The heat of transition $(\Delta H_t)$ of graphite into diamond would be,where
$C(\text{graphite}) + O_{2(g)} \to CO_{2(g)}; \Delta H = x \ kJ \ mol^{-1}$
$C(\text{diamond}) + O_{2(g)} \to CO_{2(g)}; \Delta H = y \ kJ \ mol^{-1}$

At $298 \ K$ and constant pressure, the heat of formation of $C_2H_{2(g)}$ and $C_6H_{6(g)}$ is $230 \ kJ \ mol^{-1}$ and $85 \ kJ \ mol^{-1}$ respectively. What will be the change in heat for the reaction?
$3 \ C_2H_{2(g)} \rightarrow C_6H_{6(g)}$

Heat of combustion of two isomers $x$ and $y$ are $17 \ kJ/mol$ and $12 \ kJ/mol$ respectively. From this information,it may be concluded that:

Given:
$(A) \ 2 \ CO_{(g)} + O_{2(g)} \rightarrow 2 \ CO_{2(g)} \quad \Delta H_1^\theta = -x \ kJ \ mol^{-1}$
$(B) \ C \ (\text{graphite}) + O_{2(g)} \rightarrow CO_{2(g)} \quad \Delta H_2^\theta = -y \ kJ \ mol^{-1}$
The $\Delta H^\theta$ for the reaction $C \ (\text{graphite}) + \frac{1}{2} O_{2(g)} \rightarrow CO_{(g)}$ is:

$S + \frac{3}{2} O_2 \to SO_3 + 2x \ kcal$
$SO_2 + \frac{1}{2} O_2 \to SO_3 + y \ kcal$
Find out the heat of formation of $SO_2$.

The heat change for the following reaction $C_{(s)} + 2S_{(s)} \to CS_{2(l)}$ is known as