# Calculating enthalpy change and energy required to convert lead sulphide to lead

Consider the following: \begin{alignat}{2} \ce{PbS(s) + 3/2O2(g) &-> PbO(s) + SO2(g)}\qquad&&\Delta H^\circ = -413.7~\mathrm{kJ} \\ \ce{PbO(s) + C(s) &-> Pb(s) + CO(g)}\qquad&&\Delta H^\circ = +106.8~\mathrm{kJ} \end{alignat} a. Calculate the enthalpy of the following process: $$\ce{PbS(s) + 3/2O2(g) + C(s) -> Pb(s) + SO2(g) + CO(g)}$$

b. What is the energy needed to make $454\ \mathrm g$ of $\ce{PbS}$ into $\ce{Pb}$?

My efforts:
a.

$\ce{PbS(s) + 3/2O2(g) -> PbO(s) + SO2(g)}$
$\ce{PbO(s) + C(s) -> Pb(s) + CO(g)}$
$\Delta H_\mathrm r^\circ = \Delta H_{\mathrm r(1)}^\circ + \Delta H_{\mathrm r(2)}^\circ$
$\Delta H_\mathrm r^\circ = -413.7~\mathrm{kJ~mol^{-1}} + 106.8~\mathrm{kJ~mol^{-1}} = -306.9~\mathrm{kJ~mol^{-1}}$

b. How do I find the energy needed?

Molecular mass of $\ce{PbS}$ is $239.2650~\mathrm{g~mol^{-1}}$

so we have $\frac{454~\mathrm{g}}{239.2650~\mathrm{g~mol^{-1}}} = 1.897~\mathrm{mol}$ of $\ce{PbS}$

So energy required is $1.897~\mathrm{mol} \times 306.9~\mathrm{kJ~mol^{-1}} = 582.2~\mathrm{kJ}$

It's easier to imagine \begin{equation}§§math§§[extract_tex]\ce{A + B -> C + D} \tag{1}[/extract_tex]\end{equation} \begin{equation}§§math§§[extract_tex]\ce{D + E -> F + G} \tag{2}[/extract_tex]\end{equation} \begin{equation}\end{equation} Now if we wanted to write a 3rd equation where both the reactions 1 and 2 happen, we'd get $$\ce{A + B + E -> C + F + G} \tag{3}$$ Why?
So, as the third reaction is reaction 1 and 2 happening at the same place, $\Delta H_{r_3}$ is the algebraic sum of $\Delta H_{r_2}$ and $\Delta H_{r_1}$. Thus, your final result is correct: $$\mathbf{-413.7 ~kJ \times mol^{−1} +106.8 ~kJ \times mol^{−1}=−306.9 ~kJ \times mol^{−1}}$$
We have to take a more careful look at the unit of $\Delta H$. What does it mean? It means that for each mole of the reactant ($\ce{PbS}$), one mole of product ($\ce{Pb}$) is formed. (With regards to their stoichiometric coefficients in reaction 3) So, we've got to simply 'punch' the numbers in the formula $n= \frac{m}{M}$ (Where $n$ is the number of moles, $m$ is the given mass and $M$ is the molar mass) And then multiply that number by the $\Delta H_{r_3}$. Thus, your final answer is correct too.