# What is the enthalpy change of reaction for this reaction?

## $2 \text{A" + B rightleftharpoons 2"C" + 2"D}$

Mar 24, 2017

$\Delta {H}_{\text{rxn"^@ = - "237 kJ}}$

#### Explanation:

The idea here is that we can use Hess' Law to calculate the standard enthalpy change of reaction, $\Delta {H}_{\text{rxn}}^{\circ}$, for this reaction using the standard enthalpy changes of formation, $\Delta {H}_{f}^{\circ}$, of the chemical species that take part in the reaction.

Hess' Law states that the enthalpy change of a reaction is independent of the path taken by the reaction.

$\textcolor{b l u e}{2} \text{A" + "B" rightleftharpoons color(purple)(2)"C" + color(brown)(2)"D}$

Now, the standard enthalpy change of reaction can be calculated by taking the sum of the enthalpy changes of formation of the products multiplied by their respective stoichiometric coefficients and the sum of the enthalpy changes of formation of the reactants multiplied by their respective stoichiometric coefficients.

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\Delta {H}_{\text{rxn"^@ = sum_i n xx DeltaH_"f i"^@ - sum_j m xx DeltaH_"f j}}^{\circ}}}}$

Here

• $n$ is the stoichiometric coefficient of a product
• $m$ is the stoichiometric coefficient of a reactant

So, you know that you have

• $\textcolor{b l u e}{2}$ moles of $\text{A}$ and $1$ mole of $\text{B}$ on the reactants' side
• $\textcolor{p u r p \le}{2}$ moles of $\text{C}$ and $\textcolor{b r o w n}{2}$ moles of $\text{C}$ on the products' side

You can say that for the products, you have

sum_j m xx DeltaH_"f j"^@ = color(purple)(2)color(red)(cancel(color(black)("moles"))) * (+215 "kJ"/color(red)(cancel(color(black)("mol")))) + color(brown)(2)color(red)(cancel(color(black)("moles"))) * (-523 "kJ"/color(red)(cancel(color(black)("mol"))))

= "430 kJ" + (-"1046 kJ")

$= - \text{616 kJ}$

Similarly, you can say that for the reactants', you have

sum_i n xx DeltaH_"f i"^@ = color(blue)(2)color(red)(cancel(color(black)("moles"))) * (-225 "kJ"/color(red)(cancel(color(black)("mol")))) + 1color(red)(cancel(color(black)("mole"))) * (-403 "kJ"/color(red)(cancel(color(black)("mol"))))

 = -"450 kJ" + (-"403 kJ")

$= - \text{853 kJ}$

Therefore, you can say that the standard enthalpy change of reaction is equal to

DeltaH_"rxn"^@ = -"853 kJ" - (-"616 kJ")

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\Delta {H}_{\text{rxn"^@ = -"237 kJ}}}}}$

The answer is rounded to three sig figs.