# Question 97d3a

May 25, 2015

$\Delta {H}_{\text{rxn}}$ for your reaction will be equal to -2831 kJ.

You can use bond enthalpies to calculate the change in enthalpy for a particular reaction by taking into account the energy required to break the bonds of the reactants and the energy released when the bonds of the products are formed.

Mathematically, this can be written as

DeltaH_"rxn" = sum(DeltaH_"reactants") - sum(DeltaH_"products")

In order to figure out exactly what is the energy needed to break the bonds of the reactants and the energy released when the products are formed, you need to determine the number and type of bonds that are found in the species that take part in your reaction.

• Bonds broken

Your reactants are ethane, ${C}_{2} {H}_{6}$, and oxygen, ${O}_{2}$. Each ethane molecule has 1 $\text{C-C}$ bonds and 6 $\text{C-H}$ bonds. However, you're dealing with 2 moles of ethane, so you'll need to break 2 $\text{C-C}$ bonds and 12 $\text{C-H}$ bonds.

An oxygen molecule consists of a single oxygen-oxygen double bond, $\text{O=O}$. Since you have 7 moles present, you'll need to break 7 such bonds.

• Bonds formed

Your products are carbon dioxide, $C {O}_{2}$, and water, ${H}_{2} O$.

Each carbon dioxide molecule fetures 2 $\text{C=O}$ bonds, and since you're forming 4 moles, you'll need to form 8 such bonds.

In water's case, you have 2 $\text{H-O}$ bonds per molecule, so the 6 moles produced will mean that you need to form 12 such bonds.

Now all you have to do is use the above equation

DeltaH_"rxn" = (2cancel("moles") * 6 * 413"kJ"/cancel("mol") + 2cancel("moles") * 348"kJ"/cancel("mol") + 7cancel("moles") * 495"kJ"/cancel("mol")) - (4cancel("moles") * 2 * 799"kJ"/cancel("mol") + 6cancel("moles") * 2 * 463"kJ"/cancel("mol"))

DeltaH_"rxn" = "9117 kJ" - "11948 kJ" = color(green)("-2831 kJ")#