For the complete combustion of ethanol,C_2H_5 | vedclass

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(A) The reaction is: $C_2H_5OH_{(l)} + 3O_{2(g)} \rightarrow 2CO_{2(g)} + 3H_2O_{(l)}$Given internal energy change $\Delta U = -1364.47 \ kJ \ mol^{-1

Vedclass · Accepted Answer

$-1366.95$

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(A) The reaction is: $C_2H_5OH_{(l)} + 3O_{2(g)} ightarrow 2CO_{2(g)} + 3H_2O_{(l)}$Given internal energy change $\Delta U = -1364.47 \ kJ \ mol^{-1}$ (heat released in bomb calorimeter).Temperature $T = 25 + 273 = 298 \ K$.Change in moles of gaseous species $\Delta n_g = (n_{products, gas} - n_{reactants, gas}) = 2 - 3 = -1$.Gas constant $R = 8.314 \ J \ K^{-1} \ mol^{-1} = 8.314 	imes 10^{-3} \ kJ \ K^{-1} \ mol^{-1}$.The relationship between enthalpy change and internal energy change is $\Delta H = \Delta U + \Delta n_g RT$.Substituting the values: $\Delta H = -1364.47 + (-1 	imes 8.314 	imes 10^{-3} 	imes 298)$.$\Delta H = -1364.47 - 2.477572 \approx -1366.95 \ kJ \ mol^{-1}$.

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