Q. 143.9( 42 Votes )

Calculate the sta

Answer :

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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