Question

when 0.550 g of na(s) reacts with excess f2(g) to form naf(s), 13.8 kj of heat is evolved at standard - state conditions. what is the standard enthalpy of formation (δhf°) of naf(s)?
o 567 kj/mol
o - 24.8 kj/mol
o - 7.8 kj/mol
o - 24.8 kj/mol
o - 567 kj/mol

Explanation:

Step1: Calculate the number of moles of Na

The molar mass of Na is \(M = 22.99\space g/mol\).
The number of moles \(n=\frac{m}{M}\), where \(m = 0.560\space g\).
So \(n=\frac{0.560\space g}{22.99\space g/mol}\approx0.02436\space mol\)

Step2: Relate the heat evolved to the enthalpy of formation

The reaction is \(2Na(s)+F_{2}(g)=2NaF(s)\).
If \(0.02436\space mol\) of \(Na\) reacts and \(13.8\space kJ\) of heat is evolved.
For \(2\space mol\) of \(Na\) (since the stoichiometric coefficient of \(Na\) in the formation reaction of \(NaF\) is \(2\)), let \(\Delta H_f^{\circ}\) be the standard enthalpy of formation of \(NaF\).
We set up a proportion: \(\frac{13.8\space kJ}{0.02436\space mol}=\frac{\Delta H_f^{\circ}}{1\space mol}\) (per \(2\space mol\) of \(Na\) in the reaction).
\(\Delta H_f^{\circ}=\frac{13.8\space kJ\times2}{0.02436\space mol}\approx - 567\space kJ/mol\) (negative because heat is evolved)

Answer:

\(\text{O}-567\space kJ/mol\)