Question

When 16.9 g KOH is dissolved in 90.8 g of water in a coffee-cup calorimeter, the temperature rises from 18.5 °C to 34.27 °C. What is the enthalpy change per gram (in J/g) of KOH dissolved in the water? Assume that the solution has a specific heat capacity of 4.18 J/g×K. Water has a density of 1.00 g/ml. Be sure to enter the correct sign (+/-). Enter to 1 decimal place.

Answer 1

Answer:

-431.5 J/g

Explanation:

Mass of solution = Mass of solute + mass of solvent

Solute is KOH while solvent is water.

Mass of KOH = 16.9 g

Mass of water = 90.8 g

Mass of solution = 16.9 + 90.8

                          = 107.7 g

Change in temperature (Δt) = 34.27 - 18.5

                                           = 16.2 °C

Heat required to raise the temperature of water is released by dissolving KOH.

Therefore,

Heat released by KOH = m × s× Δt

                                    = 107.7 × 4.18 × 16.2

                                    = 7293 J

Heat released by per g KOH = 7293 J/16.9 g

                                             = 431.5 J/g

As heat is released therefore, enthalpy change would be negative.

Enthalpy change of KOH = -431.5 J/g

Answer 2

The heat of reaction is 23.7 kJ/mol.

Number of moles of solute =  16.9 g /56 g/mol = 0.3 moles

Temperature rise = 34.27 °C -  18.5 °C = 15.77 °C

Mass of solution = mass of solute + mass of solvent = 16.9 g + 90.8 g = 107.7 g

Volume of solution = 90.8 mL

Using;

H = mcdT

m = mass of solution

c = specific heat capacity

dt = Temperature rise

H = 107.7 g × 4.18 J/g×K × 15.77 °C

H = 7.1 kJ

Heat of reaction =  7.1 kJ/0.3 moles

Heat of reaction = 23.7 kJ/mol

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