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.

76.20 kJ

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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