Question da5d4

Oct 25, 2016

Δ_"r"H = "+109 kJ".

Explanation:

In this case, the statement of the problem gives you the answer.

However, you could also have solved the problem by looking up the standard enthalpy of formation of each of the substances.

The formula for enthalpy of reaction is

color(blue)(bar(ul(|color(white)(a/a)Δ_rH^° = sumΔ_fH_text(products)^° - sumΔ_fH_text(reactants)^°color(white)(a/a)|)))" "

$\textcolor{w h i t e}{m m m m m m m m m} \text{Ca(OH)"_2"(s)" → "CaO(s)" + "H"_2"O(g)}$
Δ_text(f)H^°"/kJ·mol"^"-1": color(white)(l)+986.09color(white)(mmmll)"-635.0"color(white)(mll)"-241.83"

 Δ_rH = "[1(-635.0) + 1(-241.83) - 1(986.09)] kJ" = "+109.3 kJ"#