For the following reaction occurring in automobiles,the values of $\Delta H, \Delta S, \text{and } \Delta G$ are respectively:
$2C_8H_{18(g)} + 25O_2(g) \rightarrow 16CO_2(g) + 18H_2O(g)$

Assertion $(A)$: If heat of combustion of $C_2H_6$ is $X \ kJ \ mol^{-1}$,heat liberated on combustion of $150 \ g$ of $C_2H_6$ is $5X \ kJ$.
Reason $(R)$: Enthalpy is an extensive property.

For the reaction,$2 CO + O_2 \longrightarrow 2 CO_2$; $\Delta H = -560 \ kJ$. Two moles of $CO$ and one mole of $O_2$ are taken in a container of volume $1 \ L$. They completely form two moles of $CO_2$. The gases deviate appreciably from ideal behavior. If the pressure in the vessel changes from $70 \ atm$ to $40 \ atm$,find the magnitude (absolute value) of $\Delta U$ at $500 \ K$. $(1 \ L \ atm = 0.1 \ kJ)$

One mole of water is converted into steam at $373 \, K$. The heat absorbed at $1 \, atm$ pressure is $40.68 \, kJ$. If the molar volumes of water and steam are $18 \, mL$ and $30600 \, mL$ respectively,find $\Delta U$ for the process 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 ........... $kJ \, mol^{-1}$.

The enthalpy change on freezing of $1 \ mol$ of water at $5 \ ^\circ C$ to ice at $-5 \ ^\circ C$ is ..... $kJ \ mol^{-1}$.
(Given $\Delta _{fus}H = 6 \ kJ \ mol^{-1}$ at $0 \ ^\circ C$,
$C_p(H_2O, l) = 75.3 \ J \ mol^{-1} \ K^{-1}$,
$C_p(H_2O, s) = 36.8 \ J \ mol^{-1} \ K^{-1}$)