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

Answer:

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)

$\Delta S ^0_{rxn} =536 . 7 \ J/K * \frac{1 \ kJ}{1000 \ J}$

$\Delta S ^0_{rxn} =0.5367 \frac{kJ}{K}$

Given that;

at T = 25°C = ( 25 + 273) K = 298 K

$\Delta G^0 _{rxn} = \Delta H^0_{rxn} - T \Delta S^0_{rxn}$

$\Delta G^0 _{rxn} = -68 \ kJ - 298 * 0.5367 \frac{kJ}{K}$

$\Delta G^0 _{rxn} = -227. 9 \ \ \ kJ$

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.