Question #414d3

1 Answer
Ernest Z.
May 23, 2017

Δ_text(c)H = "-8686.9 kJ"

Explanation:

color(white)(mmmmmmmmll)"2B"_5"H"_9"(g)" + "12O"_2("g") → "5B"_2"O"_3("s")color(white)(l) + 9"H"_2"O(l)"
Δ_text(f)H^°"/kJ·mol⁻¹":color(white)(m)73.2color(white)(mmmmll) "0"color(white)(mmmll) "-1272.8"color(white)(mmll) "-241.83"

For most chemistry problems involving Δ_text(f)H^°, you need the equation:

color(blue)(bar(ul(|color(white)(a/a)Δ_text(c)H^° = sumΔ_text(f)H^°("p") - sumΔ_text(f)H^°("r")color(white)(a/a)|)))" "

where "p" = products and "r" = reactants.

sumΔ_text(f)H^°("p") = "[5(-1272.8) + 9(-241.83)] kJ = -8540.47 kJ"

sumΔ_text(f)H^°("r") = "2(73.2 kJ)" = "146.4 kJ"

Δ_text(c)H^° = "-8540.47 kJ - 146.4 kJ" = "-8686.9 kJ"

The standard enthalpy change for the combustion is "-8686.9 kJ".

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