# Calculate the sta

Given:

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

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

H2(g) +O2(g) H2O(l); ΔfH°= –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); ΔcH° = –393 kJ mol–1

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

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

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

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

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

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

(formation of required reaction)

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

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