At $298 \, K$,the bond energies of $C-H, C-C, C=C$,and $H-H$ are $414, 347, 615$,and $435 \, kJ \, mol^{-1}$ respectively. What will be the enthalpy change for the reaction ${H_2}C=CH_{2(g)} + H_{2(g)} \to H_3C-CH_{3(g)}$ at $298 \, K$ in $kJ \, mol^{-1}$?

Which factor affects the heat of reaction in the Kirchhoff equation?

If $3.365 \text{ g}$ of ethanol $(l)$ is burnt completely in a bomb calorimeter at $298.15 \text{ K}$,the heat produced is $99.472 \text{ kJ}$. The $|\Delta H_f^\circ|$ of ethanol at $298.15 \text{ K}$ is . . . . . . $\times 10^2 \text{ kJ mol}^{-1}$.

What is the enthalpy change for the reaction $2C(\text{graphite}) + 3H_{2(g)} \rightarrow C_2H_{6(g)}$ called?

When $600 \; mL$ of $0.2 \; M \; HNO_3$ is mixed with $400 \; mL$ of $0.1 \; M \; NaOH$ solution in a flask,the rise in temperature of the flask is $\dots \times 10^{-2} \; ^{\circ}C$. (Enthalpy of neutralisation $= 57 \; kJ \; mol^{-1}$ and Specific heat of water $= 4.2 \; J \; K^{-1} \; g^{-1}$) (Neglect heat capacity of flask)

For the allotropic change represented by the equation $C(\text{diamond}) \to C(\text{graphite})$,the enthalpy change is $\Delta H = -1.89 \ kJ$. If $6 \ g$ of diamond and $6 \ g$ of graphite are separately burnt to yield carbon dioxide,the heat liberated in the first case is: