Question

Calculate the w, q, delta U, delta H, delta S system of heating 2.4 L of methanol from 20 C to 85 C b) what would be the delta S surroundings and Delta G of the reversible condensation of methanol? is the condensation of methanol at 37 C a spontaneous process? please show formulas and tell me why youre using them because I'm lost

Answer 1

Volume of methanol = 2.4 L = 2400 mL

Temperature difference = 85 - 20

= 65 oC

Density of methanol = 0.7913 g/ mL

Mass of methanol = Density * Volume

= 0.7913 * 2400

= 1899.12 g

Moles of methanol = 1899.12 / 32.04

= 59.27 moles

Molar heat capacity of methanol = 79.5 J/(mol K)

Heat absorbed, q = m*C*T

= 59.27 * 79.5 * 65

q = 306.27 kJ

H = q / moles

= 306.27 / 59.27

H = 5.167 kJ/ mol

Since we are at constant pressure and constant volume,

w = 0

U = q + w

U = 306.27 kJ

S = q / T

= 306.27 / (85 + 273)

= 306.27 / 358

= 0.855 kJ / K

S = 855 J/K

(b). For reversible condensation of methanol -

q = - 306.27 kJ

S(surrounding) = - q / T

= - (- 306.27) / (20 + 273)

= 306.27 / 293

S_surr = 1.045 kJ / K

G = H - TS

= - 5.167 - 293*(- 1.045)

= - 5.167 + 306.185

G = 301.018 kJ/mol

At 37 oC = 310 K

S = - 306.27 / 310

= - 987.96 J/ K

Since S is negative, process will be spontaneous.