# Question 98e63

Feb 27, 2017

$\Delta {H}_{\text{rxn" = + "74 kJ mol}}^{- 1}$

#### Explanation:

You know that

${\text{A " -> " B" " " " "DeltaH_1 = -"210 kJ mol}}^{- 1}$

${\text{A " -> " C" " " " "DeltaH_2 = -"50 kJ mol}}^{- 1}$

${\text{C " -> " D" " " " "DeltaH_3 = -"86 kJ mol}}^{- 1}$

Now, your goal here is to figure out the enthalpy change associated with the reaction

"B " -> " D" " " " " DeltaH_"rxn" = ?

The first thing to do here is to make sure that you gert $\text{B}$ as a reactant by reversing the first reaction. You will have

$\text{B " -> " A"" " " "DeltaH_"1 rev}$

Keep in mind that when you reverse a reaction, you must change the sign of the enthalpy of reaction. In this case, you will have

DeltaH_"1 rev" = - (-"210 kJ mol"^(-1))

$\Delta {H}_{\text{1 rev" = + "210 kJ mol}}^{- 1}$

Notice what happens when you add the first reaction reverse and the second reaction

${\text{B " -> " A"" " " "DeltaH_"1 rev" = + "210 kJ mol}}^{- 1}$

${\text{A " -> " C" " " " "DeltaH_2 = -"50 kJ mol}}^{- 1}$
$\frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{a}}$

$\text{B" + color(red)(cancel(color(black)("A"))) -> color(red)(cancel(color(black)("A"))) + "C}$

which is equivalent to

$\text{B " -> " C}$

The enthalpy change for this reaction will be the sum of the enthalpy changes of the two reactions we've added to get to this point

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

DeltaH_"1 rev + 2" = + "210 kJ mol"^(-1) + (-"50 kJ mol"^(-1))

$\Delta {H}_{\text{1 rev + 2" = + "160 kJ mol}}^{- 1}$

Finally, to get from $\text{B}$ to $\text{D}$, add the above reaction and the third reaction

${\text{B " -> " C"" " " "DeltaH_"1 rev + 2" = + "160 kJ mol}}^{- 1}$

${\text{C " -> " D" " " " "DeltaH_2 = -"86 kJ mol}}^{- 1}$
$\frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{a}}$

$\text{B" + color(red)(cancel(color(black)("C"))) -> color(red)(cancel(color(black)("C"))) + "D}$

which is equivalent to

$\text{B " -> " D}$

Once again, the enthalpy change for this reaction will be equal to the sum of the enthalpy changes associated with the two reactions we've added to get to this point

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

DeltaH_"rxn" = + "160 kJ mol"^(-1) + (-"86 kJ mol"^(-1))#

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\Delta {H}_{\text{rxn" = + "74 kJ mol}}^{- 1}}}}$

This goes to show that the enthalpy change of a reaction is independent of the number of steps needed to make said reaction. $\to$ think Hess' Law here.

In this case, the enthalpy change of the reaction

${\text{B " -> " D"" " " "DeltaH_"rxn" = + "74 kJ mol}}^{- 1}$

is the same regardless if the reaction takes place in a single step or in the three steps we've used to get to it.