Calculate the standard enthalpy of formation of CH₃OH(l) from the following data:
CH₃OH (I) + O₂(g) → CO₂(g) + 2H₂O(l); = –726 kJ mol^–1

C(graphite) + O₂(g) → CO₂(g); = –393 kJ mol^–1

H₂(g) +O₂(g) → H₂O(l); = –286 kJ mol^–1.

Question

Philoid · Accepted Answer

Given:

CH₃OH (I) + O₂(g) → CO₂(g) + 2H₂O(l); Δ_rH^° = –726 kJ mol^–1(1)

C(graphite) + O₂(g) → CO₂(g); Δ_cH^° = –393 kJ mol^–1 (2)

H₂(g) +O₂(g) → H₂O(l); Δ_fH^°= –286 kJ mol^–1 (3)

To calculate the standard enthalpy of formation of CH₃OH(l), first we will make the formation reaction of CH₃OH by using carbon, hydrogen, oxygen.

C (s) + 2H₂ (g) + 1/2 O₂ (g)→ CH₃OH (l)

This is the required reaction

Now, we make the required reaction from the given data.

Step 1: Add the reaction (2) and reaction (3)

C(graphite) + O₂(g) → CO₂(g); Δ_cH^° = –393 kJ mol^–1

2H₂(g) +O₂(g) → 2H₂O(l); Δ_fH^°= –572 kJ mol^–1. [2× reaction (2)]

C + 2H₂(g) + 2O₂(g) → CO₂(g) + 2H₂O(l); Δ_fH^° = -965 kJ mol^-1

Step 2: Subtract reaction (1) from the above reaction, we get

C + 2H₂(g) +2O₂(g) → CO₂(g) + 2H₂O(l); Δ_fH^° = -965 kJ mol^-1

CO₂(g) + 2H₂O(l) → CH₃OH (l) + O₂(g); Δ_rH^°= –726 kJ mol^–1

C (s) + 2H₂ (g) + 1/2 O₂ (g)→ CH₃OH (l); Δ_rH^°= –239 kJ mol^–1

(formation of required reaction)

Thus, the standard enthalpy of formation of CH₃OH(l) is –239 kJ mol^–1

Calculate the standard enthalpy of formation of CH3OH(l) from the following data:CH3OH (I) + O2(g) → CO2(g) + 2H2O(l); = –726 kJ mol–1C(graphite) + O2(g) → CO2(g); = –393 kJ mol–1H2(g) +O2(g) → H2O(l); = –286 kJ mol–1.