Methanol (CH₃OH) is made industrially in two steps ...

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) + 2 H₂(g) S CH₃OH(l) ΔS° = - 332 J/K

Step 2. CH₃OH₁l₂ → CH₄(g) + 1/2 O₂(g) ΔS° = 162 J/K

(k) Calculate an overall ΔG°, ΔH°, and ΔS° for the formation of CH₄ from CO and H₂.

Accepted Answer

Identify the given reactions and their respective $$ \Delta S^\circ $$ values: Step 1: $$ 	ext{CO(g) + 2 H}_2	ext{(g) } ightarrow 	ext{ CH}_3	ext{OH(l) } \Delta S^\circ = -332 	ext{ J/K} $$, Step 2: $$ 	ext{CH}_3	ext{OH(l) } ightarrow 	ext{ CH}_4	ext{(g) + } \frac{1}{2} 	ext{O}_2	ext{(g) } \Delta S^\circ = 162 	ext{ J/K} .. To $$find the overall $$ \Delta S^\circ $$, add the $$ \Delta S^\circ $$ values of the two steps: $$ \Delta S^\circ_{	ext{overall}} = \Delta S^\circ_{	ext{Step 1}} + \Delta S^\circ_{	ext{Step 2}} .. $$Use Hess's Law to determine $$ \Delta H^\circ $$ and $$ \Delta G^\circ $$ for each step if not given, or assume standard enthalpy and Gibbs free energy changes if provided.. Calculate the overall $$ \Delta H^\circ $$ and $$ \Delta G^\circ  by $$summing the values from each step: $$ \Delta H^\circ_{	ext{overall}} = \Delta H^\circ_{	ext{Step 1}} + \Delta H^\circ_{	ext{Step 2}} $$ and $$ \Delta G^\circ_{	ext{overall}} = \Delta G^\circ_{	ext{Step 1}} + \Delta G^\circ_{	ext{Step 2}} .. $$Ensure all units are consistent and check if any additional data or assumptions are needed to complete the calculations, such as standard enthalpies or Gibbs free energies of formation.

Key Concepts

Gibbs Free Energy (ΔG°)

Enthalpy (ΔH°)

Entropy (ΔS°)