# Question 31f5b

Aug 26, 2015

155.65 KJ

#### Explanation:

The Hess law states that the final state that regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is equal to the sum of the heats of reaction for any set of reactions which in sum are equivalent to the overall reaction:. Hess Law .

Then the first thing to do is to balance the equations so you can have the final reaction by adding the individual steps given. You can invert the reactions but remember to change its sign. Also when you multiply or factor them you should do the same for the energy value.

For this specific problem the procedure will be as follows, of course like in any math problem you can swap the order of the reactions, because it is an addition:

2${N}_{2}$O $\to$ 2${N}_{2}$ $+$ ${O}_{2}$ $\Delta$H $=$$-$163.2 (No changes)

2$N {O}_{2}$ $\to$ 2NO $+$ ${O}_{2}$ $\Delta$H$=$$+$113.1 (inverted, sign changed)

2${N}_{2}$ $+$ 2${O}_{2}$ $\to$ 4NO $\Delta$H$=$$+$361.4 (multiplied by 2)

Any changes done to them is noted in the parenthesis, now you add them.

2${N}_{2}$O $+$2$N {O}_{2}$$+$ 2${N}_{2}$$+$ 2${O}_{2}$$\to$ 6NO $+$ 2${N}_{2}$ $+$ 2${O}_{2}$ $\Delta$H $=$$+$311.3

cancelling terms you get

2${N}_{2}$O $+$2$N {O}_{2}$$\to$ 6NO $\Delta$H $=$$+$311.3

factorizing by 2

${N}_{2}$O $+$$N {O}_{2}$$\to$ 3NO $\Delta$H $=$$+$155.65

Aug 26, 2015

$\Delta {H}_{\text{rxn" = +"155.7 kJ}}$

#### Explanation:

I'll start by writing down your three chemical equations

$\text{N"_ (2(g)) + "O"_ (2(g)) -> 2"NO"_ ((g))" "color(blue)((1))" "DeltaH_1 = +"180.7 kJ}$

$2 \text{NO"_ ((g)) + "O"_ (2(g)) -> 2"NO"_ (2(g))" "color(blue)((2))" "DeltaH_2 = -"113.1 kJ}$

$2 \text{N"_ 2"O"_ ((g)) -> 2"N"_ (2(g)) + "O"_ (2(g))" "color(blue)((3))" "DeltaH_3 = -"163.2 kJ}$

You need to use these three equations to derive the enthalpy change of reaction for a fourth reaction

${\text{N"_ 2"O"_ ((g)) + "NO"_ (2(g)) -> 3"NO}}_{\left(g\right)}$

Now, what Hess' Law tells you is that the total change in enthalpy for a given reaction is independent of the number of steps needed to complete that reaction.

This means that you can think of the first these reactions as the steps you need to take in order to get the target reaction.

For example, notice that the target equation has 1 mole of ${\text{NO}}_{2}$ on the reactants' side, and that equation $\textcolor{b l u e}{\left(2\right)}$ has 2 moles of ${\text{NO}}_{2}$ on the products' side.

This means that if you reverse the second equation and divide everything by 2, you can write

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} {\text{NO"_ (2(g)) -> color(red)(cancel(color(black)(2)))/color(red)(cancel(color(black)(2)))"NO"_ ((g)) + 1/2"O}}_{2 \left(g\right)}$

$\text{NO"_ (2(g)) -> "NO"_ ((g)) + 1/2"O"_ (2(g))" } \textcolor{b l u e}{\left({2}^{'}\right)}$

Now think of these changes will affect this reaction's $\Delta H$. You're practically reversing the direction of the reaction, so you go from a negative $\Delta {H}_{2}$ to a positive $\Delta {H}_{2}^{'}$.

This happens because the forward reaction is exothermic, i.e. it has a negative $\Delta H$, then the reverse reaction must be enothermic, i.e. has a positive $\Delta H$.

When you use half as much amounts of the species that take part in the reaction, you will of course have a $\Delta {H}_{2}^{'}$ that is half the value of $\Delta {H}_{2}$.

This means that you have

DeltaH_2^' = (-DeltaH_2)/2 = (-(-"113.1 kJ"))/2 = +"56.55 kJ"

Now notice that the target reaction has 1 mole of $\text{N"_2"O}$ on the reactants' side, and equation $\textcolor{b l u e}{\left(3\right)}$ has 2 moles of $\text{N"_2"O}$ on the reactants' side.

This means that you can divide everything by 2 to get

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} {\text{N"_ 2"O"_ ((g)) -> color(red)(cancel(color(black)(2)))/color(red)(cancel(color(black)(2)))"N"_ (2(g)) + 1/2"O}}_{\left(g\right)}$

$\text{N"_ 2"O"_ ((g)) -> "N"_ (2(g)) + 1/2"O"_ (2(g))" }$ $\textcolor{b l u e}{\left({3}^{'}\right)}$

The enthalpy change for this reaction will be

DeltaH_3^' = (DeltaH_3)/2 = (-"163.2 kJ")/2 = -"81.6 kJ"

Notice what you get when you add reactions $\textcolor{b l u e}{\left(1\right)}$, $\textcolor{b l u e}{\left({2}^{'}\right)}$, and $\textcolor{b l u e}{\left({3}^{'}\right)}$

$\textcolor{p u r p \le}{\cancel{\textcolor{b l a c k}{{\text{N"_ (2(g))))) + color(purple)(cancel(color(black)("O"_ (2(g))))) + "NO"_ (2(g)) + "N"_ 2"O"_ ((g)) -> 2"NO"_ ((g)) + "NO"_ ((g)) + color(purple)(cancel(color(black)(1/2"O"_ (2(g))))) + color(purple)(cancel(color(black)("N"_ (2(g))))) + color(purple)(cancel(color(black)(1/2"O}}_{2 \left(g\right)}}}}$

This is equivalent to

${\text{N"_ 2"O"_ ((g)) + "NO"_ (2(g)) -> 3"NO}}_{\left(g\right)}$

The enthalpy change of reaction will be

$\Delta {H}_{\text{rxn}} = \Delta {H}_{1} + \Delta {H}_{2}^{'} + \Delta {H}_{3}^{'}$

$\Delta {H}_{\text{rxn" = "180.7 kJ" + "56.55 kJ" - "81.6 kJ}}$

DeltaH_"rxn" = color(green)(+"155.7 kJ")#

The answer is rounded to four sig figs.