Question

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.

Answer 1

Answer on Question #79848, Chemistry/ General Chemistry

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.

Answer

2Na + 2H₂O → 2NaOH + H₂

ΔHₜₐₙ = -q

q = cm ΔT

where

- q is amount of heat absorbed/released

- c is specific heat of solution (in such calculation an assumption is made that c_solution = c_water)

- m is mass of solution

- ΔT is temperature change

Find mass of solution:

m_solution = m(H₂O) + m(Na) - m(H₂)

m(H₂O) = V(H₂O) × ρ(H₂O) = 247 cm³ × 1 g/cm³ = 247 g

m(Na) = 13 g

Find m(H₂):

n(Na) = m/M = 13 g / 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) = n × M = 0.29 mol × 2 g/mol = 0.58 g

m_solution = 247 g + 13 g - 0.58 g = 259.42 g

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 ΔHₜₐₙ = -q = -45218 J per 0.57 mol of Na

Find ΔHₜₐₙ per 1 mole of Na

ΔHₜₐₙ = -45218 J / 0.57 mol = -79331 J/mol ≈ -79 kJ/mol

Answer: -79 kJ/mol