What is the enthalpy for the reaction between Mg and HCl?    

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Mg reacts with HCl according to this equation:

`Mg_((s)) + 2 HCl_(aq) -> MgCl_2_(aq) + H_2_(g)`

If this reaction takes place under standard conditions the enthalpy change (∆Hº)  can be calculated using the standard enthalpies of formation of the products and reactants. It's calculated as follows:

∆Hº(reaction) = `Sigma` ∆Hºn(products) - `Sigma` ∆Hºn(reactants)

The n represents moles. Each molar value for ∆H is mulitplied by the corresponding coefficient from the equation.

The standard enthalpy of formation of an element in its standard state is zero. Since the standard state for hydrogen is gas and the standard state for magnesium is solid the ∆Hº of formation values for each of these elements is zero. The other values that we need are obtained from a table of standard enthalpies of formation, found in most textbooks and available online.

In the table I included under "Sources," refer to the bottom section, "Standard Enthalpy of Formation for Atomic and Molecular Ions." 

For MgCl2, we must add the enthalpy of formation for Mg (-462.0) to the enthalpy of formation of Cl (-167.4) x 2 because there are two Cl molecules. We follow the same procedure for HCl, adding the enthalpy of formation for H (0.0) to to the enthalpy of formation of Cl (-167.4).

MgCl2 (aq)  = -796.8 kJ/mol

HCl (aq) = -167.4 kJ/mol

Plugging in these values give us:

∆Hº(reaction) = [(-796.8 kJ/mol) +0] - [(0 + 2(-167.4 kJ/mol)]

∆Hº(reaction) = -462 kJ/mol

The negative sign tells us that this is an exothermic reaction.


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