Question

Ethanol (C2H5OH) melts at -114 degrees C and boils at 78 degrees C. The enthalpy of fusion of ehtinal is 5.02 kj/mol, and its enthalpy of vaporization is 38.56 kj/mol. The specific heat of solid nd liquid ethanol are 0.97 and 2.3 J/g-K, repectively.

(A) How much heat is required to convert 42.0 g of ethanol at 35 degrees C to the vapor phase at 78 degrees C?

(B) How much heat i required to convert the same amount of ethanol at -155 degrees C to the vapor phase at 78 degrees C?

(if you could show each and every step that would be awesome because my teacher was horrible last year and did't show us how to do this very well so I am pretty lost :( )

Answer 1

(A) Break down the process into the following subprocesses;

i) Ethanol (l, 35⁰C) ------> Ethanol (l, 78⁰C)

ii) Ethanol (l, 78⁰C) ------> Ethanol (v, 78⁰C)

Heat involved in step (i) = (mass of ethanol)*(specific heat of ethanol, l)*(change in temperature) = (42.0 g)*(2.3 J/g.K)*(78 – 35) K (change in temperature is the same in both Celsius and Kelvin scales) = 4153.8 J.

Molar mass of ethanol, C₂H₆O = (2*12 + 6*1 + 1*16) g/mol = 46 g/mol; therefore, mole(s) of ethanol corresponding to 42.0 g = (42.0 g)/(46.0 g/mol) = 0.9130 mole.

Heat involved in step (ii) = (moles of ethanol)*(enthalpy of vaporization of ethanol) = (0.9130 mole)*(38.56 kJ/mol) = 35.20528 kJ.

The total heat required is the sum of the heats involved in steps (i) and (ii); therefore, the total heat required = (4153.8 J) + (35.20528 kJ) = (4153.8 J)*(1 kJ/1000 J) + (35.20528 kJ) = (4.1538 kJ) + (35.20528 kJ) = 39.35908 kJ ≈ 39.360 kJ (ans).

(B) This process will involve a few additional processes.

(i) Ethanol (s, -155⁰C) ------> Ethanol (s, -114⁰C)

(ii) Ethanol (s, -114⁰C) -------> Ethanol (l, -114⁰C)

(iii) Ethanol (l, -114⁰C) ------> Ethanol (l, 78⁰C)

(iv) Ethanol (l, 78⁰C) ------> Ethanol (v, 78⁰C)

Heat involved in step (i) = (42.0 g)*(0.97 J/g.K)*[(-114⁰C) – (-155⁰C)] = 1670.34 J.

Heat involved in step (ii) = (moles of ethanol)*(enthalpy of fusion of ethanol) = (0.9130 mole)*(5.02 kJ/mol) = 4.58326 kJ.

Heat involved in step (iii) = (42.0 g)*(2.3 J/g.K)*[(78⁰C) – (-114⁰C)] = 18547.2 J.

The heat involved in step (iv) is the same as in step (ii) in part A above.

The total heat involved = 1670.34 J + 4.58326 kJ + 18547.2 J + 35.20528 kJ = (1670.34 J)*(1 kJ/1000 J) + 4.58326 kJ + (18547.2 J)*(1 kJ/1000 J) + 35.20528 kJ = 60.00618 kJ ≈ 60.006 kJ (ans).