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Standard XII
Chemistry
Hess' Law
Compute the h...
Question

Compute the heat of formation of liquid methyl alcohol (in kJ mol −1) using the following data. The heat of vaporisation of liquid methyl alcohol is 38 kJ mol−1.


The heat of formation of gaseous atoms from the elements in their standard states:
H= 218 kJ mol−1., C=715 kJ mol−1. , O=249 kJ mol−1.

Average bond energies:
C−H=415 kJ mol−1.
C−O=356 kJ mol−1.
O−H=463 kJ mol−1.

A
ΔH=−266 kJ mol−1.
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B
ΔH=+266 kJ mol−1.
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C
ΔH=−190 kJ mol−1.
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D
None of these
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Solution

The correct option is B ΔH=−266 kJ mol−1.
CH3OH(l)⟶CH3OH(g);ΔH=38 kJ mol−1. .....(1)

12 H2(g) ⟶H(g);ΔH=218 kJ mol−1. .....(2)

C(graphite)⟶C(g);ΔH=715 kJ mol−1. .....(3)

12O2(g)⟶O(g);ΔH=249 kJ mol−1. .....(4)

C−H(g)⟶C(g)+H(g);ΔH=415 kJ mol−1. .....(5)

C−O(g)⟶C(g)+O(g);ΔH=356 kJ mol−1. .....(6)

O−H(g)⟶O(g)+H(g);ΔH=463 kJ mol−1. .....(7)

C(s)+2H2(g)+12O2(g)⟶CH3OH(g) .....(8)

The following energies are required:

2H2(g)⟶4H(g)

C(graphite)⟶C(g)

12O2(g)⟶O(g)

The total energy required is

(218×4)+175+249=1836 kJ .....From equation (5), (6) and (7)

The following bonds are formed:

Three C-H bonds, one C-O bond, and O-H bond

The total energy released is

(3×415)+356+463=2064 kJ

Net energy released for CH3OH(l)=228+38=266 kJ .... From eq (1) and (2)

Therefore,

ΔHfCH3OH(l)=−266 kJ mol−1

Hence, the correct option is A


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Similar questions
Q. What is the heat of formation of liquid methyl alcohol in kJ mol−1 , use the following data.
Heat of vaporization of liquid methyl alcohol = 38 kJ mol−1 .
Heat of formation of gaseous atoms from the elements in their standard states:
H=218 kJ mol−1,C=715 kJ mol−1, O=249 kJ mol−1
Average bond energies :
C−H=415 kJ mol−1 C−O=356 kJ mol−1 O−H=463 kJ mol−1
Q. What is the heat of formation of liquid methyl alcohol in kJ mol−1 , use the following data.
Heat of vaporization of liquid methyl alcohol = 38 kJ mol−1 .
Heat of formation of gaseous atoms from the elements in their standard states:
H=218 kJ mol−1,C=715 kJ mol−1, O=249 kJ mol−1
Average bond energies :
C−H=415 kJ mol−1 C−O=356 kJ mol−1 O−H=463 kJ mol−1
Q. Compute the heat of formation of liquid methyl alcohol in kilojoules per mole using the following data.
The heat of vaporisation of liquid methyl alcohol = 38 kJ/mol.
The heats of formation of gaseous atoms from the elements is their standard states: H,218 kJ/mol; C,713 kJ/mol; O,249 kJ/mol.
Average bond energies: C−H=415 kJ/mol; C−O=365 kJ/mol; O−H=463 kJ/mol.
Q. Compute the heat of formation of liquid methyl alcohol in kilojoules per mole using the following data.
The heat of vaporisation of liquid methyl alcohol = 38 kJ/mol.
The heats of formation of gaseous atoms from the elements in their standard states: H,218 kJ/mol; C,715 kJ/mol; O,249 kJ/mol.
Average bond energies: C−H=415 kJ/mol; C−O=365 kJ/mol; O−H=463 kJ/mol.
Q. Compute the enthalpy of formation of liquid methyl alcohol in kJ mol−1, using the following data.
Enthalpy of vaporisation of liquid CH3OH=38 kJ / mol.
Enthalpy of formation of gaseous atoms from the elements in their standard states are
H→218 kJ / mol ; C→715 kJ / mol ; O→249 kJ / mol.
Average Bond energies
C−H→415 kJ / mol ; C−O→356 kJ / mol; O−H→463 kJ / mol

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