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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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In the reaction of nitrogen gas with oxygen gas to produce nitrogen oxide, what is the effect of adding more oxygen gas to the i
Alborosie
<h3>Answer:</h3>

The Equilibrium would shift to produce more NO

<h3>Explanation:</h3>

The reaction is;

N₂(g) + O₂(g) ⇆ 2NO(g)

  • When a reaction is at equilibrium then the forward reaction rate will be equivalent to the reverse reaction rate. Additionally, the concentration of the reactants and products are the same.
  • From Le Chatelier's principle, additional reactants favor the formation of more products while additional products favor the formation of more reactants.
  • For example, when more oxygen is added then more Nitrogen (II) oxide will be formed.
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3 years ago
Consider the process used to produce iron metal from its ore.
Dmitry_Shevchenko [17]

Answer:

223.4 grams of iron can be produced from 2.5 moles of Fe2O3 and 6.0 moles of CO.

Explanation:

The balanced reaction is:

Fe₂O₃ (s) + 3 CO(g) → 2 Fe(s) + 3 CO₂ (g)

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reactant and product participate in the reaction:

  • Fe₂O₃: 1 mole
  • CO: 3 moles
  • Fe: 2 moles
  • CO₂: 3 moles

Being:

  • Fe: 55.85 g/mole
  • O: 16 g/mole
  • C: 12 g/mole

the molar mass of the compounds participating in the reaction is:

  • Fe₂O₃: 2*55.85 g/mole + 3*16 g/mole= 159.7 g/mole
  • CO: 12 g/mole + 16 g/mole= 28 g/mole
  • Fe: 55.85 g/mole
  • CO₂: 12 g/mole + 2*16 g/mole= 44 g/mole

Then, by stoichiometry of the reaction, the following quantities participate in the reaction:

  • Fe₂O₃: 1 mole* 159.7 g/mole= 159.7 g
  • CO: 3 moles* 28 g/mole= 84 g
  • Fe: 2 moles* 55.85 g/mole= 111.7 g
  • CO₂: 3 moles* 44 g/mole= 132 g

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

So, first of all, you can apply the following rule of three: if by reaction stoichiometry 1 mole of Fe₂O₃ reacts with 3 moles of CO, then 2.5 moles of Fe₂O₃ react with how many moles of CO?

moles of CO=\frac{2.5 moles of Fe_{2} O_{3}*3 moles of CO }{1 mole of Fe_{2} O_{3}}

moles of CO= 7.5

But 7.5 moles of CO are not available, 6.0 moles are available. Since you have less moles than you need to react with 2.5 moles of Fe₂O₃, CO will be the limiting reagent.

Now you can apply the following rule of three: if by reaction stoichiometry 3 moles of CO produce with 111.7 grams of Fe, then 6 moles of CO will produce how much mass of Fe?

mass of Fe=\frac{6 moles of CO*111.7 grams of Fe}{3 moles of CO}

mass of Fe= 223.4 grams

<u><em>223.4 grams of iron can be produced from 2.5 moles of Fe2O3 and 6.0 moles of CO.</em></u>

5 0
2 years ago
Someone please help me
ehidna [41]
I hop this will help u

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The heat of vaporization of a liquid is 84.0 J/g. How many joules of heat would it take to completely vaporize 172 g of this liq
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Answer:

14,448 J of heat would it take to completely vaporize 172 g of this liquid at its boiling point.

Explanation:

The heat Q that is necessary to provide for a mass m of a certain substance to change phase is equal to Q = m*L, where L is called the latent heat of the substance and depends on the type of phase change.

During the evaporation process, a substance goes from a liquid to a gaseous state and needs to absorb a certain amount of heat from its immediate surroundings, which results in its cooling. The heat absorbed is called the heat of vaporization.

So, it is called "heat of vaporization", the energy required to change 1 gram of substance from a liquid state to a gaseous state at the boiling point.

In this case, being:

  • Q=?
  • m= 172 g
  • L= 84 \frac{J}{g}

and replacing in the expression Q = m*L you get:

Q=172 g*84 \frac{J}{g}

Q=14,448 J

<u><em>14,448 J of heat would it take to completely vaporize 172 g of this liquid at its boiling point.</em></u>

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