If $\Delta H_f (H_2O) = X$,then the heat of neutralization of $CH_3COOH$ and $NaOH$ will be:

$18.0 \ g$ of water completely vaporises at $100^{\circ}C$ and $1 \ bar$ pressure and the enthalpy change in the process is $40.79 \ kJ \ mol^{-1}$. What will be the enthalpy change for vaporising two moles of water under the same conditions? What is the standard enthalpy of vaporisation for water?

The value of $\Delta_{f} H^{o}$ for $NH_3$ is $-91.8 \ kJ \ mol^{-1}$. Calculate the enthalpy change for the following reaction: $2NH_{3(g)} \to N_{2(g)} + 3H_{2(g)}$

Enthalpy change for the reaction,$\frac{1}{2} H_2(g) + \frac{1}{2} Cl_2(g) \to HCl(g)$,is called:

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:

Average $C-H$ bond energy is $416 \ kJ \ mol^{-1}$. Which of the following is correct?