Question

For the following process, determine the change in entropy for the system (ΔSsys), the surrounding (ΔSsys), and the universe (ΔSuniv). Then identify the process as a spontaneous process, a nonspontaneous process, or an equilibrium process at the specified temperature. (Assume that the thermodynamic data will not vary with temperature.) PCl3(l) → PCl3(g) at 61.2°C ΔH o f ( kJ mol ) −319.7 −288.07 S o ( J K · mol ) 217.1 311.7

Answer 1

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:

`PCl_3(l)\rightarrow PCl_3(g)`

First we have to calculate the entropy of reaction
`(\Delta S^o)`.

`\Delta S^o=S_f_(product)-S_f_(reactant)`

`\Delta S^o=[n_(PCl_3(g))* \Delta S^0_((PCl_3(g)))]-[n_(PCl_3(l))* \Delta S^0_((PCl_3(l)))]`

where,

`\Delta S^o` = entropy of reaction = ?

n = number of moles

`\Delta S^0_((PCl_3(g)))` = standard entropy of formation of gaseous
`PCl_3` = 311.7 J/K.mol

`\Delta S^0_((PCl_3(l)))` = standard entropy of formation of liquid
`PCl_3` = 217.1 J/K.mol

Now put all the given values in this expression, we get:

`\Delta S^o=[1mole* (311.7J/K.mol)]-[1mole* (217.1J/K.mol)]`

`\Delta S^o=94.6J/K.mol`

Entropy of reaction =
`\Delta S^o` = 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
`(\Delta H^o)`.

`\Delta H^o=H_f_(product)-H_f_(reactant)`

`\Delta H^o=[n_(PCl_3(g))* \Delta H^0_((PCl_3(g)))]-[n_(PCl_3(l))* \Delta H^0_((PCl_3(l)))]`

where,

`\Delta H^o` = enthalpy of reaction = ?

n = number of moles

`\Delta H^0_((PCl_3(g)))` = standard enthalpy of formation of gaseous
`PCl_3` = -319.7 kJ/mol

`\Delta H^0_((PCl_3(l)))` = standard enthalpy of formation of liquid
`PCl_3` = -288.07 kJ/mol

Now put all the given values in this expression, we get:

`\Delta H^o=[1mole* (-319.7kJ/mol)]-[1mole* (-288.07kJ/mol)]`

`\Delta H^o=-31.63kJ/mol`

Now we have to calculate the Gibbs free energy.

As we know that,

`\Delta G^o=\Delta H^o-T\Delta S^o`

where,

`\Delta G^o` = standard Gibbs free energy = ?

`\Delta H^o` = standard enthalpy = -31.63 kJ = -31630 J

`\Delta S^o` = standard entropy = -94.6 J/K

T = temperature of reaction =
`61.2^oC=273+61.2=334.2K`

Now put all the given values in the above formula, we get:

`\Delta G^o=(-31630J)-(334.2K* -94.6J/K)`

`\Delta G^o=-14.68J`

A reaction to be spontaneous when
`\Delta G<0`

A reaction to be non-spontaneous when
`\Delta G>0`

For the reaction to be spontaneous, the Gibbs free energy of the reaction
`\Delta G` is negative or we can say that the value of
`\Delta G` is less than zero.

As the value of
`\Delta G` is less than zero that means the reaction is spontaneous.