$18 \ g$ glucose is completely combusted in a bomb calorimeter of heat capacity $1400 \ kJ/K$. The temperature changes from $27 \ ^\circ C$ to $27.2 \ ^\circ C$. Calculate the magnitude of the standard enthalpy of combustion of glucose in $kJ/mol$. $[R = 8.314 \ J/mol-K]$

$1 \ g$ of graphite is burnt in a bomb calorimeter in excess of oxygen at $298 \ K$ and $1 \ atm$ atmospheric pressure according to the equation:
$C \ (graphite) + O_{2(g)} \rightarrow CO_{2(g)}$
During the reaction,the temperature rises from $298 \ K$ to $299 \ K$. If the heat capacity of the bomb calorimeter is $20.7 \ kJ \ K^{-1}$,what is the enthalpy change for the above reaction at $298 \ K$ and $1 \ atm$?

For the reaction $2 Cl ( g ) \rightarrow Cl _{2}( g )$,the correct option is

The standard enthalpies of formation of $1,3-butadiene(g)$,$CO_{2(g)}$,and $H_2O_{(l)}$ at $298 \ K$ are $-30$,$-94$,and $-68 \ kcal/mol$ respectively. If the magnitude of resonance enthalpies of $1,3-butadiene$ and $CO_2$ are $10$ and $20 \ kcal/mol$ respectively,the enthalpy of combustion of $1,3-butadiene(g)$ at $298 \ K$ is $........ \ kcal/mol$. (Enthalpy of vaporization of $H_2O_{(l)}$ at $298 \ K = 10 \ kcal/mol$)

The $\Delta H$ and $\Delta S$ for a reaction at $1 \ \text{atm}$ pressure are $+30.558 \ \text{kJ}$ and $0.066 \ \text{kJ K}^{-1}$ respectively. The temperature at which the free energy change will be zero and below this temperature the nature of the reaction would be:

How much energy in $kJ$ is required to convert $54 \ g$ of ice at $0 \ ^\circ C$ to water at $27 \ ^\circ C$? $\left( \Delta H_{fusion} = 6.01 \ kJ \ mol^{-1}, C_{p(liquid)} = 4.18 \ J \ K^{-1} \ g^{-1} \right)$