Calculate $\Delta H_f^o$ of $SiH_2$ from the following reactions:
$Si_2H_{6(g)} + H_{2(g)} \to 2SiH_{4(g)}, \Delta H = -11.7 \ kJ/mol$
$SiH_{4(g)} \to SiH_{2(g)} + H_{2(g)}, \Delta H = +239.7 \ kJ/mol$
$\Delta H_f^o(Si_2H_{6(g)}) = 80.3 \ kJ/mol$

The internal energy change (in $J$) when $90 \ g$ of water undergoes complete evaporation at $100^{\circ} C$ is........
(Given: $\Delta H_{vap}$ for water at $373 \ K = 41 \ kJ/mol$,$R = 8.314 \ J \ K^{-1} mol^{-1}$)

The heat of combustion of $CO$ at $290 \ K$ and constant volume is $-280.5 \ kJ$. What will be the heat of combustion at constant pressure in $kJ$?

At $298 \ K$,$\Delta_r U^{\ominus}$ and $\Delta_r S^{\ominus}$ for the following reaction are $-10.5 \ kJ$ and $+44.1 \ J \ K^{-1} \ mol^{-1}$ respectively. The reaction is: $2 \ X_{(g)} + Y_{(g)} \longrightarrow 2 \ Z_{(g)}$. What is $\Delta_r G^{\ominus}$ (in $kJ$) for this reaction? $(R = 8.314 \ J \ K^{-1} \ mol^{-1})$

$6 \ g$ of graphite is burnt in a bomb calorimeter at $25^{\circ} C$ and $1 \ atm$ pressure. The temperature of water increased from $25^{\circ} C$ to $31^{\circ} C$. If $\Delta H$ of this reaction is $-248 \ kJ \ mol^{-1}$,find out $C_V$ (in $kJ \ K^{-1}$) of the bomb calorimeter.

The enthalpy of neutralisation of $KOH$ by a weak acid $(HA)$ is $-12.2 \ kcal \ mol^{-1}$. If the weak acid is $10 \%$ ionised,then the enthalpy of dissociation of $1 \ mole$ of $HA$ is. Given that the enthalpy of neutralisation of $KOH$ by $HCl$ is $-13.7 \ kcal \ mol^{-1}$.