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

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Part 1:

It's easier to imagine \begin{equation}\ce{A + B -> C + D} \tag{1} \end{equation} \begin{equation}\ce{D + E -> F + G} \tag{2}\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 \begin{equation}\ce{A + B + E -> C + F + G} \tag{3}\end{equation} Why?

  1. D is both a reactant and a product. So, it's a reaction intermediate. It's like it's produced in 1 and again consumed in 2. So, if both reactions take place, it's neither on the reactants' side nor the products.
  2. The conventional way of writing reactions doesn't show the mechanism of the reaction. You have to image the reaction happening, to completely understand why D isn't written.

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: $$\mathbf{-413.7 ~kJ \times mol^{−1} +106.8 ~kJ \times mol^{−1}=−306.9 ~kJ \times mol^{−1}}$$

Part 2:

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.

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