# Question f11be

Feb 26, 2016

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

#### Explanation:

Part 1. Balance the equation.

$\text{PbS" + "O"_2 → "SO"_2 + "PbO}$

Start with the most complicated formula. Put a $1$ in front of ${\text{SO}}_{2}$.

$\text{PbS" + "O"_2 → color(red)(1)"SO"_2 + "PbO}$

Balance $\text{S}$. Put a $1$ in front of "PbS.

$\textcolor{b l u e}{1} \text{PbS" + "O"_2 → color(red)(1)"SO"_2 + "PbO}$

Balance $\text{Pb}$. Put a $1$ in front of $\text{PbO}$.

$\textcolor{b l u e}{1} \text{PbS" + "O"_2 → color(red)(1)"SO"_2 + color(green)(1)"PbO}$

Balance $\text{O}$. Put $1.5$ in front of ${\text{O}}_{2}$.

$\textcolor{b l u e}{1} \text{PbS" + color(orange)(1.5)"O"_2 → color(red)(1)"SO"_2 + color(green)(1)"PbO}$

Remove fractions. Multiply by $2$.

$\text{2PbS" + "3O"_2 → "2SO"_2 + 2PbO}$

Part 2. Calculate ΔH_"rxn".

$\textcolor{w h i t e}{m m m m m m m m m} \text{2PbS(s)" + "3O"_2("s") → "2SO"_2("g")color(white)(l) + "2PbO(s)}$
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 $\text{p}$ = products and $\text{r}$ = 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 $\text{-827.5 kJ}$.