Question

Methanol (CH3OH) is made industrially in two steps from CO and H2. It is so cheap to make that it is being considered for use as a precursor to hydrocarbon fuels such as methane (CH4): Step 1. CO(g)+2H2(g)→CH3OH(l) ΔS∘ = -332J/K Step 2. CH3OH(l)→CH4(g)+1/2O2(g) ΔS∘ = 162J/K

a) Calculate an overall ΔG∘ for the formation of CH4 from CO and H2H2.

b) Calculate an overall ΔH∘ for the formation of CH4 from CO and H2.

c) Calculate an overall ΔS∘ for the formation of CH4 from CO and H2

Answer 1

Step I :     CO(g)   + 2H2(g) → CH3OH(l)         ΔS∘ = -332J/K

Step II:     CH3OH(l)   →   CH4(g) + 1/2O2(g)    ΔS∘ = 162J/K

(c) Combining step I & II =>

CO(g)   + 2H2(g) →   CH4(g) + 1/2O2(g)      dS = -170 J/K   ……………….(1)

Hence, dS for the formation of CH4 from CO and H2 is -170 J/K.

(a) Using the standard value of dG of formation for the reaction (1) =>

dG = dGf(CH4) + ½ dGf(O2) – dGf(CO) – 2dGf(H2)

      = -50.72 + 0 + 137.17 – 0 = 86.45 kJ/mol

Hence, dG for the formation of CH4 from CO and H2 is 86.45 kJ/mol.

(b) Using the standard value of dH of formation for the reaction (1) =>

dH = dHf(CH4) + ½ dHf(O2) – dHf(CO) – 2dHf(H2)

      = -74.81 + 0 + 110.53 – 0 = 35.72 kJ/mol

Hence, dH for the formation of CH4 from CO and H2 is 35.72 kJ/mol.