# Question b4e08

Jun 19, 2015

You use the standard enthalpies of formation.

#### Explanation:

To calculate the enthalpy change for a reaction, in your case for the combustion of ethane, you can use the standard enthalpies of formation of the species involved in the reaction.

More specifically, you can calculate the enthalpy change of a reaction by

DeltaH_"rxn" = sum(n * DeltaH_"f prod"^@) - sum(m * DeltaH_"f react"^@), where

$n$, $m$ - the number of moles of the compound;
$\Delta {H}_{\text{f prod}}^{\circ}$ - the standard enthlapy of formation of a product;
$\Delta {H}_{\text{f react}}^{\circ}$ - the standard enthalpy of formation of a reactant.

You can get the values for the standard enthalpies of formations from here:

So, the balanced chemical equation looks like this

${C}_{2} {H}_{6 \left(g\right)} + \frac{7}{2} {O}_{2 \left(g\right)} \to 2 C {O}_{2 \left(g\right)} + 3 {H}_{2} {O}_{\left(l\right)}$

Keeping in mind that the standard enthalpy of formation for an element in its natural state is zero, you will have

DeltaH_"rxn" = (2 * DeltaH_("f "CO_2)^@ + 3 * DeltaH_("f "H_2O)^@) - (1 * DeltaH_("f "C_2H_6)^@ + 7/2 * underbrace(DeltaH_("f "O_2)^@)_(color(blue)("=0)))

Using the values from the table will get you

DeltaH_"rxn" = [2cancel("moles") * (-393.5"kJ"/cancel("mole")) + 3cancel("moles") * (-285.8"kJ"/cancel("mole"))] - [1cancel("mole") * (-83.7"kJ"/cancel("mole")) + 0]

$\Delta {H}_{\text{rxn" = -"787 kJ" -"857.4 kJ" + "83.7 kJ}}$

DeltaH_"rxn" = color(green)(-"1560.7 kJ")#