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Question

What is the heat of formation of liquid methyl alcohol in kJ mol1 , use the following data.
Heat of vaporization of liquid methyl alcohol = 38 kJ mol1 .
Heat of formation of gaseous atoms from the elements in their standard states:
H=218 kJ mol1,C=715 kJ mol1, O=249 kJ mol1
Average bond energies :
CH=415 kJ mol1 CO=356 kJ mol1 OH=463 kJ mol1

A
256 kJ mol1
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B
+266 kJ mol1
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C
266 kJ mol1
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D
+256 kJ mol1
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Solution

The correct option is C 266 kJ mol1
We are given:
CH3OH (g) CH3OH (l); ΔH=38 kJ mol1

C (g)+4H (g)+O (g)CH3OH (g);ΔH=(3×415+356+463)=2064 kJ mol1

C (graphite)C (g);ΔH=715 kJ mol1

2H2 (g)4H (g);
ΔH=2×2×218=872 kJ mol1

12O2 (g)O(g); ΔH=249 kJ mol1

Adding all the above equations we get our desired equation.: C (graphite)+2H2 (g)+12O2 (g)CH3OH (l) ΔH=Sum of all =ΔHs266 kJ mol1




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