Question

(3) (8 pts total) An important industrial process to obtain hydrogen gas is by steam reforming of natural gas at temperatures 1000-1350 K. An important chemical equilibrium in this process occurs between methane and water vapor with carbon monoxide and hydrogen gas.

(a) (2pts)Writethechemicalequilibriumbetweenmethaneandwatervaporasreactantsandcarbonmonoxide and hydrogen gas as products (include phases), using the smallest integer coefficients.

(b) (2 pts) Evaluate ?H? (kJ), ?S? (J/K) at 298.15 K for the equilibrium in part (a).

(c) (2pts)Assumingthat?H?and?S?donotchangesignificantlywithtemperature, for what temperatures (in

K) is the equilibrium in part (a) spontaneous from left-to-right?

(d) (2 pts) Using the values of ?H? and ?S? from part (a), estimate the values of the equilibrium constant for the equilibrium in part (a) at 298 K and 1200 K.

Answer 1

(a) The chemical equilibrium can be represented as

CH₄ (g) + H₂O (g) <======> CO (g) + 3 H₂ (g)

(b) The enthalpy of the reaction can be obtained by using the relation

Δ_rH⁰ = ∑n.Δ_fH⁰(products) - ∑n.Δ_fH⁰(reactants)

where n denotes the number of moles of the reactant/product and Δ_fH⁰ represents the standard enthalpy of formation of a substance. The values of Δ_fH⁰ are easily obtained from a table of thermodynamic parameters.

For the given reaction,

Δ_rH⁰ = [(1 mole CO)*Δ_fH⁰ (CO, g) + (3 moles H₂)*Δ_fH⁰ (H₂, g)] –

[(1 mole CH₄)*Δ_fH⁰ (CH₄, g) + (1 mole H₂O)*Δ_fH⁰ (H₂O, g)]

= [(1 mole)*(-110.525 kJ/mol) + (3 moles)*(0 kJ/mol)] – [(1 mole)*(-74.81 kJ/mol) + (1 mole)*(-241.818 kJ/mol)]

= (-110.525 kJ) – (-316.628 kJ)

= 206.103 kJ (ans).

The entropy of the reaction is given as

Δ_rS⁰ = ∑n.S⁰(products) - ∑n.S⁰(reactants)

where S⁰ represents the standard entropy of formation of a substance. The values of S⁰ are easily obtained from a table of thermodynamic parameters.

For the given reaction,

Δ_rS⁰ = [(1 mole CO)*S⁰ (CO, g) + (3 moles H₂)*S⁰ (H₂, g)] –

[(1 mole CH₄)*S⁰ (CH₄, g) + (1 mole H₂O)*S⁰ (H₂O, g)]

= [(1 mole)*(197.674 J/mol.K) + (3 moles)*(130.684 J/mol.K)] – [(1 mole)*(186.264 J/mol.K) + (1 mole)*(188.825 J/mol.K)]

= (589.726 J/K) – (375.089 J/K)

= 214.637 J/K (ans).

The values of Δ_fH⁰ and S⁰ are obtained from

http://www.ars-chemia.net/Permanent_Files/Tables/ThermoData.pdf

(c) The free energy of the reaction is given as

Δ_rG⁰ = Δ_rH⁰ – T*Δ_rS⁰

where T is the temperature of the reaction. A reaction is said to be spontaneous when Δ_rG⁰ 0. We determine the temperature when Δ_rG⁰ is zero. Therefore,

0 = (206.103 kJ) – T*(214.637 J/K)

=====> 206.103 kJ = T*(214.637 J/K)

=====> (206.103 kJ)*(1000 J)/(1 kJ) = T*(214.637 J/K)

=====> T = (206103 J)/(214.637 J/K)

=====> T = 960.2398 K ≈ 960.24 K

The reaction is spontaneous at temperatures below 960.24 K (ans).

(d) At T = 298.15 K, we have,

Δ_rG⁰ = Δ_rH⁰ – T*Δ_rS⁰

= (206.103 kJ) – (298.15 K)*(214.637 J/K)

= (206.103 kJ)*(1000 J)/(1 kJ) – (63997.02155 J)

= 206103 J – 63997.02155 J

= 142105.9785 J.

The equilibrium constant is related to Δ_rG⁰ as

Δ_rG⁰ = -R*T*ln K

======> 142105.9785 J = -(8.314 J/mol.K)*(298.15 K)*ln K

======> 142105.9785 J = -(2478.8191 J/mol)*ln K

======> ln K = -(142105.9785 J)/(2478.8191 J/mol)

======> ln K = -57.3281 (ignore unit)

======> K = exp(-57.3281)

======> K = 1.2668*10^-25

======> K ≈ 1.27*10^-25 (ans).

At T = 1200 K,

Δ_rG⁰ = Δ_rH⁰ – T*Δ_rS⁰

= (206.103 kJ) – (1200 K)*(214.637 J/K)

= (206.103 kJ)*(1000 J)/(1 kJ) – (257564.4 J)

= 206103 J – 257564.4 J

= -51461.4 J.

The equilibrium constant is related to Δ_rG⁰ as

Δ_rG⁰ = -R*T*ln K

======> -51461.4 J = -(8.314 J/mol.K)*(1200 K)*ln K

======> -51461.4 J = -(9976.8 J/mol)*ln K

======> ln K = (51461.4 J)/(9976.8 J/mol)

======> ln K = 5.1581 (ignore unit)

======> K = exp(5.1581)

======> K = 173.8338

======> K ≈ 173.83 (ans).