For the reaction $2C_6H_6(l) + 15O_2(g) \rightarrow 12CO_2(g) + 6H_2O(l)$ at $25 \ ^\circ C$,calculate the difference between the heat of reaction at constant volume and constant pressure in $kJ$.

Given
$C_{(graphite)} + O_{2(g)} \rightarrow CO_{2(g)};$
$\Delta_rH^o = -393.5 \, kJ \, mol^{-1}$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_2O_{(l)};$
$\Delta_rH^o = -285.8 \, kJ \, mol^{-1}$
$CO_{2(g)} + 2H_2O_{(l)} \rightarrow CH_{4(g)} + 2O_{2(g)};$
$\Delta_rH^o = + 890.3 \, kJ \, mol^{-1}$
Based on the above thermochemical equations,the value of $\Delta_rH^o$ at $298 \, K$ for the reaction
$C_{(graphite)} + 2H_{2(g)} \rightarrow CH_{4(g)}$ will be ........... $kJ \, mol^{-1}$.

$10 \, mol$ of an ideal gas expands isothermally and reversibly from a pressure of $10 \, atm$ to $1 \, atm$ at $300 \, K$. What is the largest mass (in $kg$) which can be lifted through a height of $100 \, m$ by the energy obtained in this process (in $, kg$)?

For a pure substance, if $T_B$ is the melting point and $T_A$ is the freezing point, which graph correctly represents the relationship between the change in entropy $(\Delta S)$ and temperature $(T)$?

At $300 \, K$ temperature,the standard heat of formation of benzoic acid $(s)$,$CO_{2(g)}$,and $H_2O_{(l)}$ are $-408$,$-393$,and $-286 \, kJ \, mol^{-1}$ respectively. At constant pressure,the heat of combustion for benzoic acid will be $..... \, kJ$.

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 $\frac{5}{2} R$)