Based on the bond enthalpy $(B.E.)$ values given,the standard enthalpy of formation $(\Delta_fH^o)$ of $N_2H_{4(g)}$ is ...... $kJ\ mol^{-1}$.
Given: $B.E.(N-N) = 159\ kJ\ mol^{-1}$,$B.E.(H-H) = 436\ kJ\ mol^{-1}$,$B.E.(N \equiv N) = 941\ kJ\ mol^{-1}$,$B.E.(N-H) = 398\ kJ\ mol^{-1}$.

The molar neutralization heat for $KOH$ and $HNO_3$ as compared to the molar neutralization heat of $NaOH$ and $HCl$ is:

The heat of formation of methane $C_{(s)} + 2H_{2(g)} \to CH_{4(g)}$ at constant pressure is $-18500 \ cal$ at $25 \ ^oC$. The heat of reaction at constant volume would be (in $cal$)

Considering the reaction $3O_2 \rightarrow 2O_3$; $\Delta H = +ve$,what can we conclude about the reaction?

The heat of solution of $Na_2SO_{4(s)}$ and $Na_2SO_4 \cdot 10H_2O_{(s)}$ are $-2.34 \ kJ \ mol^{-1}$ and $78.87 \ kJ \ mol^{-1}$ respectively. The heat of hydration of $Na_2SO_{4(s)}$ is:

At standard conditions,if the change in the enthalpy for the following reaction is $-109 \; kJ \ mol^{-1}$:
$H_{2(g)} + Br_{2(g)} \rightarrow 2HBr_{(g)}$
Given that bond energy of $H_2$ and $Br_2$ is $435 \; kJ \ mol^{-1}$ and $192 \; kJ \ mol^{-1}$,respectively,what is the bond energy (in $kJ \ mol^{-1}$) of $HBr$?