Question

Be sure to answer all parts. The standard enthalpy of formation and the standard entropy of gaseous benzene are 82.93 kJ/mol and 269.2 J/K·mol, respectively. Calculate ΔH o , ΔS o , and ΔG o for the following process at 25.00°C. C6H6(l) → C6H6(g) ΔH o = kJ/mol ΔS o = J/K·mol ΔG o = kJ/mol Is the reaction spontaneous at 25.00°C?

Answer 1

The reaction is not spontaneous at 25.00°C

Further explanation

Gibbs free energy is the maximum possible work given by chemical reactions at constant pressure and temperature. Gibbs free energy can be used to determine the spontaneity of a reaction

If the Gibbs free energy value is <0 (negative) then the chemical reaction occurs spontaneously. If the change in free energy is zero, then the chemical reaction is at equilibrium, if it is> 0, the process is not spontaneous

Free energy of reaction (G) is the sum of its enthalpy (H) plus the product of the temperature and the entropy (S) of the system

Can be formulated: (at any temperature)

$\large {\boxed {\bold {\Delta G = \Delta H-T. \Delta S}}}$

or at (25 Celsius / 298 K, 1 atm = standard)

ΔG ° reaction = ΔG ° f (products) - ΔG ° f (reactants)

Under standard conditions:

∆G ° = ∆H ° - T∆S °

The value of °H ° can be calculated from the change in enthalpy of standard formation:

∆H ° (reaction) = ∑Hf ° (product) - ∑ Hf ° (reagent)

The value of ΔS ° can be calculated from standard entropy data

∆S ° (reaction) = ∑S ° (product) - ∑ S ° (reagent)

We complete the existing data from the reaction

C6H6 (l) → C6H6 (g)

∑Hf ° C6H6 (l): 49.04 kJ mol⁻¹

∑ Hf ° C6H6 (g): 82.93kJ mol⁻¹

∑S ° [C6H6 (l)]: 172.8 J mol⁻¹

∑S ° [C6H6 (g)]: 269.2 J mol⁻¹

so that:

∆H ° (reaction) = ∑ Hf ° C6H6 (g) -∑Hf ° C6H6 (l)

∆H ° (reaction) = 82.93kJ mol⁻¹ - 49.04 kJ mol⁻¹

∆H ° (reaction) = 33.89kJ mol⁻¹

∆S ° (reaction) = ∑S ° [C6H6 (g)] - ∑S ° [C6H6 (l)]

∆S ° (reaction) = 269.2J mol⁻¹ - 172.8 J mol⁻¹

∆S ° (reaction) = 96.4 J mol⁻¹ = 96.4 .10-3 kJ mol⁻¹

∆G ° at 298 K

∆G ° = ∆H ° - T∆S °

∆G ° = 33.89kJ mol⁻¹ - 298.96.4 .10-3 kJ mol⁻¹

∆G ° = 5.16 kJ mol⁻¹

Because the value of ∆G ° is positive, the reaction is not spontaneous

Learn more  

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Keywords: the standard gibbs free energy of formation,nonspontaneous

Answer 2

Answer : The values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $33.89kJ,95.94J/K\text{ and }5.299kJ/mol$ respectively.

Explanation :

The given balanced chemical reaction is,

$C_6H_6(l)\rightarrow C_6H_6(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$.

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{C_6H_6(g)}\times \Delta H_f^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta H_f^0_{(C_6H_6(l))}]$

where,

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

n = number of moles

$\Delta H_f^0_{(C_6H_6(g))}$ = standard enthalpy of formation  of gaseous benzene = 82.93 kJ/mol

$\Delta H_f^0_{(C_6H_6(l))}$ = standard enthalpy of formation  of liquid benzene = 49.04 kJ/mol

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

$\Delta H^o=[1mole\times (82.93kJ/mol)]-[1mole\times (49.04J/mol)]$

$\Delta H^o=33.89kJ/mol=33890J/mol$

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

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{C_6H_6(g)}\times \Delta S^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta S^0_{(C_6H_6(l))}]$

where,

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

n = number of moles

$\Delta S^0_{(C_6H_6(g))}$ = standard entropy of formation  of gaseous benzene = 269.2 J/K.mol

$\Delta S^0_{(C_6H_6(l))}$ = standard entropy of formation  of liquid benzene = 173.26 J/K.mol

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

$\Delta S^o=[1mole\times (269.2J/K.mol)]-[1mole\times (173.26J/K.mol)]$

$\Delta S^o=95.94J/K.mol$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$.

As we know that,

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

At room temperature, the temperature is $25^oC\text{ or }298K$.

$\Delta G^o=(33890J)-(298K\times 95.94J/K)$

$\Delta G^o=5299.88J/mol=5.299kJ/mol$

Therefore, the values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $33.89kJ,95.94J/K\text{ and }5.299kJ/mol$ respectively.