When 2.0 g of sucrose is oxidized to form CO | vedclass

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Question

(A) The balanced chemical equation for the combustion of sucrose is: $C_{12}H_{22}O_{11(s)} + 12O_{2(g)} \rightarrow 12CO_{2(g)} + 11H_2O(\ell)$.Calcu

Vedclass · Accepted Answer

$-4104 \ kJ \ mol^{-1}$

Vedclass · Answer

(A) The balanced chemical equation for the combustion of sucrose is: $C_{12}H_{22}O_{11(s)} + 12O_{2(g)} ightarrow 12CO_{2(g)} + 11H_2O(\ell)$.Calculate the change in the number of moles of gas: $\Delta n_g = n_{p(g)} - n_{r(g)} = 12 - 12 = 0$.The relationship between enthalpy change and internal energy change is: $\Delta H = \Delta U + \Delta n_gRT$.Since $\Delta n_g = 0$,$\Delta H = \Delta U = -24 \ kJ$ for $2.0 \ g$ of sucrose.To find $\Delta H$ for $1 \ mol$ $(342 \ g)$: $\Delta H = (\frac{-24 \ kJ}{2.0 \ g}) 	imes 342 \ g \ mol^{-1} = -4104 \ kJ \ mol^{-1}$.

When $2.0 \ g$ of sucrose is oxidized to form $CO_{2(g)}$ and $H_2O(\ell)$,the internal energy changes by $-24 \ kJ$. Calculate the value of $\Delta H$ at $298 \ K$ in $kJ \ mol^{-1}$. (Molar mass of sucrose $= 342 \ g \ mol^{-1}$)

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