Find the enthalpy change per mole of sodium when sodium reacts with water. 13 grams of sodium reacts with 247 cm3 of water, producing a temperature change from 298 k to 339.7 k. the specific heat capacity of water is 4.18 j/k g.

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

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