The standard enthalpy of atomization of ethan | vedclass

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Question

(D) The standard enthalpy of atomization $(\Delta_{a}H^{\circ})$ is the sum of the bond dissociation enthalpies of all bonds in the molecule. For etha

Vedclass · Accepted Answer

$82$

Vedclass · Answer

(D) The standard enthalpy of atomization $(\Delta_{a}H^{\circ})$ is the sum of the bond dissociation enthalpies of all bonds in the molecule. For ethane $(C_2H_6)$,there are $6$ $C-H$ bonds and $1$ $C-C$ bond. Thus,$\Delta_{a}H^{\circ} = (6 	imes \Delta_{C-H}H^{\circ})   \Delta_{C-C}H^{\circ}$. Given: $\Delta_{a}H^{\circ} = 622 \ kJ \ mol^{-1}$ and $\Delta_{C-H}H^{\circ} = 90 \ kJ \ mol^{-1}$. Substituting the values: $622 = (6 	imes 90)   \Delta_{C-C}H^{\circ}$. $622 = 540   \Delta_{C-C}H^{\circ}$. $\Delta_{C-C}H^{\circ} = 622 - 540 = 82 \ kJ \ mol^{-1}$.

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