Enthalpy of Mg + 2HCl --> MgCl2 + H2?

Question

What is the enthalpy of that equation? It's been a while since I've done those and I can't find it/figure it out.

Answer 1

we need a text with the dHf's of HCl(aq) & MgCl2(aq)

some references will give MgCl2(s) but we need it in solution,

when it can't be found, we add it up ourselves if we can find the dHf's of Mg+2(aq) & 2 Cl- (aq)

when I get home, I'll get back to you with an edit of web sites

but I have a text @ school with this:

dHf's of the elements Mg & H2 are zero

dHf of HCl(aq) = -167.2kJ

dHf Mg+2 (aq) = -466.85kJ & dHf Cl- (aq) = -167.2 kJ each

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Mg(s) + 2HCl(aq) --> MgCl2(aq) + H2(g)

dHreaction = dHf products - reactants

dH rxn = dHf MgCl2(aq) - dHf HCl (aq)

dH rxn = [( -466.85kJ & (2) (-167.2 kJ) ] - (-167.2kJ)

dH rxn = -801.25 + 167.2

your answer is: -634kJ