Question

5.00 grams of sodium hydroxide was placed in 100.0 grams of water in a coffee cup calorimeter. The temperature of the water increased from 25.0 deg C to 35.2 deg C. Find the heat of solution (ΔHsoln) for sodium hydroxide in kJ/mol. (Specific heat of water = 4.184J/g degC)?

Answer 1

The heat of solution, ΔHsoln, of NaOH is -34.139 kJ/mol

Step-by-step explanation:

Here we have

Mass of sodium hydroxide = 5.00 g

Mass of water = 100.0 g

Initial temperature of water, T₁ = 25 °C

Final temperature of water, T₂ = 35.2 °C

Temperature rise of water, ΔT= T₂ - T₁ = 35.2 - 25 = 10.2 °C

Therefore, from the equation of heat capacity of a medium, we have

Q = m·c·ΔT

Where;

Q = Heat acquired by water

m = Mass of water present = 100.0 g

ΔT = 10.2 °C

c = Specific heat capacity of water = 4.184 J/(g·°C)

Therefore, by plugging in the values we have;

Q = 100 g × 4.184 J/(g·°C) × 10.2 °C = -4267.68 J

Molar mass of NaOH = 39.997 g/mol

Therefore number of moles of NaOH present in 5.00 g = 5/39.997 = 0.125 moles

Which means that the heat of solution, ΔHsoln, of 0.125 moles of NaOH = -4267.68 J

Therefore, the heat of solution of 1 mole of NaOH =
`(-4267.68 J )/(0.125\, moles)`

= -34138.9 J/mole =

`-34138.9 \, J/mole * ( 1 \, kJ)/(1000\, J)` = -34.139 kJ/mol

The heat of solution, ΔHsoln, of NaOH = -34.139 kJ/mol.