The answer is
-79 kJ/mol
The explanation:
The reaction equation is:
2Na + 2H2O 2NaOH + H2
when ΔHrxn= -q
and q=cmΔT
where:
-q is amount of heat absorbed/released
-and c is specific heat of solution (in such calculation an assumption is made that c solution = c water)
- and m is mass of solution
- and ΔT is temperature change
1- now we need to Find mass of solution:
m solution = m(H2O)+ m(Na)-m(H2)m(H2O)
m H2O= V(H2O)*ρ(H2O)
= 247 cm3*1g/cm3 = 247 gm
and m(Na)= 13 g
2- now we need to Find m(H2):
n(Na)(moles Na)= mass/Molar mass = 13g/23 g/mol = 0.57 mol.
According to equation mole ratio n(Na):n(H2)= 2:1,
then n(H2)=n(Na)/2
=0.57/2=0.29 mol.
∴m(H2)= moles * molar mass
=0.29 mol *2g/mol
= 0.58 g
∴m solution= 247 g + 13 g – 0.58 g = 259.42 g
and when q=4.18 J/K g* 259.42 g *(339.7 K – 298 K)= 45218 J
The temperature of solution increased because heat was absorbed by the solution (q>0).
Then ΔHrxn= -q = -45218 J per 0.57 mol of Na
3- now we need to Find ΔHrxn per 1 mole of Na
ΔHrxn = -45218J/0.57 mol = -79331 J/mol ≈ -79 kJ/mol