Calculate the Delta_fH^{circ} of N_2H_{4(g)} | vedclass

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(B) The formation reaction for $N_2H_{4(g)}$ is:$N_{2(g)} + 2H_{2(g)} \rightarrow N_2H_{4(g)}$The enthalpy of formation $\Delta_fH^{\circ}$ is calcula

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$62 \ kJ \ mol^{-1}$

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(B) The formation reaction for $N_2H_{4(g)}$ is:$N_{2(g)} + 2H_{2(g)} ightarrow N_2H_{4(g)}$The enthalpy of formation $\Delta_fH^{\circ}$ is calculated using bond energies as:$\Delta_fH^{\circ} = \sum B.E.(reactants) - \sum B.E.(products)$$\Delta_fH^{\circ} = [B.E.(N \equiv N) + 2 	imes B.E.(H-H)] - [B.E.(N-N) + 4 	imes B.E.(N-H)]$Substituting the given values:$\Delta_fH^{\circ} = [941 + 2(436)] - [159 + 4(398)]$$\Delta_fH^{\circ} = [941 + 872] - [159 + 1592]$$\Delta_fH^{\circ} = 1813 - 1751$$\Delta_fH^{\circ} = 62 \ kJ \ mol^{-1}$

Calculate the $\Delta_fH^{\circ}$ of $N_2H_{4(g)}$ given the following bond energies:
$B.E.(N-N) = 159 \ kJ \ mol^{-1}$
$B.E.(H-H) = 436 \ kJ \ mol^{-1}$
$B.E.(N \equiv N) = 941 \ kJ \ mol^{-1}$
$B.E.(N-H) = 398 \ kJ \ mol^{-1}$

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Calculate the $\Delta_fH^{\circ}$ of $N_2H_{4(g)}$ given the following bond energies:$B.E.(N-N) = 159 \ kJ \ mol^{-1}$$B.E.(H-H) = 436 \ kJ \ mol^{-1}$$B.E.(N \equiv N) = 941 \ kJ \ mol^{-1}$$B.E.(N-H) = 398 \ kJ \ mol^{-1}$