The standard enthalpy of atomization of ethan | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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: $622 = (6 	imes 90)   \Delta_{C-C}H^{\circ}$. $622 = 540   \Delta_{C-C}H^{\circ}$. $\Delta_{C-C}H^{\circ} = 622 - 540 = 82 \ kJ \ mol^{-1}$.

Similar Questions

Hess's law is applicable for the determination of heat of

In the reaction $C + 2S \to CS_2 + \Delta H$,$\Delta H$ is the

The enthalpy of formation of ammonia is $-46.0 \ kJ \ mol^{-1}$. The enthalpy change for the reaction $2NH_{3(g)} \rightarrow N_{2(g)} + 3H_{2(g)}$ is ............... $kJ \ mol^{-1}$.

The value of heat generated when $36.5 \, g$ of $HCl$ and $40 \, g$ of $NaOH$ react during neutralization is.....$kcal$.

State and prove Hess's law of constant heat summation.

Vedclass Test Series

Exam Paper Generator

Online Exam Module