Question

C2Cl3F3 has a normal boiling point of 47.6°C. The specific heats of and C2Cl3F3(g) are 0.91 J/gK and 0.67 J/gK, respectively. The heat of vaporization of the compound is 27.49 kJ/mol. The heat required to convert 50.0 g of the compound from the liquid at to the gas at 65.0°C is ________ kJ.

a. 1454

b. 8.46

c. 9.63

d. 29.78

e. 2301

Answer 1

The question is missing a value i.e. liquid at 10.0⁰C to the gas at 65.0⁰C

T_f = 47.6⁰C = (47.6 + 273 ) K= 320.6 K , T_i = 10.0⁰C =( 10.0 +273)K = 283 K

q = mcT

=> q= (50.0 g) x (0.91 J/gK) x (320.6 - 283) K

=> q = 1710.8 J is required to warm the liquid to its boiling point

(27.49 KJ/mol) x (50.0 g C₂Cl₃F₃ / (187.3756 g C₂Cl₃F₃/mol)

= 7.3355 KJ

= 7.3355 KJ x ( 1000 J / 1 KJ)

= 7335.5 J is required to vaporize the liquid.

Again,

T_f = 65.0⁰C = (65 +273) K = 338 K , T_i = 47.6⁰C =( 47.6 + 273) K = 320.6 K

q = mcT

=> q = (50.0 g) x (0.67 J/gK) x (338 - 320.6) K

=> q= 582.9 J is required to heat the vapor to 65.0°C

Now,

Total heat required will be---

1710.8 J + 7335.5 J + 582.9 J

= 9629.2 J

=9629 J x ( 1 KJ / 1000 J )

= 9.629 KJ

= 9.63 KJ

The required answer will be C) 9.63