Question

A table of enthalpies of solvation in water is given below for various substances at 298 K and 1 bar.

Which of these substances release heat when dissolved (exothermic) and which absorb heat when dissolved (endothermic)?

If you dissolved 5 mol of each of these in water at 298 K, how much heat would be absorbed or released? Assume the enthalpy of solvation is constant for all solute concentrations.

Calculate the temperature change if 1 L of liquid water at 298 K at 1 atm in an insulated container gained or lost the amount of heat calculated for each substance in b). Would the water freeze or boil in any of these cases? cP,ice=38.1 J K-1 mol-1, cP,liquid water=75.3 J K-1 mol-1.

Questions for you to think about (optional, no points). Which of these substances would you prefer to use in a hot pack? Cold pack? To melt the snow outside your house? Would there be any safety or environmental risks? Does NaCl seem like an odd choice for melting snow/ ice on the sidewalk?

	urea	NaOH	HCl	NH4Cl	CaCl2	NaCl
ΔH°dissolution, water [kJ/mol]	13.8	-44.51	-74.8	14.8	-81.7	3.9

Answer 1

Q.1: For exothermic reaction,  H^o_dissolution is negative.

Hence NaOH, HCl and CaCl₂ release heat when dissolved

For endothermic reaction,  H^o_dissolution is positive.

Hence urea, NH₄Cl, and NaCl absorb heat when dissolved.

Q.2: (i): Heat absorbed by 5 mole of urea = 13.8 kJ / mole * 5 mole = 69 kJ

(ii): Heat released by 5 mole NaOH = - 44.51 kJ/mol * 5 mole = - 222.55 kJ

(iii):Heat released by 5 mole HCl = - 74.8 kJ/mol * 5 mole = - 374 kJ

(iv): Heat absorbed by 5 mole of NH₄Cl = 14.8 kJ / mole * 5 mole = 74 kJ

(v):Heat released by 5 mole CaCl₂ = - 81.7 kJ/mol * 5 mole = - 408.5 kJ

(vi): Heat absorbed by 5 mole of NaCl = 3.9 kJ / mole * 5 mole = 19.5 kJ

Q.3:(i) Heat absorbed Q = 69 kJ * (1000 J / 1 kJ) = 69000 J

mass water = 1000 mL * (1g / 1mL) = 1000 g

=> Moles of water = 1000 g * (1 mol / 18.015 g) = 55.51 mole

Q = n*Cp*T

=> 69000 J = 55.51 mol * 75.3 J*K^-1mol^-1*T

=> T = 16.51 K (answer)

=> Tf - 298K = 16.51K

=> Tf = 298+16.51 = 314.51 K

Since 314.51 K is less than 373.15 K, it will not boil.

(ii): Heat released, Q = -222.55 kJ * (1000 J / 1 kJ) = - 222550 J

mass water = 1000 mL * (1g / 1mL) = 1000 g

=> Moles of water = 1000 g * (1 mol / 18.015 g) = 55.51 mole

Q = n*Cp*T

=> - 222550 J = 55.51 mol * 75.3 J*K^-1mol^-1*T

=> T = - 53.24 ^oC (answer)

=> Tf - 298K = - 53.24 K

=> Tf = 298 - 53.24 = 244.76 K

Since 244.76 K is less than the freezing point temperature 273.15, it will freeze.

(iii): Heat released, Q = - 374 kJ * (1000 J / 1 kJ) = - 374000 J

mass water = 1000 mL * (1g / 1mL) = 1000 g

=> Moles of water = 1000 g * (1 mol / 18.015 g) = 55.51 mole

Q = n*Cp*T

=> - 374000 J = 55.51 mol * 75.3 J*K^-1mol^-1*T

=> T = - 89.48 K (answer)

=> Tf - 298K = - 89.48 K

=> Tf = 298 - 89.48 = 208.52 K

Since 208.52 K is less than the freezing point temperature 273.15, it will freeze.

(iv): Heat absorbed Q = 74 kJ * (1000 J / 1 kJ) = 74000 J

mass water = 1000 mL * (1g / 1mL) = 1000 g

=> Moles of water = 1000 g * (1 mol / 18.015 g) = 55.51 mole

Q = n*Cp*T

=> 74000 J = 55.51 mol * 75.3 J*K^-1mol^-1*T

=> T = 17.7 K (answer)

=> Tf - 298K = 17.7 K

=> Tf = 298+17.7 = 315.7 K

Since 315.7 K is less than the boiling point temperature 373.15, it will not boil.

(v): Heat released, Q = - 408.5 kJ * (1000 J / 1 kJ) = - 408500 J

mass water = 1000 mL * (1g / 1mL) = 1000 g

=> Moles of water = 1000 g * (1 mol / 18.015 g) = 55.51 mole

Q = n*Cp*T

=> - 408500 J = 55.51 mol * 75.3 J*K^-1mol^-1*T

=> T = - 97.73 K (answer)

=> Tf - 298K = - 97.73 K

=> Tf = 298 - 97.73 = 200.27 K

Since 200.27 K is less than the freezing point temperature 273.15, it will freeze.

(vi): Heat absorbed Q = 19.5 kJ * (1000 J / 1 kJ) = 19500 J

mass water = 1000 mL * (1g / 1mL) = 1000 g

=> Moles of water = 1000 g * (1 mol / 18.015 g) = 55.51 mole

Q = n*Cp*T

=> 19500 J = 55.51 mol * 75.3 J*K^-1mol^-1*T

=> T = 4.67 ^oC (answer)

Tf = 302.67 K

This will not boil.