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: $B.E.(N-N) = 159\ kJ\ mol^{-1}$,$B.E.(H-H) = 436\ kJ\ mol^{-1}$,$B.E.(N \equiv N) = 941\ kJ\ mol^{-1}$,$B.E.(N-H) = 398\ kJ\ mol^{-1}$.

Calculate the heat of formation for propene $(C_3H_6)$ using the following thermochemical equations:
$(i) C_{(s)} + O_{2(g)} \to CO_{2(g)}; \Delta H_1 = -94.05 \ k.cal/mole$
$(ii) H_{2(g)} + \frac{1}{2} O_{2(g)} \to H_2O_{(l)}; \Delta H_2 = -68.32 \ k.cal/mole$
$(iii) C_3H_{6(g)} + \frac{9}{2} O_{2(g)} \to 3 CO_{2(g)} + 3 H_2O_{(l)}; \Delta H_3 = -499.7 \ k.cal/mole$
(Note: The original question provided propane combustion data; assuming propene combustion data for consistency).

Which of the following is an endothermic reaction?

Given the following thermochemical equations:
$(1) \ S + O_2 \rightarrow SO_2 ; \Delta H = -298.2 \ kJ$
$(2) \ SO_2 + \frac{1}{2} O_2 \rightarrow SO_3 ; \Delta H = -98.7 \ kJ$
$(3) \ SO_3 + H_2O \rightarrow H_2SO_4 ; \Delta H = -130.2 \ kJ$
$(4) \ H_2 + \frac{1}{2} O_2 \rightarrow H_2O ; \Delta H = -287.3 \ kJ$
Calculate the enthalpy of formation of $H_2SO_4$ at $298 \ K$ in $kJ$.

Calculate the enthalpy of formation of carbon monoxide $(CO).$ Given: $C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)}, \Delta H = -393.3 \ kJ \ mol^{-1}$ and $CO_{(g)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{2(g)}, \Delta H = -282.2 \ kJ \ mol^{-1}.$

The enthalpy of solution of anhydrous $CuSO_4$ is $-15.9 \, kcal$ and that of $CuSO_4 \cdot 5H_2O$ is $2.8 \, kcal$. The enthalpy of hydration of $CuSO_4$ is ..... $kcal$. (in $.7$)