Question

The U.S. requires automobile fuels to contain a renewable component. The fermentation of glucose from corn produces ethanol, which is added to gasoline to fulfill this requirement: C6H12O6(s) → 2 C2H5OH(l) + 2 CO2(g) Calculate ΔH o , ΔS o and ΔG o for the reaction at 25°C. Is the spontaneity of this reaction dependent on temperature?

Answer 1

Step-by-step explanation:

The chemical equation for the reaction is :

`C_6H_(12)O_6_((s)) \to 2C_2H_5OH _((l)) + 2CO_(2(g))`

The standard enthalpy of formation
`\Delta H ^0_f` of the above equation is as follows:

`\Delta H^0_{f, C_6H_(12)O_6}` = -1274.4 kJ/mol

`\Delta H ^0_{f, C_2H_(5)OH` = -277.7 kJ/mol

`\Delta H ^0_(f, CO_2)` = -393.5 kJ/mol

`\Delta H ^0_(rxn )= \sum n_p \Delta H ^0_(f,p) - \sum n_r \Delta H ^0_(f,r)`

where ;

`n_p` = stochiometric coefficients of products

`n_r=` stochiometric coefficients of reactants

`\Delta H^0_(f.p)` = formation standard enthalpy of products

`\Delta H^0_(f.r)` = formation standard enthalpy of reactants

`\Delta H ^0_(rxn )= (2 \ mol* -2777.7 \ kJ/mol + 2 \ mol * - 393.5 \ kJ/mol) - (1\ mol *(-1274.4 \ kJ/mol)`

`\Delta H ^0_(rxn )= -68 \ kJ`

For
`\Delta S ^0` ;

The standard enthalpy of formation of
`\Delta S ^0_f` of the reactant and the products are :

`\Delta S ^0 _{f \ C_6H_(12)O_6} = 212 \ \ J/Kmol`

`\Delta S ^0 _(f \ C_2H_5O_H) = 160.7 \ \ J/Kmol`

`\Delta S ^0 _(f \ CO_2) = 213.63 \ \ J/Kmol`

The
`\Delta S ^0_(rxn)` is as follows:

`\Delta S ^0_(rxn) = \sum n_p \Delta S^0_(f.p) - \sum n_r \Delta S^0_(f.r)`

`\Delta S ^0_(rxn) = (2 \ mol *160.7 \ \ J/Kmol + 2 \ mol *213.63 \ \ J/Kmol) -(1*212 J/Kmol)`

`\Delta S ^0_(rxn) =536 . 7 \ J/K` (to kJ/K)

`\Delta S ^0_(rxn) =536 . 7 \ J/K * (1 \ kJ)/(1000 \ J)`

`\Delta S ^0_(rxn) =0.5367 (kJ)/(K)`

Given that;

at T = 25°C = ( 25 + 273) K = 298 K

`\Delta G^0 _(rxn) = \Delta H^0_(rxn) - T \Delta S^0_(rxn)`

`\Delta G^0 _(rxn) = -68 \ kJ - 298 * 0.5367 (kJ)/(K)`

`\Delta G^0 _(rxn) = -227. 9 \ \ \ kJ`

As
`\Delta G^0 _(rxn)` is negative; the reaction is spontaneous

`\Delta H^0 _(rxn)` = negative

`\Delta S^0 _(rxn)` = positive; Therefore , the reaction is spontaneous at all temperature , We can then say that the spontaneity of this reaction
`\Delta G^0 _(rxn) = \Delta H^0_(rxn) - T \Delta S^0_(rxn)` is dependent on temperature.