# Question 7ee50

Feb 22, 2017

$\Delta {H}_{\text{rxn" = - "132 kJ}}$

#### Explanation:

Your goal here is to figure out the enthalpy change of reaction for

"D + F " -> " G + M", " "DeltaH_"rxn" = ? color(white)(aaaaa)color(red)("(*)")

by using the thermochemical equations that describe the following reactions

$\text{G + C " -> " A + B" color(white)(aaaa)DeltaH_1 = +"277 kJ"" " " } \textcolor{b l u e}{\left(1\right)}$

$\text{C + F " -> " A" color(white)(aaaaaaaa)DeltaH_2 = +"303 kJ"" " " } \textcolor{b l u e}{\left(2\right)}$

$\textcolor{w h i t e}{a a a .} \text{D " ->" B + M" color(white)(aa.a)DeltaH_3 = - "158 kJ"" " " } \textcolor{b l u e}{\left(3\right)}$

Now, Hess' Law tells you that the enthalpy change of a reaction is independent of the number of steps needed to get from the products to the reactants.

Simply put, it doesn't matter how you get from the products to the reactants, the enthalpy change will always be the same for a given reaction.

Notice that in your target equation, you need $\text{G}$ as a product, but that equation $\textcolor{b l u e}{\left(1\right)}$ has $\text{G}$ as a reactant. What you can is flip equation $\textcolor{b l u e}{\left(1\right)}$ to have $\text{G}$ as a product.

When you flip a chemical equation, you must change sign of the enthalpy change of reaction!

In this case, $\text{G}$ is consumed in an endothermic reaction, so it will be produced in an exothermic reaction.

$\text{Flip" color(white)(.)color(blue)((1)):" " "A + B " -> " G + C" color(white)(aaaa)DeltaH_"1 flip" = -"277 kJ}$

Now notice what happens when you add equation $\textcolor{b l u e}{\left(1\right)}$ flipped and equation $\textcolor{b l u e}{\left(2\right)}$

$\text{A + B " -> " G + C" color(white)(aaaa)DeltaH_"1 flip" = -"277 kJ}$

$\text{C + F " -> " A" color(white)(aaaaaaaaaa)DeltaH_2 = +"303 kJ}$
$\frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{a}}$
$\textcolor{red}{\cancel{\textcolor{b l a c k}{\text{A"))) + "B" + color(red)(cancel(color(black)("C"))) + "F " -> " G" + color(red)(cancel(color(black)("C"))) + color(red)(cancel(color(black)("A}}}}$

which is equivalent to

$\text{B + F " -> " G}$

Since you've added the two reactions, you must also 8add8 the enthalpy changes of reaction

$\Delta {H}_{\text{1 flip + 2" = DeltaH_"1 flip}} + \Delta {H}_{2}$

$\Delta {H}_{\text{1 flip + 2" = -"277 kJ" + "303 kJ}}$

$\Delta {H}_{\text{1 flip + 2" = +"26 kJ}}$

This means that I can get from $\text{B}$ and $\text{F}$ directly to $\text{G}$ by either using two steps that have $\Delta {H}_{\text{1 flip}}$ and $\Delta {H}_{2}$, respectively or by using a single step that has $\Delta {H}_{\text{1 flip + 2}}$.

Notice that in order to get $\text{D}$ as a reactant and $\text{M}$ as a product, I can add the above single-step equation and equation $\textcolor{b l u e}{\left(3\right)}$

$\text{B + F " -> " G" color(white)(aaaaaaaa)DeltaH_"1 flip + 2" = +"26 kJ}$

$\textcolor{w h i t e}{a a a a} \text{D " ->" B + M" color(white)(aaaaaaaa)DeltaH_3 = - "158 kJ}$
$\frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{a}}$
color(red)(cancel(color(black)("B"))) + "D + F " -> " G" + color(red)(cancel(color(black)("B"))) + "M"

which is equivalent to equation $\textcolor{red}{\text{(*)}}$

$\text{D + F " -> " G + M}$

The enthalpy change for this reaction will be

$\Delta {H}_{\text{rxn" = DeltaH_"1 flip + 2}} + \Delta {H}_{3}$

DeltaH_"rxn" = + "26 kJ" + (-"158 kJ")#

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\Delta {H}_{\text{rxn" = -"132 kJ}}}}}$