When 0.560 g of Na(s) reacts with excess F_2(g) to form NaF(s), 13.8 kJ of heat is evolved at standard-slate conditions. What is the standard enthalpy of formation?

Dec 18, 2015

$\Delta {H}_{\text{f"^@ = -"567 kJ/mol}}$

Explanation:

Your starting point here will be the balanced chemical equation for this synthesis reaction

$2 {\text{Na"_text((s]) + "F"_text(2(g]) -> 2"NaF}}_{\textrm{\left(s\right]}}$

Now, the standard enthalpy of formation is always given for the formation of one mole of a substance. In this case, the chemical equation that describes the formation of one mole of sodium fluoride looks like this

${\text{Na"_text((s]) + 1/2"F"_text(2(g]) -> "NaF}}_{\textrm{\left(s\right]}}$

Now, notice that you have a $1 : 1$ mole ratio between sodium and sodium fluoride. This means that in order for the reaction to produce one mole of sodium fluoride, one mole of sodium must take part in the reaction.

You know that when $\text{0.560 g}$ of sodium react, the reaction gives off $\text{13.8 kJ}$ of heat. Use sodium's molar mass to determine how many moles of sodium would give off this much heat

0.560 color(red)(cancel(color(black)("g"))) * "1 mole Na"/(23.0color(red)(cancel(color(black)("g")))) = "0.02435 moles Na"

So, if $0.02435$ moles of sodium are needed to give off $\text{13.8 kJ}$ of heat, how much heat will be given off when one mole of sodium will react?

1 color(red)(cancel(color(black)("mole Na"))) * "13.8 kJ"/(0.02435color(red)(cancel(color(black)("moles Na")))) = "566.7 kJ"

Now, when heat is given off, the standard enthalpy change of formation carries a negative sign. This means that the enthalpy change of formation for sodium fluoride will be

DeltaH_"f"^@ = - color(green)("567 kJ/mol")

The answer is rounded to three sig figs.

The listed value for sodium fluoride's standard enthalpy of formation is $- \text{569 kJ/mol}$, so this is a very good result.

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