What is the difference between $\Delta H$ and $\Delta U$ for the combustion of methane at $25\,^oC$?

For the reaction $\frac{1}{2}X_2 + \frac{3}{2}Y_2 \to XY_3$,$\Delta H = -30 \ kJ/mol$. Given $\Delta S_{X_2} = 60 \ J/mol \cdot K$,$\Delta S_{Y_2} = 40 \ J/mol \cdot K$,and $\Delta S_{XY_3} = 50 \ J/mol \cdot K$,calculate the temperature at equilibrium in $K$.

The molar enthalpy change for $H_2O_{(l)} \rightleftharpoons H_2O_{(g)}$ at $373 \ K$ and $1 \ atm$ is $41 \ kJ \ mol^{-1}$. Assuming ideal behavior,the internal energy change for vaporization of $1 \ mol$ of water at $373 \ K$ and $1 \ atm$ in $kJ \ mol^{-1}$ is:

The enthalpy of combustion of methane at $25\,^{\circ}C$ is $890\,kJ$. The heat liberated when $3.2\,g$ of methane is burnt in air is.....$kJ$

One mole of an ideal gas is allowed to expand reversibly and adiabatically from a temperature of $27\,^oC$. If the work done during the process is $3\,kJ$,the final temperature will be equal to $(C_v = 20\,J\,K^{-1} \, mol^{-1})$

The $\Delta E^o$ of combustion of $2-$methylpropene is $-X \ kJ \ mol^{-1}$. The value of $\Delta H^o$ is