When $128 \, g$ of oxygen gas is heated from $0 \, ^oC$ to $100 \, ^oC$,the average values of $C_v$ and $C_p$ are $5 \, cal \, mol^{-1} \, K^{-1}$ and $7 \, cal \, mol^{-1} \, K^{-1}$ respectively. Find the values of $\Delta U$ and $\Delta H$.

The ratio of heats liberated at $298 \ K$ from the combustion of one $kg$ of coke and by burning water gas obtained from $1 \ kg$ of coke is. (Assume coke to be $100 \%$ carbon.) (Given enthalpies of combustion of $CO_{2}, CO$ and $H_{2}$ as $393.5 \ kJ/mol, 283.5 \ kJ/mol, 285.5 \ kJ/mol$ respectively all at $298 \ K$.) (in $: 1$)

The reaction of cyanamide,$NH_2CN_{(s)}$,with dioxygen was carried out in a bomb calorimeter,and $\Delta U$ was found to be $-742.7 \ kJ \ mol^{-1}$ at $298 \ K$. Calculate the enthalpy change for the reaction at $298 \ K$.
$NH_2CN_{(s)} + \frac{3}{2}O_{2_{(g)}} \to N_{2_{(g)}} + CO_{2_{(g)}} + H_2O_{(l)}$

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

Considering the reaction $C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)} + 393.5 \ kJ$,the signs of $\Delta H, \Delta S$,and $\Delta G$ respectively are:

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: