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Consider the following reaction: CH3OH(g)⇌CO(g)+2H2(g) Part A Calculate ΔG for this reaction at 25 ∘C under the following conditions: PCH3OH= 0.850 atm PCO= 0.140 atm PH2= 0.180 atm

Answer :

Answer: The [tex]\Delta G[/tex] of the reaction at given temperature is -12.964 kJ/mol.

Explanation:

For the given chemical reaction:

[tex]CH_3OH(g)\rightleftharpoons CO(g)+2H_2(g)[/tex]

The expression of [tex]K_p[/tex] for the given reaction:

[tex]K_p=\frac{(p_{CO})\times (p_{H_2}^2)}{p_{CH_3OH}}[/tex]

We are given:

[tex]p_{CO}=0.140atm\\p_{H_2}=0.180atm\\p_{CH_3OH}=0.850atm[/tex]

Putting values in above equation, we get:

[tex]K_p=\frac{(0.140)\times (0.180)^2}{0.850}\\\\K_p=5.34\times 10^{-3}[/tex]

To calculate the Gibbs free energy of the reaction, we use the equation:

[tex]\Delta G=\Delta G^o+RT\ln K_p[/tex]

where,

[tex]\Delta G[/tex] = Gibbs' free energy of the reaction = ?

[tex]\Delta G^o[/tex] = Standard gibbs' free energy change of the reaction = 0 J (at equilibrium)

R = Gas constant = [tex]8.314J/K mol[/tex]

T = Temperature = [tex]25^oC=[25+273]K=298K[/tex]

[tex]K_p[/tex] = equilibrium constant in terms of partial pressure = [tex]5.34\times 10^{-3}[/tex]

Putting values in above equation, we get:

[tex]\Delta G=0+(8.314J/K.mol\times 298K\times \ln(5.34\times 10^{-3}))\\\\\Delta G=-12963.96J/mol=-12.964kJ/mol[/tex]

Hence, the [tex]\Delta G[/tex] of the reaction at given temperature is -12.964 kJ/mol.

The study of carbon with hydrogen is called hydrocarbon. The Gibbs free energy is a thermodynamic potential that can be used to calculate the maximum reversible work that may be performed by a thermodynamic system at a constant temperature and pressure

The correct answer to the question is -12.964.

What is partial pressure?

  • The pressure applied by the gas molecule in the mixture to another gas molecule is called partial pressure.

The balanced reaction is as follows:-

[tex]CH_3OH(g)<---->CO(g)+2H_2[/tex]

The expression which is used to solve the question is as follows:-

[tex]k_p =\frac{P_{co}*P_2H^2}{Pch_3oh} [/tex]

The data is given as follows:-

  • PCH3OH= 0.850 atm
  • PCO= 0.140 atm
  • PH2= 0.180 atm

Placed all the values to the equation

[tex]K_p =\frac{[0.140]*[0.180]^2}{0.850} [/tex]

[tex]K_p =5.34*10^{-3}[/tex].

The Gibbs free equation is as follows:-

[tex]G = G^o+RTlnK_P[/tex]

where,

  • G= Gibbs' free energy of the reaction
  • [tex]G^o[/tex]= Standard gibbs' free energy change of the reaction = 0 J (at equilibrium)
  • R = Gas constant
  • T = Temperature
  • [tex]K_P[/tex]= equilibrium constant in terms of partial pressure

Placed all the value in the equation,

[tex]G = 0+(8.314*298*In(5.34*10^{-3})\\ =-12.946kj[/tex]

Hence, the correct answer is -12.946

For more information about Gibbs free energy, refer to the link:-

https://brainly.com/question/9552459

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