# How do you use Hess's law to find the enthalpy of reaction for these reactions?

## a) $\text{NaOH"(g) + "CO"_2(g) -> "Na"_2"CO"_3(s) + "H"_2"O} \left(g\right)$ b) ${\text{C"_2"H"_2(g) + "H"_2(g) -> "C"_2"H}}_{4} \left(g\right)$ c) ${\text{NO"_2(g) rightleftharpoons "N"_2"O}}_{4} \left(g\right)$

Jul 28, 2016

I'm actually going to go out of order for this, since it seems that (c) is easier than (b), which is easier than (a).

The general idea is, enthalpy is a state function, so we only need to think about the initial and final states of the reaction.

Therefore, we can treat the standard enthalpies of formation of each reactant and product stoichiometrically as initial and final states.

$\setminus m a t h b f \left(\Delta {H}_{\text{rxn}}^{\circ} = {\sum}_{P} {\nu}_{P} \Delta {H}_{f , P}^{\circ} - {\sum}_{R} {\nu}_{R} \Delta {H}_{f , R}^{\circ}\right)$

$= \stackrel{\text{Sum of Products' Enthalpies of Formation")overbrace([nu_(P_1)DeltaH_(f,P_1) + nu_(P_2)DeltaH_(f,P_2) + . . . + nu_(P_n)DeltaH_(f,P_n)]) - stackrel("Sum of Reactants' Enthalpies of Formation}}{\overbrace{\left[{\nu}_{{R}_{1}} \Delta {H}_{f , {R}_{1}} + {\nu}_{{R}_{2}} \Delta {H}_{f , {R}_{2}} + . . . + {\nu}_{{R}_{n}} \Delta {H}_{f , {R}_{n}}\right]}}$

where $\nu$ is the stoichiometric coefficient in the balanced chemical reaction.
$R$ / $P$ means reactants / products.

ENTHALPIES OF REACTION: SINGLE REACTANT/PRODUCT

c) For the reaction

$\setminus m a t h b f \left(\textcolor{red}{2} \text{NO"_2(g) -> "N"_2"O"_4} \left(g\right)\right)$,

which must be balanced first, it's important to realize that ${\text{N"_2"O}}_{4}$ can either be a gas or liquid at standard conditions, so you have to make sure you look at the right number.

My textbook lists:

DeltaH_(f,"NO"_2(g))^@ = "33.2 kJ/mol"

DeltaH_(f,"N"_2"O"_4(g))^@ = "9.16 kJ/mol"

So you just have:

color(blue)(DeltaH_"rxn"^@) = [nu_("N"_2"O"_4(g))DeltaH_(f,"N"_2"O"_4(g))^@] - [nu_("NO"_2(g))DeltaH_(f,"NO"_2(g))^@]

$= \left[\left(1\right) \left(\text{9.16 kJ/mol")] - [(color(red)(2))("33.2 kJ/mol}\right)\right]$

$= \textcolor{b l u e}{- \text{57.2 kJ/mol}}$

ENTHALPIES OF REACTION: ELEMENTAL STATES

b) For elements in their elemental form, $\Delta {H}_{f}^{\circ} = 0$, like for ${\text{H}}_{2} \left(g\right)$ (and for ${\text{O}}_{2} \left(g\right)$, ${\text{F}}_{2} \left(g\right)$, ${\text{Br}}_{2} \left(l\right)$, ${\text{I}}_{2} \left(s\right)$, ${\text{N}}_{2} \left(g\right)$, and ${\text{Cl}}_{2} \left(g\right)$). After all, they naturally formed that way. Besides that, my textbook lists:

DeltaH_(f,"C"_2"H"_2(g))^@ = "227.4 kJ/mol"

DeltaH_(f,"C"_2"H"_4(g))^@ = "52.4 kJ/mol"

for the already-balanced reaction written as

$\setminus m a t h b f \left({\text{C"_2"H"_2(g) + "H"_2(g) -> "C"_2"H}}_{4} \left(g\right)\right)$.

Same deal as before. Just note that you can ignore ${\text{H}}_{2} \left(g\right)$ because its standard enthalpy of formation is 0.

color(blue)(DeltaH_"rxn"^@) = [nu_("C"_2"H"_4(g))DeltaH_(f,"C"_2"H"_4(g))^@] - [nu_("C"_2"H"_2(g))DeltaH_(f,"C"_2"H"_2(g))^@ + nu_("H"_2(g))DeltaH_(f,"H"_2(g))^@]

$= \left[\left(1\right) \left(\text{52.4 kJ/mol")] - [(1)("227.4 kJ/mol") + (1)("0 kJ/mol}\right)\right]$

$= \textcolor{b l u e}{- \text{175 kJ/mol}}$

ENTHALPIES OF REACTION: MULTIPLE REACTANTS/PRODUCTS

a) This one would be the one where you could possibly mess up, since there is more than one reactant/product, and so, you could mess up your signs.

$\setminus m a t h b f \left(\textcolor{red}{2} \text{NaOH"(s) + "CO"_2(g) -> "Na"_2"CO"_3(s) + "H"_2"O} \left(g\right)\right)$,

my textbook lists:

DeltaH_(f,"NaOH"(s))^@ = -"425.8 kJ/mol"

DeltaH_(f,"CO"_2(g))^@ = -"393.5 kJ/mol"

DeltaH_(f,"Na"_2"CO"_3(s))^@ = -"1130.7 kJ/mol"

DeltaH_(f,"H"_2"O"(g))^@ = -"241.8 kJ/mol"

So, don't use the value for ${\text{CO}}_{2} \left(a q\right)$. Also, you should make sure you have parentheses:

color(blue)(DeltaH_"rxn"^@) = [nu_("Na"_2"CO"_3(s))DeltaH_(f,"Na"_2"CO"_3(s))^@ + nu_("H"_2"O"(g))DeltaH_(f,"H"_2"O"(g))^@] - [nu_("NaOH"(s))DeltaH_(f,"NaOH"(s))^@ + nu_("CO"_2(g))DeltaH_(f,"CO"_2(g))^@]

$= \left[\left(1\right) \left(- \text{1130.7 kJ/mol") + (1)(-"241.8 kJ/mol")] - [(color(red)(2))(-"425.8 kJ/mol") + (1)(-"393.5 kJ/mol}\right)\right]$

$= \textcolor{b l u e}{- \text{127.4 kJ/mol}}$