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cluponka [151]
3 years ago
12

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?
Chemistry
2 answers:
SIZIF [17.4K]3 years ago
7 0

The reaction <u>is not spontaneous</u> at 25.00°C

<h3>Further explanation </h3>

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:

<h3>∆G ° = ∆H ° - T∆S ° </h3>

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

<h3>Learn more   </h3>

Delta H solution  

brainly.com/question/10600048  

an exothermic reaction  

brainly.com/question/1831525  

as endothermic or exothermic  

brainly.com/question/11419458  

an exothermic dissolving process  

brainly.com/question/10541336  

Keywords: the standard gibbs free energy of formation,nonspontaneous

skelet666 [1.2K]3 years ago
6 0

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.

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