Question

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

A) How much heat is required to convert 26.5 g of ethanol at 34 ∘C to the vapor phase at 78 ∘C?

B) How much heat is required to convert 26.5 g of ethanol at -160 ∘C to the vapor phase at 78 ∘C?

Answer 1

We know that, heat required to change temperature of material is given as , q = m x C x T

Where, q is a heat absorbed or given out by material,  m is a mass of material, C is a specific heat capacity of material, T is a change in temperature of material.

Heat required to change in phase is given as, q= n x H

Where, n is a no. of moles of material, H is a enthalpy change.

Heating of ethanol from 34 ⁰ C to 78 ⁰ C involves following steps.

1) Heating of liquid ethanol from 34 ⁰ C to 78 ⁰ C.

2) Boiling of ethanol at 78 ⁰ C.

First calculate number of moles of ethanol.

We have, No of moles = Mass / Molar Mass

Here Mass of ethanol = 26.5 g

Molar Mass of C2H5OH = ( 2 x 12.01) + ( 6 x 1.0079) + ( 16.00) = 46.07 g / mol

No. of moles of ethanol = 26.5 g / 46.07 g / mol = 0.575 mol

Hence, total heat change is given as

q _total = ( m x C x T ) _liquid +   n x   H _{( Vap.)}

q _total = ( 26.5 g x 2.3 J / g ⁰ C x ( 78 ⁰ C - (34 ⁰ C) ) + 0.575 mol x 38560 J / mol

q _total = 2681.8 J + 22172.0 J

q _total = 24853.8 J = 24.85 kJ

PART B

Heating of ethanol from - 160 ⁰ C to 78 ⁰ C involves following steps.

1) ) Heating of solid ethanol from - 160 ⁰ C to -114 ⁰ C

2) Melting of solid ethanol

3) Heating of liquid ethanol from -114 ⁰ C to 78 ⁰ C.

4) Boiling of ethanol at 78 ⁰ C.

Hence, total heat change is given as

q _total = ( m x C x T ) _solid +  n x   H _{( fusion)} + ( m x C x T ) _liquid +   n x   H _{( Vap.)}

q _total = ( 26.5 g x 0.97 J / g K x ( -114 ⁰ C - ( - 160 ⁰ C) ) + 0.575 mol x 5020 J/mol + ( 26.5 g x 2.3 J / g ⁰ C x ( 78 ⁰ C - (-114 ⁰ C) ) + 0.575 mol x 38560 J / mol

q _total = 1182.4 J + 2886.5 J + 11702.4 J + 22172 J

q _total = 37943.3 J = 37.94 kJ