The molar enthalpies of combustion of $C_2H_{2(g)},$ $C$ (graphite) and $H_{2(g)}$ are $-1300,$ $-394$ and $-286 \ kJ \ mol^{-1},$ respectively. The standard enthalpy of formation of $C_2H_{2(g)}$ is.......$kJ \ mol^{-1}$

Which of the following reactions satisfies the following conditions?
$(a)$ Heat of combustion reaction
$(b)$ Heat of formation reaction
$(c)$ An exothermic reaction
$(d)$ Not a neutralization reaction

Given the reactions:
$C + \frac{1}{2}O_2 \to CO : \Delta H = -12 \ kJ$
$CO + \frac{1}{2}O_2 \to CO_2 : \Delta H = -10 \ kJ$
For the reaction $C + O_2 \to CO_2 : \Delta H = x \ kJ$,the value of $x$ is: (in $kJ$)

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:

The heat of combustion of $C$,$S$ and $CS_2$ are $-393.3 \ kJ$,$-293.7 \ kJ$ and $-1108.76 \ kJ$ respectively. What will be the heat of formation of $CS_2$ in $kJ$?

Given the following thermochemical equations:
$(1) \ S + O_2 \rightarrow SO_2 ; \Delta H = -298.2 \ kJ$
$(2) \ SO_2 + \frac{1}{2} O_2 \rightarrow SO_3 ; \Delta H = -98.7 \ kJ$
$(3) \ SO_3 + H_2O \rightarrow H_2SO_4 ; \Delta H = -130.2 \ kJ$
$(4) \ H_2 + \frac{1}{2} O_2 \rightarrow H_2O ; \Delta H = -287.3 \ kJ$
Calculate the enthalpy of formation of $H_2SO_4$ at $298 \ K$ in $kJ$.