The reaction $NH_2CN_{(s)} + \frac{3}{2}O_{2(g)} \to N_{2(g)} + CO_{2(g)} + H_2O_{(l)}$ is carried out in a bomb calorimeter. The heat evolved is $743 \ kJ \ mol^{-1}$. Calculate the value of $\Delta H$ at $300 \ K$ in $kJ \ mol^{-1}$.

For the reaction at $25^{\circ} C$, $X_2O_{4(l)} \longrightarrow 2 XO_{2(g)}$, $\Delta U$ and $\Delta S$ are $2.1 \ kCal$ and $20 \ cal \ K^{-1}$ respectively. What is $\Delta G$ for the reaction at the same temperature? $(R = 2 \ cal \ K^{-1} \ mol^{-1})$

If the standard enthalpy change $\left(\Delta_{r} H^{\theta}\right)$ for a certain reaction at $298 \ K$ and constant pressure is $-1860 \ kJ \ mol^{-1}$,and the standard entropy change $\left(\Delta_{\text{sys}} S^{\theta}\right)$ of the same reaction is $-550 \ J \ K^{-1} \ mol^{-1}$,which one of the following statements is correct?

An ideal gas in a thermally insulated vessel at internal pressure $= P_1$,volume $= V_1$ and absolute temperature $= T_1$ expands irreversibly against zero external pressure,as shown in the diagram. The final internal pressure,volume and absolute temperature of the gas are $P_2, V_2$ and $T_2$,respectively. For this expansion,
$(A) \ q = 0$
$(B) \ T_2 = T_1$
$(C) \ P_2 V_2 = P_1 V_1$
$(D) \ P_2 V_2^\gamma = P_1 V_1^\gamma$

Change in entropy for an ideal gas in a reversible isothermal process is given by:

Enthalpy of formation of $CO_{2(g)}$,$H_2O_{(l)}$ and $C_6H_{12}O_{6(s)}$ are $-393$,$-286$ and $-1170 \ kJ \ mol^{-1}$ respectively. The quantity of heat liberated when $18 \ g$ of $C_6H_{12}O_{6(s)}$ is burnt completely in oxygen is (in $kJ$)