Given that Delta H_f(H) = 218 kJ/mol,express | vedclass

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(C) The enthalpy of formation of $H$ atoms is given as $\Delta H_f(H) = 218 \ kJ/mol$. This corresponds to the reaction: $\frac{1}{2} H_2(g) \longrigh

Vedclass · Accepted Answer

$104$

Vedclass · Answer

(C) The enthalpy of formation of $H$ atoms is given as $\Delta H_f(H) = 218 \ kJ/mol$. This corresponds to the reaction: $\frac{1}{2} H_2(g) \longrightarrow H(g) ; \Delta H = 218 \ kJ/mol$. For the dissociation of one mole of $H_2$ molecules into $H$ atoms: $H_2(g) \longrightarrow 2H(g) ; \Delta H = 2 	imes 218 = 436 \ kJ/mol$. To convert this energy into $kcal/mol$,we use the conversion factor $1 \ kcal = 4.18 \ kJ$. Bond energy $= \frac{436 \ kJ/mol}{4.18 \ kJ/kcal} \approx 104.3 \ kcal/mol$. Thus,the $H-H$ bond energy is approximately $104 \ kcal/mol$.

Given that $\Delta H_f(H) = 218 \ kJ/mol$,express the $H-H$ bond energy in $kcal/mol$.

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