Answer:
Explanation:
The chemical equation for the reaction is :
![C_6H_{12}O_6_{(s)} \to 2C_2H_5OH _{(l)} + 2CO_{2(g)}](https://tex.z-dn.net/?f=C_6H_%7B12%7DO_6_%7B%28s%29%7D%20%5Cto%202C_2H_5OH%20_%7B%28l%29%7D%20%2B%202CO_%7B2%28g%29%7D)
The standard enthalpy of formation
of the above equation is as follows:
= -1274.4 kJ/mol
= -277.7 kJ/mol
= -393.5 kJ/mol
![\Delta H ^0_{rxn }= \sum n_p \Delta H ^0_{f,p} - \sum n_r \Delta H ^0_{f,r}](https://tex.z-dn.net/?f=%5CDelta%20H%20%20%5E0_%7Brxn%20%7D%3D%20%5Csum%20n_p%20%5CDelta%20H%20%20%5E0_%7Bf%2Cp%7D%20-%20%5Csum%20n_r%20%5CDelta%20H%20%20%5E0_%7Bf%2Cr%7D)
where ;
= stochiometric coefficients of products
stochiometric coefficients of reactants
= formation standard enthalpy of products
= 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)](https://tex.z-dn.net/?f=%5CDelta%20H%20%20%5E0_%7Brxn%20%7D%3D%20%282%20%5C%20mol%2A%20-2777.7%20%5C%20kJ%2Fmol%20%2B%202%20%5C%20mol%20%2A%20-%20393.5%20%5C%20kJ%2Fmol%29%20-%20%281%5C%20mol%20%2A%28-1274.4%20%5C%20kJ%2Fmol%29)
![\Delta H ^0_{rxn }= -68 \ kJ](https://tex.z-dn.net/?f=%5CDelta%20H%20%20%5E0_%7Brxn%20%7D%3D%20-68%20%5C%20kJ)
For
;
The standard enthalpy of formation of
of the reactant and the products are :
![\Delta S ^0 _{f \ C_6H_{12}O_6} = 212 \ \ J/Kmol](https://tex.z-dn.net/?f=%5CDelta%20S%20%5E0%20_%7Bf%20%5C%20C_6H_%7B12%7DO_6%7D%20%3D%20212%20%5C%20%5C%20%20J%2FKmol)
![\Delta S ^0 _{f \ C_2H_5O_H} = 160.7 \ \ J/Kmol](https://tex.z-dn.net/?f=%5CDelta%20S%20%5E0%20_%7Bf%20%5C%20C_2H_5O_H%7D%20%3D%20160.7%20%5C%20%5C%20J%2FKmol)
![\Delta S ^0 _{f \ CO_2} = 213.63 \ \ J/Kmol](https://tex.z-dn.net/?f=%5CDelta%20S%20%5E0%20_%7Bf%20%5C%20CO_2%7D%20%3D%20213.63%20%5C%20%5C%20J%2FKmol)
The
is as follows:
![\Delta S ^0_{rxn} = \sum n_p \Delta S^0_{f.p} - \sum n_r \Delta S^0_{f.r}](https://tex.z-dn.net/?f=%5CDelta%20S%20%5E0_%7Brxn%7D%20%3D%20%5Csum%20n_p%20%5CDelta%20S%5E0_%7Bf.p%7D%20-%20%5Csum%20n_r%20%20%5CDelta%20S%5E0_%7Bf.r%7D)
![\Delta S ^0_{rxn} = (2 \ mol *160.7 \ \ J/Kmol + 2 \ mol *213.63 \ \ J/Kmol) -(1*212 J/Kmol)](https://tex.z-dn.net/?f=%5CDelta%20S%20%5E0_%7Brxn%7D%20%3D%20%282%20%5C%20mol%20%2A160.7%20%5C%20%5C%20J%2FKmol%20%2B%202%20%5C%20mol%20%2A213.63%20%5C%20%5C%20J%2FKmol%29%20-%281%2A212%20J%2FKmol%29)
(to kJ/K)
![\Delta S ^0_{rxn} =536 . 7 \ J/K * \frac{1 \ kJ}{1000 \ J}](https://tex.z-dn.net/?f=%5CDelta%20S%20%5E0_%7Brxn%7D%20%3D536%20.%207%20%5C%20J%2FK%20%2A%20%5Cfrac%7B1%20%5C%20kJ%7D%7B1000%20%5C%20J%7D)
![\Delta S ^0_{rxn} =0.5367 \frac{kJ}{K}](https://tex.z-dn.net/?f=%5CDelta%20S%20%5E0_%7Brxn%7D%20%3D0.5367%20%5Cfrac%7BkJ%7D%7BK%7D)
Given that;
at T = 25°C = ( 25 + 273) K = 298 K
![\Delta G^0 _{rxn} = \Delta H^0_{rxn} - T \Delta S^0_{rxn}](https://tex.z-dn.net/?f=%5CDelta%20G%5E0%20_%7Brxn%7D%20%3D%20%5CDelta%20H%5E0_%7Brxn%7D%20-%20T%20%5CDelta%20S%5E0_%7Brxn%7D)
![\Delta G^0 _{rxn} = -68 \ kJ - 298 * 0.5367 \frac{kJ}{K}](https://tex.z-dn.net/?f=%5CDelta%20G%5E0%20_%7Brxn%7D%20%3D%20%20-68%20%5C%20kJ%20-%20298%20%2A%200.5367%20%5Cfrac%7BkJ%7D%7BK%7D)
![\Delta G^0 _{rxn} = -227. 9 \ \ \ kJ](https://tex.z-dn.net/?f=%5CDelta%20G%5E0%20_%7Brxn%7D%20%3D%20%20-227.%209%20%5C%20%5C%20%5C%20kJ)
As
is negative; the reaction is spontaneous
= negative
= positive; Therefore , the reaction is spontaneous at all temperature , We can then say that the spontaneity of this reaction
is dependent on temperature.
Q=M×C×T
..that's the equation
Answer:
The boiling point of the fluoromethane (CH3F) is higher than that of fluorine (F2).