Question

Calculate the standard enthalpy change for the following reaction at 25 °C. Standard enthalpy of formation values can be found in this table. MgCl2(l) +H2O(l)=MgO(s)+ 2HCL(g)

Answer 1

Concepts and reason

The concept used is to calculate the enthalpy change for the reaction.

Fundamentals

Enthalpy is the heat content of a system. Enthalpy changes for a reaction may be exothermic or endothermic.

The enthalpy change for the reaction is the amount of energy gained or lost during the reaction. It is represented as $\Delta H^\circ$ .

For a reaction, the enthalpy change is calculated as follows:

$\Delta H{^\circ _{rxn}} = \sum n {\left( {\Delta H{^\circ _f}} \right)_{{\rm{products}}}} - \sum n {\left( {\Delta H{^\circ _f}} \right)_{{\rm{reactants}}}}$

The given reaction is as follows:

${\rm{MgC}}{{\rm{l}}_2}\left( l \right) + {{\rm{H}}_2}{\rm{O}}\left( l \right)\longrightarrow{{}}{\rm{MgO}}\left( s \right) + 2{\rm{HCl}}\left( g \right)$

The expression for the enthalpy change for the reaction is as follows:

$\Delta H{^\circ _{rxn}} = \left[ {\Delta H{^\circ _f}\left( {{\rm{MgO}}} \right) + 2\Delta H{^\circ _f}\left( {{\rm{HCl}}} \right)} \right] - \left[ {\Delta H{^\circ _f}\left( {{\rm{MgC}}{{\rm{l}}_2}} \right) + \Delta H{^\circ _f}\left( {{{\rm{H}}_2}{\rm{O}}} \right)} \right]$

$\begin{array}{l}\\\Delta H{^\circ _f}\left( {{\rm{MgC}}{{\rm{l}}_2}} \right) = - 641.3{\rm{ kJ/mol}}\\\\\Delta H{^\circ _f}\left( {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right) = - 285.8{\rm{ kJ/mol}}\\\\\Delta H{^\circ _f}\left( {{\rm{MgO}}} \right) = - 601.6{\rm{ kJ/mol}}\\\\\Delta H{^\circ _f}\left( {{\rm{HCl}}} \right) = - 92.3{\rm{ kJ/mol}}\\\end{array}$

The enthalpy for the reaction is calculated as follows:

$\begin{array}{l}\\\Delta H{^\circ _{rxn}} = \left[ {\left( { - 601.6} \right) + 2\left( { - 92.3} \right){\rm{ kJ/mol}}} \right] - \left[ {\left( { - 641.3} \right) + \left( { - 285.8} \right){\rm{ kJ/mol}}} \right]\\\\{\rm{ = }}140.9{\rm{ kJ/mol}}\\\end{array}$

Ans:

Therefore, the enthalpy change for the reaction is $140.9{\rm{ kJ/mol}}$ .