Answer:
Explanation:
The chemical equation for the reaction is :
[tex]C_6H_{12}O_6_{(s)} \to 2C_2H_5OH _{(l)} + 2CO_{2(g)}[/tex]
The standard enthalpy of formation [tex]\Delta H ^0_f[/tex] of the above equation is as follows:
[tex]\Delta H^0_{f, C_6H_{12}O_6}[/tex] = -1274.4 kJ/mol
[tex]\Delta H ^0_{f, C_2H_{5}OH[/tex] = -277.7 kJ/mol
[tex]\Delta H ^0_{f, CO_2}[/tex] = -393.5 kJ/mol
[tex]\Delta H ^0_{rxn }= \sum n_p \Delta H ^0_{f,p} - \sum n_r \Delta H ^0_{f,r}[/tex]
where ;
[tex]n_p[/tex] = stochiometric coefficients of products
[tex]n_r=[/tex] stochiometric coefficients of reactants
[tex]\Delta H^0_{f.p}[/tex] = formation standard enthalpy of products
[tex]\Delta H^0_{f.r}[/tex] = formation standard enthalpy of reactants
[tex]\Delta H ^0_{rxn }= (2 \ mol* -2777.7 \ kJ/mol + 2 \ mol * - 393.5 \ kJ/mol) - (1\ mol *(-1274.4 \ kJ/mol)[/tex]
[tex]\Delta H ^0_{rxn }= -68 \ kJ[/tex]
For [tex]\Delta S ^0[/tex] ;
The standard enthalpy of formation of [tex]\Delta S ^0_f[/tex] of the reactant and the products are :
[tex]\Delta S ^0 _{f \ C_6H_{12}O_6} = 212 \ \ J/Kmol[/tex]
[tex]\Delta S ^0 _{f \ C_2H_5O_H} = 160.7 \ \ J/Kmol[/tex]
[tex]\Delta S ^0 _{f \ CO_2} = 213.63 \ \ J/Kmol[/tex]
The [tex]\Delta S ^0_{rxn}[/tex] is as follows:
[tex]\Delta S ^0_{rxn} = \sum n_p \Delta S^0_{f.p} - \sum n_r \Delta S^0_{f.r}[/tex]
[tex]\Delta S ^0_{rxn} = (2 \ mol *160.7 \ \ J/Kmol + 2 \ mol *213.63 \ \ J/Kmol) -(1*212 J/Kmol)[/tex]
[tex]\Delta S ^0_{rxn} =536 . 7 \ J/K[/tex] (to kJ/K)
[tex]\Delta S ^0_{rxn} =536 . 7 \ J/K * \frac{1 \ kJ}{1000 \ J}[/tex]
[tex]\Delta S ^0_{rxn} =0.5367 \frac{kJ}{K}[/tex]
Given that;
at T = 25°C = ( 25 + 273) K = 298 K
[tex]\Delta G^0 _{rxn} = \Delta H^0_{rxn} - T \Delta S^0_{rxn}[/tex]
[tex]\Delta G^0 _{rxn} = -68 \ kJ - 298 * 0.5367 \frac{kJ}{K}[/tex]
[tex]\Delta G^0 _{rxn} = -227. 9 \ \ \ kJ[/tex]
As [tex]\Delta G^0 _{rxn}[/tex] is negative; the reaction is spontaneous
[tex]\Delta H^0 _{rxn}[/tex] = negative
[tex]\Delta S^0 _{rxn}[/tex] = positive; Therefore , the reaction is spontaneous at all temperature , We can then say that the spontaneity of this reaction [tex]\Delta G^0 _{rxn} = \Delta H^0_{rxn} - T \Delta S^0_{rxn}[/tex] is dependent on temperature.
