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The standard enthalpy of formation (\Delta H^{\circ}_{f}) at 298 K for methane, CH_{4(g)} is  –74.8 kJ mol-1 . The additional information required to determine the average energy for C – H bond formation would be  

  • Option 1)

    the dissociation energy of H_2 and enthalpy of sublimation of carbon

  • Option 2)

    latent heat of vaporisation of methane

  • Option 3)

    the first four ionisation energies of carbon and electron gain enthalpy of hydrogen

  • Option 4)

    the dissociation energy of hydrogen molecule, H_2.

 

Answers (1)

best_answer

As we learnt in

Bond dissociation enthalpy -

It is the average of enthalpy required to dissociate the said bond present in different gaseous compound in to free atoms in gaseous state.

- wherein

N_{2}+Bond\, Energy\rightarrow 2N

 

Enthalpy of Sublimation -

Amount of enthalpy change to sublimise 1 mole solid into 1 mole vapour at a temperature below its melting point

- wherein

H_{2}O_{(s)}\rightarrow H_{2}O_{(g)}

\Delta H_{sublimation}= 46.6\, kj/mol

 

 To calculate average enthappy of C - H bond in methane following information are needed.

i) dissociation energy of H2 i.e

\frac{1}{2}H_{2}\rightarrow H(g), \Delta H = x (suppose)

ii) Sublimation energy of C (graphite) to C (g)

C (graphite) \rightarrow C (g), \rightarrow \DeltaH = y (suppose)

Given 

C (graphite) + 2H2(g)\rightarrow CH4(g), \DeltaH = 75 KJmol-1


Option 1)

the dissociation energy of H_2 and enthalpy of sublimation of carbon

this is correct option

Option 2)

latent heat of vaporisation of methane

this is incorrect option

Option 3)

the first four ionisation energies of carbon and electron gain enthalpy of hydrogen

this is incorrect option

Option 4)

the dissociation energy of hydrogen molecule, H_2.

this is incorrect option

Posted by

divya.saini

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