Question

5.Calculate the entropy changes for the following processes:(a)Melting of one mole of tin at its melting point, 213 ᵒC; ΔHfus = 7.029 kJ/mol(b)Evaporation of one mole of liquid carbon dioxide at its boiling point, 216.6 K. ΔHvap = 15.326 kJ/mol

Answer 1

Answer:

a) ΔS = 14.46 J/k

b) ΔS = 70.76 J/k

Explanation:

The general formula to calculate the entropy change accompanied with a process is:

ΔS = ΔQ/T

where,

ΔS = entropy change for the process

ΔQ = Heat Transfer during the process

T = Absolute Temperature during the process

a)

In this case the heat transfer will be given as:

ΔQ = (ΔHfus)(N)

where,

ΔHfus = Molar Heat of Fusion of Tin = 7.029 KJ/mol

N = No. of moles of tin = 1 mol

Therefore,

ΔQ = (7.029 KJ/mol)(1 mol)

ΔQ = 7.029 KJ = 7029 J

and the absolute temperature is:

T = 213°C +273 = 486 k

using these values in the entropy formula, we get:

ΔS = 7029 J/486 k

ΔS = 14.46 J/k

b)

In this case the heat transfer will be given as:

ΔQ = (ΔHvap)(N)

where,

ΔHvap = Molar Heat of Vaporization of Carbon Dioxide = 15.326 KJ/mol

N = No. of moles of Carbon Dioxide = 1 mol

Therefore,

ΔQ = (15.326 KJ/mol)(1 mol)

ΔQ = 15.326 KJ = 15326 J

and the absolute temperature is:

T = 216.6 k

using these values in the entropy formula, we get:

ΔS = 15326 J/216.6 k

ΔS = 70.76 J/k