Question

By calculating delta S at each temperature, Determine if the melting of 1 mole of NaCl is spontaneous at 500 and 700 degrees Celsius.

Answer 1

The melting of 1 mole of NaCl is non spontaneous at 500 and 700 degrees Celsius.

How we calculate the ΔS (universe)?

ΔS (universe) for any reaction is the sum of the entropy of the system and entropy of the surrounding.

Given reaction is :

NaCl(s) ⇄ NaCl(l)

Entropy of the reaction system will be calculated as:

ΔS = S°[NaCl(l)] - S°[NaCl(s)]

Standard entropy of NaCl(l) = 95.06 J/mol.K

Standard entropy of NaCl(s) = 72.11 J/mol.K

ΔS = 95.06 - 72.11 = 22.95 J/mol.K

Entropy of the surrounding will be calculated as:

ΔS = - (ΔH°/T)

ΔH° for fusion = 27950 J/mol.K

At 500 degree celsius or 773K-

ΔS = -(27950/773) = -36.157 J/mol.K

At 700 degree celsius or 973K-

ΔS = -(27950/973) = -28.72 J/mol.K

ΔS (universe) at 500 degree celsius is:

ΔS (universe) = 22.95 - 36.157 = -13.207 J/mol.K

ΔS (universe) at 700 degree celsius is:

ΔS (universe) = 22.95 - 28.72 = -5.77 J/mol.K

In both reaction entropy of the reaction is less than zero means it is non spontaneous.

To know more about entropy, visit the below link:

https://brainly.com/question/419265