Question #f11be

1 Answer
Feb 26, 2016

"2PbS(s)" + "3O"_2("g") → "2SO"_2("g") + "2PbO(s)"; ΔH_f^° = "-827.5 kJ".

Explanation:

Part 1. Balance the equation.

"PbS" + "O"_2 → "SO"_2 + "PbO"

Start with the most complicated formula. Put a 1 in front of "SO"_2.

"PbS" + "O"_2 → color(red)(1)"SO"_2 + "PbO"

Balance "S". Put a 1 in front of "PbS.

color(blue)(1)"PbS" + "O"_2 → color(red)(1)"SO"_2 + "PbO"

Balance "Pb". Put a 1 in front of "PbO".

color(blue)(1)"PbS" + "O"_2 → color(red)(1)"SO"_2 + color(green)(1)"PbO"

Balance "O". Put 1.5 in front of "O"_2.

color(blue)(1)"PbS" + color(orange)(1.5)"O"_2 → color(red)(1)"SO"_2 + color(green)(1)"PbO"

Remove fractions. Multiply by 2.

"2PbS" + "3O"_2 → "2SO"_2 + 2PbO"

Part 2. Calculate ΔH_"rxn".

color(white)(mmmmmmmmm)"2PbS(s)" + "3O"_2("s") → "2SO"_2("g")color(white)(l) + "2PbO(s)"
ΔH_f^°"/kJ·mol⁻¹":color(white)(m)"-100.4"color(white)(mmm) "0"color(white)(mmmm) "-296.83"color(white)(mm) "-217.32"

For most chemistry problems involving ΔH_f^°, you need the equation:

ΔH_(rxn)^° = ΣΔH_f^°("p") - ΣΔH_f^°("r"),

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

ΣΔH_f^°("p") = "[2(-296.83) + 2(-217.32)] kJ = -1028.30 kJ"

ΣΔH_f^°("r") = "2(-100.4 kJ)" = "-200.8 kJ"

ΔH_("rxn")^° = ΣΔH_f^°("p") - ΣΔH_f^°("r") = "-1028.30 kJ + 200.8 kJ" = "-827.5 kJ"

The standard enthalpy change is "-827.5 kJ".