For the reaction $N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$,the relationship between $\Delta H$ and $\Delta U$ is:

For the reaction $C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l)$,the value of $\Delta H - \Delta U$ is:

For combustion of one mole of magnesium in an open container at $300 \ K$ and $1 \ bar$ pressure,$\Delta_{C}H^{\ominus} = -601.70 \ kJ \ mol^{-1}$,the magnitude of change in internal energy for the reaction is $.... \ kJ$. (Nearest integer)
(Given : $R = 8.3 \ J \ K^{-1} \ mol^{-1}$ )

The enthalpy of vapour is $10 \ kcal/mole$ higher than liquid at $350 \ K$ and $2 \ atm$. Under given conditions,the difference in internal energy of vapour and liquid will be $...... \ cal$.

If $\Delta E$ is the heat of reaction for $C_2H_5OH_{(l)} + 3O_{2(g)} \longrightarrow 2CO_{2(g)} + 3H_2O_{(l)}$ at constant volume,the $\Delta H$ (heat of reaction at constant pressure) at constant temperature is

$A$ system performs mechanical work equal to $15 \ kJ$ and loses $2 \ kJ$ of heat to the surroundings. What is the change in internal energy of the system (in $kJ$)?