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Eduardwww [97]
3 years ago
15

At 298 K, what is the Gibbs free energy change (ΔG) for the following reaction?

Chemistry
1 answer:
9966 [12]3 years ago
7 0

Answer:

(a). The Gibbs free energy change is 2.895 kJ and its positive.

(b). The Gibbs free energy change is 34.59 J/mole

(c). The pressure is 14924 atm.

(d). The Gibbs free energy of diamond relative to graphite is 4912 J.

Explanation:

Given that,

Temperature = 298 K

Suppose, density of graphite is 2.25 g/cm³ and density of diamond is 3.51 g/cm³.

\Delta H\ for\ diamond = 1.897 kJ/mol

\Delta H\ for\ graphite = 0 kJ/mol

\Delta S\ for\ diamond = 2.38 J/(K mol)

\Delta S\ for\ graphite = 5.73 J/(K mol)

(a) We need to calculate the value of \Delta G for diamond

Using formula of Gibbs free energy change

\Delta G=\Delta H-T\Delta S

Put the value into the formula

\Delta G= (1897-0)-298\times(2.38-5.73)

\Delta G=2895.3

\Delta G=2.895\ kJ

The Gibbs free energy  change is positive.

(b). When it is compressed isothermally from 1 atm to 1000 atm

We need to calculate the change of Gibbs free energy of diamond

Using formula of gibbs free energy

\Delta S=V\times\Delta P

\Delta S=\dfrac{m}{\rho}\times\Delta P

Put the value into the formula

\Delta S=\dfrac{12\times10^{-6}}{3.51}\times999\times10130

\Delta S=34.59\ J/mole

(c). Assuming that graphite and diamond are incompressible

We need to calculate the pressure

Using formula of Gibbs free energy

\beta= \Delta G_{g}+\Delta G+\Delta G_{d}

\beta=V(-\Delta P_{g})+\Delta G+V\Delta P_{d}

\beta=\Delta P(V_{d}-V_{g})+\Delta G

Put the value into the formula

0=\Delta P(\dfrac{12\times10^{-6}}{3.51}-\dfrac{12\times10^{-6}}{2.25})\times10130+2895.3

0=-0.0194\Delta P+2895.3

\Delta P=\dfrac{2895.3}{0.0194}

\Delta P=14924\ atm

(d). Here, C_{p}=0

So, The value of \Delta H and \Delta S at 900 k will be equal at 298 K

We need to calculate the Gibbs free energy of diamond relative to graphite

Using formula of Gibbs free energy

\Delta G=\Delta H-T\Delta S

Put the value into the formula

\Delta G=(1897-0)-900\times(2.38-5.73)

\Delta G=4912\ J

Hence, (a). The Gibbs free energy change is 2.895 kJ and its positive.

(b). The Gibbs free energy change is 34.59 J/mole

(c). The pressure is 14924 atm.

(d). The Gibbs free energy of diamond relative to graphite is 4912 J.

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Answer:

The mass of water vapor is 44.46 grams

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Volume of O2 = 22.8 L

Pressure = 2.00 atm

Step 2: The balanced equation

2CH3OH + 3O2 → 2CO2 + 4H2O

Step 3: Calculate moles CH3OH

Moles CH3OH = mass CH3OH / molar mass CH3OH

Moles CH3OH = 42.5 grams / 32.04 g/mol

Moles CH3OH = 1.326 moles

Step 4: Calculate moles O2

p*V = n*R*T

⇒with p = the pressure = 2.00 atm

⇒with V = the volume of O2 = 22.8 L

⇒with n = the moles of O2  = TO BE DETERMINED

⇒with R = the gas constant = 0.08206 L*Atm/mol*K

⇒with T = the temperature = 27 °C = 300 K

n = (p*V) / (R*T)

n = (2.00 * 22.8) / (0.08206*300)

n = 1.85 moles

Step 5: Calculate the limiting reactant

For 2 moles CH3OH we need 3 moles O2 to produce 2 moles CO2  and 4 H2O

O2 is the limiting reactant. It will completely be consumed ( 1.85 moles). CH3OH is in excess. There will react 2/3*1.85 = 1.233 moles. There will remain  1.326 - 1.233 = 0.093 moles

Step 6: Calculate moles products

For 2 moles CH3OH we need 3 moles O2 to produce 2 moles CO2  and 4 H2O

For 1.85 moles O2 we'll have 1.233 moles CO2 and 2.467 moles H2O

Step 7: Calculate mass H2O

Mass H2O = moles H2O * molar mass H2O

Mass H2O = 2.467 moles * 18.02 g/mol

Mass H2O = 44.46 grams

Step 8: Calculate volume H2O

p*V = n*R*T

⇒with p = the pressure = 2.00 atm

⇒with V = the volume of H2O = TO BE DETERMINED

⇒with n = the moles of H2O  = 2.467 moles

⇒with R = the gas constant = 0.08206 L*Atm/mol*K

⇒with T = the temperature = 27 °C = 300 K

V = (n*R*T)/p

V = (2.467 * 0.08206 * 300) / 2.00

V = 30.37 L

The mass of water vapor is 44.46 grams

The volume of water is 30.37 L

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