The heat of neutralisation will be highest in

Enthalpy of a compound is equal to its

Given below are two statements:
Assertion $(A)$: The enthalpy of formation of graphite is taken as zero.
Reason $(R)$: Graphite is the thermodynamically most stable allotrope of carbon.
The correct answer is:

Equal volumes of $1 \, M \, HCl$ and $1 \, M \, H_2SO_4$ are neutralized by a dilute $NaOH$ solution,and $x$ and $y \, kcal$ of heat are liberated,respectively. Which of the following is correct?

The standard enthalpy of formation of $NH_3$ is $-46.0 \ kJ \ mol^{-1}.$ If the enthalpy of formation of $H_2$ from its atoms is $-436 \ kJ \ mol^{-1}$ and that of $N_2$ is $-712 \ kJ \ mol^{-1},$ the average bond enthalpy of $N-H$ bond in $NH_3$ is ................ $kJ \ mol^{-1}$

The enthalpy changes for the following processes are listed below:
$Cl_{2(g)} \to 2Cl_{(g)}$,$\Delta H = 242.3 \ kJ \ mol^{-1}$
$I_{2(g)} \to 2I_{(g)}$,$\Delta H = 151.0 \ kJ \ mol^{-1}$
$ICl_{(g)} \to I_{(g)} + Cl_{(g)}$,$\Delta H = 211.3 \ kJ \ mol^{-1}$
$I_{2(s)} \to I_{2(g)}$,$\Delta H = 62.76 \ kJ \ mol^{-1}$
Given that the standard states for iodine and chlorine are $I_{2(s)}$ and $Cl_{2(g)}$,the standard enthalpy of formation for $ICl_{(g)}$ is .............. $kJ \ mol^{-1}$