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Enthalpy of Atomisation and bond dissociation

Question

Mean bond enthalpy of $C - H$ in ${CH}_{4}$ is $X Kj / mole$ .Heat of Atomization of ${CH}_{4}$ is $Y Kj / mole .$ Correct relation is

Moderate

Solution

$For poly atomic molecules; Mean bond enthalpy = \frac{Heat of Atomisation of molecule}{Total number of similar types of bonds}; {CH}_{4} (g) \to {CH}_{3} (g) + H (g) . . . ∆ H_{1} = q_{1}; {CH}_{3} (g) \to {CH}_{2} (g) + H (g) . . . ∆ H_{2} = q_{2}; {CH}_{2} (g) \to CH (g) + H (g) . . . . . ∆ H_{3} = q_{3}; CH (g) \to C (g) + H (g) . . . . . . . ∆ H_{4} = q_{4}; Average of the above reactions is mean bond enthalpy; Mean bond enthalpy (X) = \frac{q_{1} + q_{2} + q_{3} + q_{4}}{4}; . . . . (1) {CH}_{4} (g) \to C (g) + 4 H (g) . . . ∆ H_{atm} = q_{1} + q_{2} + q_{3} + q_{4}; Net ∆ H is called enthalpy of atomization =' Y'; . . . . (2) Comparing (1) and (2) X = \frac{Y}{4};$

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