Answer: The standard enthalpy change of the reaction is coming out to be -16.3 kJ
Explanation:
Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as [tex]\Delta H[/tex]
The equation used to calculate enthalpy change is of a reaction is:
[tex]\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)][/tex]
For the given chemical reaction:
[tex]Mg(OH)_2(s)+2HCl(g)\rightarrow MgCl_2(s)+2H_2O(g)[/tex]
The equation for the enthalpy change of the above reaction is:
[tex]\Delta H_{rxn}=[(1\times \Delta H_f_{(MgCl_2(s))})+(2\times \Delta H_f_{(H_2O(g))})]-[(1\times \Delta H_f_{(Mg(OH)_2(s))})+(2\times \Delta H_f_{(HCl(g))})][/tex]
We are given:
[tex]\Delta H_f_{(Mg(OH)_2(s))}=-924.5kJ/mol\\\Delta H_f_{(HCl(g))}=-92.30kJ/mol\\\Delta H_f_{(MgCl_2(s))}=-641.8kJ/mol\\\Delta H_f_{(H_2O(g))}=-241.8kJ/mol[/tex]
Putting values in above equation, we get:
[tex]\Delta H_{rxn}=[(1\times (-641.8))+(2\times (-241.8))]-[(1\times (-924.5))+(2\times (-92.30))]\\\\\Delta H_{rxn}=-16.3kJ[/tex]
Hence, the standard enthalpy change of the reaction is coming out to be -16.3 kJ
