Question

Calculate the standard enthalpy change for the reaction at 25 ∘ C. Standard enthalpy of formation values can be found in this list of thermodynamic properties. MgCl 2 ( s ) + H 2 O ( l ) ⟶ MgO ( s ) + 2 HCl ( g )

Answer 1

Answer: The standard enthalpy change of the reaction is 358.1 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.

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]$

For the given chemical reaction:

$MgCL_2(s)+H_2O(l)\rightarrow MgO(s)+2HCl(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(1\times \Delta H_f_{(MgO(s))})+(2\times \Delta H_f_{(HCl(g))})]-[(1\times \Delta H_f_{(MgCl_2(s))})+(1\times \Delta H_f_{(H_2O(l))})]$

We are given:

$\Delta H_f_{(H_2O(l))}=-285.8kJ/mol\\\Delta H_f_{(MgCl_2(g))}=-641.8kJ/mol\\\Delta H_f_{(MgO(g))}=-384.9kJ/mol\\\Delta H_f_{(HCl(g))}=-92.30kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(1\times (-384.9))+(2\times (-92.30))]-[(1\times (-641.8))+(1\times (-285.8))]\\\\\Delta H_{rxn}=358.1kJ$

Hence, the standard enthalpy change of the reaction is 358.1 kJ