calculate the standard enthalpy change for the reaction at 25*C
Mg(OH)2 (s) + 2HCl (g) -----> MgCl2 (s) + 2H2O (g)
Mg(OH)2( s ) + 2HCl ( g ) ==> MgCl2( s ) + 2H2O(g)
Look up the ∆Hfº of reactants and products. Then ∑produts - ∑reactants = ∆Hreaction
These are the values I found:
Mg(OH)2(s) = -924.7 kJ/mol
HCl(g) = -92.5 kJ/mol
MgCl2(s) = -641.8 kJ/mol
H2O(g) = -241.8 kJ/mol
(-641.8 + 2x-241.8) - (-924.7 + -92.5) = ∆Hreaction
-1125.4 - (-1017.2) = ∆Hreaction
∆H reaction = -108.2kJ
Comments
Leave a comment