# Carbon and oxygen react to give carbon dioxide. The reaction of 4.49 g C(s) with 9.21 g O_2(g)releases 113.2 kJ of heat. What is the enthalpy of formation of CO_2(g)?

Apr 14, 2016

$\Delta {H}_{\text{f" = -"393 kJ mol}}^{- 1}$

#### Explanation:

A compound's enthalpy of formation, $\Delta {H}_{\text{f}}$, tells you the enthalpy change that accompanies the formation of one mole of that compound from its constituent elements in their stable form.

In this case, elemental carbon will react with oxygen gas to form carbon dioxide

${\text{C"_ ((s)) + "O"_ (2(g)) -> "CO}}_{2 \left(g\right)}$

Your strategy here will be to determine how many moles of carbon dioxide are formed when $\text{4.49 g}$ of carbon and $\text{9.21 g}$ of oxygen gas react, then use the known enthalpy change for this reaction to find $\Delta {H}_{\text{f}}$.

So, use the molar masses of elemental carbon and oxygen gas to determine how many moles of each are being mixed

4.49 color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011color(red)(cancel(color(black)("g")))) = "0.3738 moles C"

9.21 color(red)(cancel(color(black)("g"))) * "1 mole O"_2/(32.0color(red)(cancel(color(black)("g")))) = "0.2878 moles O"_2

Notice that the two reactants are consumed in a $1 : 1$ mole ratio, but that you have fewer moles of oxygen gas than you have of carbon.

This means that the oxygen gas will act as a limiting reagent, i.e. it will be completely consumed before all the moles of carbon get the chance to react.

Since the reaction produces carbon dioxide in a $1 : 1$ mole ratio with both reactants, you can say that you'll get

0.2878 color(red)(cancel(color(black)("moles O"_2))) * "1 mole CO"_2/(1color(red)(cancel(color(black)("mole O"_2)))) = "0.2878 moles CO"_2

So, you know that $\text{113.2 kJ}$ of heat are being released when $0.2878$ moles of carbon dioxide are formed. This means that the formation of one mole of carbon dioxide will give off

1 color(red)(cancel(color(black)("mole CO"_2))) * "113.2 kJ"/(0.2878color(red)(cancel(color(black)("moles CO"_2)))) = "393.3 kJ"

Now, heat given off corresponds to a negative enthalpy change of reaction. As a result,m the enthalpy of formation for carbon dioxide will be

DeltaH_"f" = color(green)(|bar(ul(color(white)(a/a)-"393 kJ mol"^(-1)color(white)(a/a)|)))

The answer is rounded to three sig figs.

It's worth noting that the listed value for carbon dioxide's standard enthalpy of formation, $\Delta {H}_{\text{f}}^{\circ}$, which is imply the enthalpy of formation measured at ${25}^{\circ} \text{C}$ and $\text{1 atm}$, is equal to

$\Delta {H}_{\text{f"^@ = - "393.5 kJ mol}}^{- 1}$

which confirms that the result is very accurate.

http://nshs-science.net/chemistry/common/pdf/R-standard_enthalpy_of_formation.pdf