Question

Sodium hydroxide is dissolved in enough water to make up a 20.0 mole% solution.

(a) If the NaOH and water are initially at 77°F (25°C), how much heat (Btu/lb product solution) must be removed for the solution also to be at 77°F. Assume the process is carried out at constant pressure, so that Q=ΔH, and use Table B.11 to evaluate ΔH^s.

(b) If the dissolution is done adiabatically, estimate the final temperature of the solution. Assume that the heat capacity of the solution is approximately that of pure liquid water.

(c) If the process of Part b were actually carried out, the final temperature would be less than the value calculated. Why? (Neglect errors caused by the assumptions of adiabatic dissolution and a solution heat capacity equal to that of pure water.)

Answer 1

a) given x_b= 0.20

x_b = n_b/n_b+n_w

0.20=  n_b/n_b+n_w

Rearrange

0.20n_b + 0.20n_w= n_b

0.20n_w = 0.80n_b

n_w/n_b = 0.80/0.20

n_w/n_b = r = 4.00

now from table B11, for r= 4.00

we have

Now 1 mol NaOH= 40 grams

and 1 gram = 0.00220462 lb

So 40 grams = 0.0022*40 lb = 0.088 lb NaOH

also 1 J = 0.00095 Btu

= -0.3608 Btu/lb NaOH

- ve sign only indicates that process is exothermic.

This is the total heat that will be resposnsible for temperature rise of the solution so, if the products to remain at 77 F the same amount (0.3608 Btu/lb) of heat must be removed.

b) Q= 0

=

+33.43 J/mol = 75.28 J/mol K * (T₂ - 298)

T₂ = (+33.43/75.28) + 298 = 298.44 K

C) if the process is carried out as in the b, the final temperature will be less that calulated.

Reason is that during actual process heat is transfred to surrounding of the mixing vessel. This is where part of heat will go to the surrounding.