Question

When a 3.23 g sample of solid sodium hydroxide was dissolved in a calorimeter in 100.0 g of water, the temperature rose from 23.9 °C to 32.0 °C. Calculate ΔH (in kJ/mol NaOH) for the following solution process: NaOH(s) →Na+(aq)+ OH−(aq)

Answer 1

Answer:

-41,9kJ/mol NaOH

Explanation:

For the solution process:

NaOH(s) →Na⁺(aq) + OH⁻(aq)

The released heat is:

Q = -C×m×ΔT

Where Q is the released heat, C is specific heat of the solution (4,18J/g°C), m is the mass of water (100,0g) and ΔT is (32,0°C-23,9°C)

Replacing:

Q = -3385,8J

This heat is released per 3,23g of NaOH. Now, the heat released (ΔH) per mole of NaOH is:

$\frac{-3385,8J}{3,23gNaOH} *\frac{40g}{1mol}$= -41929J/molNaOH ≡

-41,9kJ/mol NaOH

I hope it helps!