18 g glucose is completely combusted in a bo | vedclass

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(C) $1$. Moles of glucose $(n)$ = $\frac{18 \ g}{180 \ g/mol} = 0.1 \ mol$. $2$. Heat released $(q)$ = $C 	imes \Delta T = 1400 \ kJ/K 	imes (27.2 -

Vedclass · Accepted Answer

$2800$

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(C) $1$. Moles of glucose $(n)$ = $\frac{18 \ g}{180 \ g/mol} = 0.1 \ mol$. $2$. Heat released $(q)$ = $C 	imes \Delta T = 1400 \ kJ/K 	imes (27.2 - 27) \ K = 1400 	imes 0.2 = 280 \ kJ$. $3$. Internal energy of combustion $(\Delta_{c}U^{\circ})$ = $-\frac{q}{n} = -\frac{280 \ kJ}{0.1 \ mol} = -2800 \ kJ/mol$. $4$. The combustion reaction is $C_{6}H_{12}O_{6(s)} + 6O_{2(g)} ightarrow 6CO_{2(g)} + 6H_{2}O_{(l)}$. $5$. $\Delta n_{g} = n_{p(g)} - n_{r(g)} = 6 - 6 = 0$. $6$. Since $\Delta n_{g} = 0$,$\Delta_{c}H^{\circ} = \Delta_{c}U^{\circ} + \Delta n_{g}RT = \Delta_{c}U^{\circ} = -2800 \ kJ/mol$. $7$. The magnitude is $2800 \ kJ/mol$.

$18 \ g$ glucose is completely combusted in a bomb calorimeter of heat capacity $1400 \ kJ/K$. The temperature changes from $27 \ ^\circ C$ to $27.2 \ ^\circ C$. Calculate the magnitude of the standard enthalpy of combustion of glucose in $kJ/mol$. $[R = 8.314 \ J/mol-K]$

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