For the complete combustion of ethene,C_2H_{4 | vedclass

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(A) The combustion reaction is: $C_2H_{4(g)} + 3O_{2(g)} \rightarrow 2CO_{2(g)} + 2H_2O_{(l)}$.Given that the heat measured in a bomb calorimeter is t

Vedclass · Accepted Answer

$1411$

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(A) The combustion reaction is: $C_2H_{4(g)} + 3O_{2(g)} ightarrow 2CO_{2(g)} + 2H_2O_{(l)}$.Given that the heat measured in a bomb calorimeter is the change in internal energy,$\Delta U = -1406 \ kJ \ mol^{-1}$.The change in the number of gaseous moles is $\Delta n_g = (2) - (1 + 3) = -2$.The relation between enthalpy change and internal energy change is $\Delta H = \Delta U + \Delta n_g RT$.Substituting the values: $\Delta H = -1406 \ kJ \ mol^{-1} + (-2 	imes 8.3 	imes 10^{-3} \ kJ \ K^{-1} \ mol^{-1} 	imes 300 \ K)$.$\Delta H = -1406 - 4.98 = -1410.98 \ kJ \ mol^{-1}$.At equilibrium,$\Delta G = 0$,which implies $\Delta H - T \Delta S = 0$,or $T \Delta S = \Delta H$.Therefore,$T \Delta S = -1410.98 \ kJ \ mol^{-1} \approx -1411 \ kJ \ mol^{-1}$.

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