Question

Hydrogen gas, H₂, reacts explosively with gaseous chlorine, $\mathbf{C}{\mathbf{l}}_{\mathbf{2}}$, to form hydrogen chloride, HCl(g). What is the enthalpy change for the reaction of 1 mole of ${\mathbf{H}}_{\mathbf{2}}$(g) with 1 mole of $\mathbf{C}{\mathbf{l}}_{\mathbf{2}}$(g) if both the reactants and products are at standard state conditions? The standard enthalpy of formation of HCl(g) is −92.3 kJ/mol.

Short answer

$\mathrm{\Delta }\mathrm{H}{\mathrm{f}}^{\circ }$for two moles of HCl = -184. 6 kJ/mol.

Step-by-step solution

Given reaction

The reaction of 1 mole of ${\mathrm{H}}_2$(g) with 1 mole of $\mathrm{C}{\mathrm{l}}_2$(g) is given by:

${\mathrm{H}}_2\left(\mathrm{g}\right)+\mathrm{C}{\mathrm{l}}_2\left(\mathrm{g}\right)\to 2\mathrm{H}\mathrm{C}\mathrm{l}\left(\mathrm{g}\right)\mathrm{\Delta }\mathrm{H}{\mathrm{f}}^{\circ }=-92.3\mathrm{k}\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l}$

The change of enthalpy of one mole of HCl is -92.3 kJ/ mol is used to find the heat produced when one mole of HCl is formed.

Enthalpy change

Enthalpy change refers to the heat released inside or outside a system in a reaction.

The enthalpy change is evaluated as:

$\begin{array}{l}\mathrm{\Delta }\mathrm{H}{\mathrm{f}}^{\circ }\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{o}\mathrm{n}\mathrm{e}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{H}\mathrm{C}\mathrm{l}=-92.3\mathrm{k}\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l}\\ \begin{array}{c}\mathrm{\Delta }\mathrm{H}{\mathrm{f}}^{\circ }\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{t}\mathrm{w}\mathrm{o}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{o}\mathrm{f}\mathrm{H}\mathrm{C}\mathrm{l}=2\times 92.3\\ =-184.6\mathrm{k}\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l}\end{array}\end{array}$