 \times 2 \times 290)$
$Q_{v} = - 651, 000 - 3480$
$Q_{v} = - 654, 480 c a l$

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