# Write the reaction which has a heat of reaction equal to heat of formation for HCl(g) ?

## Write the reaction which has a heat of reaction equal to heat of formation for HCl(g) ?

May 4, 2016

$\frac{1}{2} {\text{H"_ (2(g)) + 1/2"Cl"_ (2(g)) -> "HCl}}_{\left(g\right)}$

#### Explanation:

A compound's standard heat of formation, or standard enthalpy of formation, $\Delta {H}_{f}^{\circ}$, represents the change in enthalpy that accompanies the formation of one mole of that compound from its constituent elements in their standard state.

In your case, the chemical reaction that has a change in enthalpy equal to $\Delta {H}_{f}^{\circ}$ will describe the formation of one mole of hydrogen chloride, $\text{HCl}$, from its constituent elements in their standard state.

Now, one molecule of hydrogen chloride contains one atom of hydrogen, $\text{H}$, and one atom of chlorine, $\text{Cl}$.

The key now is to remember that hydrogen and chlorine exist as diatomic elements in their standard state. This means that you can write

${\text{H"_ (2(g)) + "Cl"_ (2(g)) -> 2"HCl}}_{\left(g\right)}$

This represents one form of the balanced chemical equation that describes the synthesis of hydrogen chloride.

In order for the enthalpy change of reaction to be equal to that of the standard enthalpy change of formation, you must divide everything by $2$. This will get you

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\frac{1}{2} {\text{H"_ (2(g)) + 1/2"Cl"_ (2(g)) -> "HCl}}_{\left(g\right)}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

This equation will have

$\Delta {H}_{\text{rxn}}^{\circ} = \Delta {H}_{f}^{\circ}$

because it describes the formation of one mole of hydrogen chloride from its constituent elements in their standard state.