The standard enthalpy of formation (Delta_fH^ | vedclass

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(C) The standard enthalpy of formation of $CH_4$ is given by the reaction: $C_{(s)} + 2H_{2(g)} \rightarrow CH_{4(g)}$ To calculate the average $C-H$

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the dissociation energy of $H_2$ and enthalpy of sublimation of carbon

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(C) The standard enthalpy of formation of $CH_4$ is given by the reaction: $C_{(s)} + 2H_{2(g)} \rightarrow CH_{4(g)}$ To calculate the average $C-H$ bond energy,we need to consider the atomization of the reactants: $1.$ Enthalpy of sublimation of carbon: $C_{(s)} \rightarrow C_{(g)}$ $2.$ Dissociation energy of hydrogen: $2H_{2(g)} \rightarrow 4H_{(g)}$ By using Hess's Law,we combine these with the enthalpy of formation to find the total bond dissociation energy of $CH_4$,which is then divided by $4$ to get the average $C-H$ bond energy. Therefore,the dissociation energy of $H_2$ and the enthalpy of sublimation of carbon are required.

The standard enthalpy of formation $(\Delta_fH^o)$ 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

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Based on the values of $B.E.$ given,calculate the $\Delta_fH^o$ of $N_2H_4\, (g)$ in $kJ\, mol^{-1}$.
Given : $N-N = 159\, kJ\, mol^{-1};\, H-H = 436\, kJ\, mol^{-1}$
$N \equiv N = 941\, kJ\, mol^{-1};\, N-H = 398\, kJ\, mol^{-1}$

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