# Enthalpy question?

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:

$\text{CaO" (s) + 2"HCl"(aq)->"CaCl"_2(aq)+"H"_2"O"(l), " " DeltaH=-194.2 " kJ}$

$\text{CaO"(s)+"CO"_2(g)->"Ca"color(red)"C""O"_3(s)," " DeltaH=-178.4 " kJ}$

You should get $\left(1\right) - \left(2\right)$:

cancel("CaO" (s)) + 2"HCl"(aq)->"CaCl"_2(aq)+"H"_2"O"(l)
${\text{CaCO"_3(s) -> cancel("CaO"(s))+"CO}}_{2} \left(g\right)$
$\text{----------------------------------------------------------------------}$
${\text{CaCO"_3(s) + 2"HCl"(aq) -> "CaCl"_2(aq) + "H"_2"O"(l) + "CO}}_{2} \left(g\right)$

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:

$\Delta {H}_{1} + \Delta {H}_{2} = \Delta {H}_{r x n}$

And the second step was reversed, thus reversing the sign of its $\Delta H$.

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

$= \textcolor{b l u e}{- \text{15.8 kJ}}$