# Question 7fabf

Nov 6, 2016

$- {\text{370 kJ mol}}^{- 1}$

#### Explanation:

The idea here is that you need to take every molecule shown in the diagram as being equivalent to one mole of molecules that take part in the reaction.

Now, to make things simpler, I'll use the following notation for these molecules

• ${\text{A " -> DeltaH_f = -"110 kJ mol}}^{- 1}$
• ${\text{B " -> DeltaH_f = +"90 kJ mol}}^{- 1}$
• ${\text{C " -> DeltaH_f = -"390 kJ mol}}^{- 1}$
• ${\text{D " -> DeltaH_f = "0 kJ mol}}^{- 1}$

The first thing to notice here is that you're dealing with a limiting reagent situation because you have

• $4$ moles of $\text{B}$ on the reactants' side
• $2$ moles of $\text{B}$ on the products' side

This implies that only $2$ moles of $\text{B}$ take part in the reaction, the other $2$ moles being in excess. In other words, compound A" is acting as a limiting reagent.

You can thus say that the balanced chemical equation that describes this reaction will be

$2 \text{A" + 2"B" -> 2"C" + "D}$

Now all you have to do is use Hess' Law to calculate the enthalpy change of reaction, $\Delta {H}_{\text{rxn}}$, by looking at the enthalpy change associated with the formation of the products and the enthalpy change associated with the formation of the reactants.

color(blue)(bar(ul(|color(white)(a/a)color(black)(DeltaH_"rxn" = (sum_i n xx DeltaH_"f i") - (sum_j m xx DeltaH_"f j"))color(white)(a/a)|)))

Here

$n$ - the number of moles of a given product that take part in the reaction
$\Delta {H}_{\text{f i}}$ - the enthalpy of formation os that product
$m$ - the number of moles of a given reactant that take part in the reaction
$\Delta {H}_{\text{f j}}$ - the enthalpy of formation of that reactant

You will have

DeltaH_"rxn" = [2 color(red)(cancel(color(black)("moles C"))) * (-390 "kJ"/(1color(red)(cancel(color(black)("mole C"))))) + 1 color(red)(cancel(color(black)("mole D"))) * 0 "kJ"/(1color(red)(cancel(color(black)("mole D")))) ] - [2 color(red)(cancel(color(black)("moles A"))) * (-110"kJ"/(1color(red)(cancel(color(black)("mole A"))))) + 2 color(red)(cancel(color(black)("moles B"))) * (+90"kJ"/(1color(red)(cancel(color(black)("mole B")))))]

which is equal to

$\Delta {H}_{\text{rxn" = -"740 kJ}}$

Now, notice that the problem wants you to figure out the enthalpy change of reaction kilojoules per mole, ${\text{kJ mol}}^{- 1}$.

This basically means that you must figure out the enthalpy change associated with this version of the reaction

$\text{A" + "B" -> "C" + 1/2"D}$

Without having to make other calculations, you can say that the enthalpy change that occurs when $1$ mole of $\text{A}$ reacts with $1$ mole of $\text{B}$ to form $1$ mole of $\text{C}$ and $\frac{1}{2}$ moles of $\text{D}$ is equal to

$\textcolor{g r e e n}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\Delta {H}_{\text{rxn" = -"740 kJ"/2 = -"370 kJ mol}}^{- 1}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

This can be seen by using any of the four chemical species as the base for the calculation

1 color(red)(cancel(color(black)("mole A"))) * overbrace((-"740 kJ")/(2color(red)(cancel(color(black)("moles A")))))^(color(blue)("what we've calculated before")) = -"370 kJ"

1/2 color(red)(cancel(color(black)("moles D"))) * overbrace((-"740 kJ")/(1color(red)(cancel(color(black)("mole D")))))^(color(blue)("what we've calculated before")) = -"370 kJ"#