Respuesta :
Answer :
Entropy of system = 94.6 J/K.mol
Entropy of surrounding = -94.6 J/K.mol
Entropy of universe = 0 J/K.mol
The process is a spontaneous process.
Explanation :
The given chemical reaction is:
[tex]PCl_3(l)\rightarrow PCl_3(g)[/tex]
First we have to calculate the entropy of reaction [tex](\Delta S^o)[/tex].
[tex]\Delta S^o=S_f_{product}-S_f_{reactant}[/tex]
[tex]\Delta S^o=[n_{PCl_3(g)}\times \Delta S^0_{(PCl_3(g))}]-[n_{PCl_3(l)}\times \Delta S^0_{(PCl_3(l))}][/tex]
where,
[tex]\Delta S^o[/tex] = entropy of reaction = ?
n = number of moles
[tex]\Delta S^0_{(PCl_3(g))}[/tex] = standard entropy of formation of gaseous [tex]PCl_3[/tex] = 311.7 J/K.mol
[tex]\Delta S^0_{(PCl_3(l))}[/tex] = standard entropy of formation of liquid [tex]PCl_3[/tex] = 217.1 J/K.mol
Now put all the given values in this expression, we get:
[tex]\Delta S^o=[1mole\times (311.7J/K.mol)]-[1mole\times (217.1J/K.mol)][/tex]
[tex]\Delta S^o=94.6J/K.mol[/tex]
Entropy of reaction = [tex]\Delta S^o[/tex] = Entropy of system = 94.6 J/K.mol
Entropy of system = -Entropy of surrounding = - 94.6 J/K.mol
As, we know that:
Entropy of universe = Entropy of system + Entropy of surrounding
Entropy of universe = 94.6 J/K.mol + (-94.6 J/K.mol)
Entropy of universe = 0
Now we have to calculate the enthalpy of reaction [tex](\Delta H^o)[/tex].
[tex]\Delta H^o=H_f_{product}-H_f_{reactant}[/tex]
[tex]\Delta H^o=[n_{PCl_3(g)}\times \Delta H^0_{(PCl_3(g))}]-[n_{PCl_3(l)}\times \Delta H^0_{(PCl_3(l))}][/tex]
where,
[tex]\Delta H^o[/tex] = enthalpy of reaction = ?
n = number of moles
[tex]\Delta H^0_{(PCl_3(g))}[/tex] = standard enthalpy of formation of gaseous [tex]PCl_3[/tex] = -319.7 kJ/mol
[tex]\Delta H^0_{(PCl_3(l))}[/tex] = standard enthalpy of formation of liquid [tex]PCl_3[/tex] = -288.07 kJ/mol
Now put all the given values in this expression, we get:
[tex]\Delta H^o=[1mole\times (-319.7kJ/mol)]-[1mole\times (-288.07kJ/mol)][/tex]
[tex]\Delta H^o=-31.63kJ/mol[/tex]
Now we have to calculate the Gibbs free energy.
As we know that,
[tex]\Delta G^o=\Delta H^o-T\Delta S^o[/tex]
where,
[tex]\Delta G^o[/tex] = standard Gibbs free energy = ?
[tex]\Delta H^o[/tex] = standard enthalpy = -31.63 kJ = -31630 J
[tex]\Delta S^o[/tex] = standard entropy = -94.6 J/K
T = temperature of reaction = [tex]61.2^oC=273+61.2=334.2K[/tex]
Now put all the given values in the above formula, we get:
[tex]\Delta G^o=(-31630J)-(334.2K\times -94.6J/K)[/tex]
[tex]\Delta G^o=-14.68J[/tex]
A reaction to be spontaneous when [tex]\Delta G<0[/tex]
A reaction to be non-spontaneous when [tex]\Delta G>0[/tex]
For the reaction to be spontaneous, the Gibbs free energy of the reaction [tex]\Delta G[/tex] is negative or we can say that the value of [tex]\Delta G[/tex] is less than zero.
As the value of [tex]\Delta G[/tex] is less than zero that means the reaction is spontaneous.
