Enthalpy question?

1 Answer
Sep 11, 2017

No, it's not correct. Does carbonic acid dissolve in water (could it ever be a solid?), or does it decompose in water? Furthermore, is the second reaction balanced? (No, it is not. Where is carbon?)


The corrected reactions are:

#"CaO" (s) + 2"HCl"(aq)->"CaCl"_2(aq)+"H"_2"O"(l), " " DeltaH=-194.2 " kJ"#

#"CaO"(s)+"CO"_2(g)->"Ca"color(red)"C""O"_3(s)," " DeltaH=-178.4 " kJ"#

You should get #(1) - (2)#:

#cancel("CaO" (s)) + 2"HCl"(aq)->"CaCl"_2(aq)+"H"_2"O"(l)#
#"CaCO"_3(s) -> cancel("CaO"(s))+"CO"_2(g)#
#"----------------------------------------------------------------------"#
#"CaCO"_3(s) + 2"HCl"(aq) -> "CaCl"_2(aq) + "H"_2"O"(l) + "CO"_2(g)#

which is due to correcting and then reversing the second reaction mechanistic step.

The enthalpy of reaction then comes from the additive property of state functions:

#DeltaH_1 + DeltaH_2 = DeltaH_(rxn)#

And the second step was reversed, thus reversing the sign of its #DeltaH#.

#color(blue)(DeltaH_"rxn") = -"194.2 kJ" + (-(-"178.4 kJ"))#

#= color(blue)(-"15.8 kJ")#