If the heat of formation of $CO_2$ is $-393 \ kJ/mol$,the amount of heat evolved in the formation of $0.156 \ kg$ of $CO_2$ is.....$kJ$.

In the reaction $CO_{2(g)} + H_{2(g)} \to CO_{(g)} + H_2O_{(g)}; \Delta H = 80 \ kJ$,$\Delta H$ is known as

The enthalpy of reaction for the reaction: $2 H_{2(g)} + O_{2(g)} \to 2 H_{2}O_{(l)}$ is $\Delta_{r}H^{\theta} = -572 \ kJ \ mol^{-1}$. What will be the standard enthalpy of formation of $H_{2}O_{(l)}$?

Consider the following processes:

Process	$\Delta H \ (kJ/mol)$
$I. \ \frac{1}{2} A \rightarrow B$	$+150$
$II. \ 3B \rightarrow 2C + D$	$-125$
$III. \ E + A \rightarrow 2D$	$+350$

For $B + D \rightarrow E + 2C, \Delta H$ will be ............. $kJ/mol$

What will be the enthalpy of formation of benzene in $kJ$ for the reaction: $6C_{(s)} + 3H_{2(g)} \rightarrow C_6H_{6(l)}$? Given that the enthalpy of combustion of benzene is $-3268 \ kJ$, the enthalpy of formation of $CO_{2(g)}$ is $-393.5 \ kJ$, and the enthalpy of formation of $H_2O_{(l)}$ is $-285.8 \ kJ$.

Combustion of $1$ $mol$ of benzene is expressed as:
$C_6H_{6(l)} + \frac{15}{2} O_{2(g)} \rightarrow 6CO_{2(g)} + 3H_2O_{(l)}$
The standard enthalpy of combustion of $2$ $mol$ of benzene is $-x$ $kJ$.
$x = . . . . . . . . . .$
$(1)$ Standard enthalpy of formation of $1$ $mol$ of $C_6H_{6(l)}$ is $48.5$ $kJ \ mol^{-1}$.
$(2)$ Standard enthalpy of formation of $1$ $mol$ of $CO_{2(g)}$ is $-393.5$ $kJ \ mol^{-1}$.
$(3)$ Standard enthalpy of formation of $1$ $mol$ of $H_2O_{(l)}$ is $-286$ $kJ \ mol^{-1}$.