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 $kJ$)

Consider the reaction at $300 \ K$:
$C_6H_{6(l)} + \frac{15}{2}O_{2(g)} \longrightarrow 6CO_{2(g)} + 3H_2O_{(l)} ; \Delta H = -3271 \ kJ$
What is $\Delta U$ for the combustion of $1.5 \ mol$ of benzene at $27 \ ^\circ C$? .....$kJ$

Calculate the standard internal energy change for $OF_{2(g)} + H_2O_{(g)} \longrightarrow 2 HF_{(g)} + O_{2(g)}$ at $300 \ K$,if $\Delta_{f} H^{\circ}$ of $OF_{2(g)}$,$H_2O_{(g)}$,and $HF_{(g)}$ are $20$,$-250$,and $-270 \ kJ \ mol^{-1}$ respectively. $[R = 8.314 \ J \ K^{-1} \ mol^{-1}]$ (in $kJ$)

Which of the following $p-V$ curves represents the maximum work done?

The reaction of cyanamide,$NH_2CN_{(s)}$,with dioxygen was carried out in a bomb calorimeter,and $\Delta U$ was found to be $-742.7 \ kJ \ mol^{-1}$ at $298 \ K$. Calculate the enthalpy change for the reaction at $298 \ K$.
$NH_2CN_{(s)} + \frac{3}{2}O_{2_{(g)}} \to N_{2_{(g)}} + CO_{2_{(g)}} + H_2O_{(l)}$

One mole of an ideal gas at $900 \ K$ undergoes two reversible processes,$I$ followed by $II$,as shown in the graph. If the work done by the gas in the two processes is the same,the value of $\ln \frac{V_3}{V_2}$ is. . . . . . . . ($U$: internal energy,$S$: entropy,$p$: pressure,$V$: volume,$R$: gas constant). (Given: molar heat capacity at constant volume,$C_{V, m}$ of the gas is $\frac{5}{2} R$)