If $1 \ mole$ of aqueous nitric acid is formed,calculate the total heat released based on the following reactions:

Reaction	$\Delta H \ (kJ)$
$(i) \ 4NH_{3(g)} + 5O_{2(g)} \to 4NO_{(g)} + 6H_2O_{(l)}$	$-904$
$(ii) \ 2NO_{(g)} + O_{2(g)} \to 2NO_{2(g)}$	$-112$
$(iii) \ 3NO_{2(g)} + H_2O_{(l)} \to 2HNO_{3(aq)} + NO_{(g)}$	$-140$

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 ........... $kJ \, mol^{-1}$.

$A$ cylinder of gas supplied by Bharat Petroleum is assumed to contain $14 \ kg$ of butane. If a normal family requires $20,000 \ kJ$ of energy per day for cooking,how many days will the butane gas in the cylinder last? ($\Delta H_c$ of $C_4H_{10} = -2658 \ kJ/mole$)

The heat liberated when $1.89 \ g$ of benzoic acid is burnt in a bomb calorimeter at $25^{\circ} C$ increases the temperature of $18.94 \ kg$ of water by $0.632^{\circ} C$. If the specific heat of water at $25^{\circ} C$ is $0.998 \ cal / (g^{\circ} C)$,then find the heat of combustion of benzoic acid.

For the reaction $2Cl_{(g)} \rightarrow Cl_{2(g)}$,the signs of $\Delta H$ and $\Delta S$ are respectively:

Following data is known about the melting of a compound $AB$: $\Delta H = 9.2 \ kJ \ mol^{-1}$,$\Delta S = 0.008 \ kJ \ K^{-1} \ mol^{-1}$. Its melting point is: