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

(A) the dissociation energy of $\mathrm{H}_{2}$ and enthalpy of sublimation of carbon

(B) latent heat of vapourization of methane

(C) the first four ionization energies of carbon and electron gain enthalpy of hydrogen

(D) the dissociation energy of hydrogen molecule, $\mathrm{H}_{2}$

JEE Main · 2006 · 30 Apr 2006 · Offline · Chemical Thermodynamics

The standard enthalpy of formation (

Δ_{f} H^{\circ}

) 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

(A)the dissociation energy of $H_{2}$ and enthalpy of sublimation of carbon✓ Correct
(B)latent heat of vapourization of methane
(C)the first four ionization energies of carbon and electron gain enthalpy of hydrogen
(D)the dissociation energy of hydrogen molecule, $H_{2}$

Answer: (A)

Solution

The average

C - H

bond energy is obtained from a thermochemical cycle built around

C (s) + 2 H_{2} (g) \to C H_{4} (g)

. To reduce the reaction to free atoms you must first turn solid carbon into gaseous atoms, which calls for the enthalpy of sublimation of carbon, and split the hydrogen molecules into atoms, which calls for the dissociation energy of

H_{2}

. With the given enthalpy of formation and these two quantities the energy of forming the four

C - H

bonds can be found. The additional data required are therefore the dissociation energy of

H_{2}

and the enthalpy of sublimation of carbon.

JEE Main · 2006 · 30 Apr 2006 · Offline · Single correct option

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JEE Main · 2006 · 30 Apr 2006 · Offline · Chemical Thermodynamics

Solution

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