# What is the enthalpy change for the reaction?

## 2A+B⇌2C+2D Use the following data: Substance ΔH∘f (kJ/mol) A -231 B -393 C 187 D -475

Jan 14, 2018

$+ \text{277 kJ}$

#### Explanation:

All you really need to do here is to use Hess' Law, which allows you to calculate the standard enthalpy change of the reaction by using the standard enthalpy changes of the reactants and of the products.

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\Delta {H}_{\text{rxN"^@ = sum DeltaH_"f, products"^@ - sum DeltaH_"f, reactants}}^{\circ}}}}$

Now, you know that for the given reaction

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

you have $2$ moles of $\text{A}$ reacting with $1$ mole of $\text{B}$ in order to produce $2$ moles of $\text{C}$ and $2$ moles of $\text{D}$, so you can say that the sum of the standard enthalpy changes of the products will be

sum DeltaH_"f, products"^@ = 2 color(red)(cancel(color(black)("moles C"))) * "187 kJ"/(1color(red)(cancel(color(black)("mole C")))) + 2 color(red)(cancel(color(black)("moles D"))) * (-"475 kJ")/(1color(red)(cancel(color(black)("mole D"))))

$\sum \Delta {H}_{\text{f, products"^@ = -"576 kJ}}$

Similarly, the sum of standard enthalpy changes of the reactants will be

sum DeltaH_"f, reactants"^@ = 2color(red)(cancel(color(black)("moles A"))) * (-"231 kJ")/(1color(red)(cancel(color(black)("mole A")))) + 1 color(red)(cancel(color(black)("mole B"))) * (-"391 kJ")/(1color(red)(cancel(color(black)("mole B"))))

$\sum \Delta {H}_{\text{f, reactants"^@ = -"853 kJ}}$

You can thus say that the standard enthalpy change of the reaction will be

DeltaH_"rxn"^@ = -"576 kJ" - (-"853 kJ")

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\Delta {H}_{\text{rxn"^@ = +"277 kJ}}}}}$