Question

3. The dissolution of NaCl has a standard enthalpy of +3.87 kJ mol-1 at 25.0ºC and 1bar. Salt water has 35g/kg salt, and the heat capacity of sea water is 3993 J kg−1 K−1 at 25.0ºC.

a. (20 pts) How much would water heat up or cool down by adding enough NaCl to make sea water? b. (20 pts) How much heat would be released by making 1.00kg of sea water?

c. (20 pts) Assuming that 1.00kg of sea water equals 1.00L in volume, how much osmotic work would be required to pump this water to a final salt concentration of 10mM at 25.0ºC?

Show work and ill rate!

Answer 1

dH_diss = positive, i.e., the reaction is endothermic. Sea water will cool down when the salt is dissolved.

Amount of salt in 1kg sea water = 35 g = 35 g/58.5 g.mol^-1

= 0.598 moles

Q = - msdT           where, Q = heat change, m = mass of water, s = specific heat of water, dT = change of                                              temperature

0.598 mole * (+3.87 kJ mol-1) = - 1kg * (3993 J kg−1 K−1) * dT

or, dT = -0.58 K

or, T₂ - 298 K = -0.58 K

or, = 297.42 K

So water will cool down to 297.42 K by adding enough NaCl to make sea water.

b. As heat of dissolution of NaCl to water is positive, heat will be absorbed. Amount of heat absorbed

= 0.598 mole * 3.87 kJ mol^-1

= 2.314 kJ

c. Concentration of sea water = 0.598 M [as 0.598 moles of salt is dissolved in 1 L of sea water.]

Final volume of sea water when fresh water is pumped to 1 L of sea water to decreases its concentration to 10 mM or 0.010 M

= 1 L * 0.598 M / 0.010 M

= 59.8 L

Volume of water pumped

= (59.8 - 1 ) L

= 58.8 L

Osmotic pressure of sea water

= 0.598*2 moles.lit^-1 * 0.082 lit.atm.mol^-1.K^­1 * 298 K     [As salt is dissociated 100% in the solution                 Concentration is 0.598*2]

= 29.2 atm

Osmotic work required

= PdV

= 29.2 atm * 58.8 L

= 1716.96 L.atm

= 1716.96*101.325 J

= 173.971 kJ