Question

QUESTION 1

CH₃OH _(l)  →  CH₃OH _(g)

Although the ΔH^o and ΔS^o values do change slightly with temperature, assumethat the change is not large enough to significantly affect the outcome of the following calculation.

Use the ΔH^o and ΔS^o values to determine the temperature (in ^oC) at which this phase change occurs (find the boiling point).

QUESTION 2

N_{2 (g)} + 3 H_{2 (g)}  →  2 NH_{3 (g)}

Although the ΔH^o and ΔS^o values do change slightly with temperature, assume that the change is not large enough to significantly affect the outcome of the following calculation.

Use the ΔH^o and ΔS^o values to determine the temperature (in K) at which this reaction will become nonspontaneous.

QUESTION 3

Calculate the Entropy change (in J/K) that occurs when 25.0 grams H₂O_(l)vaporizes at 100^oC.

ΔH_vaporization = 40.67 kJ/mole

Answer 1

1) CH3OH (l) → CH3OH (g)

DH0rxn = dH0fgas - liquid

         = (-201.08)-(-239.03)

         = 37.95 kj/mol

DS0rxn = DS0gas - liquid

         = (239.7 - 127.23)
      
         = 112.47 j/mol.k

the process, to be spontaneous

DG = 0

DG0 = dH0-TDS0

     0 = (37.95*10^3)-(T*112.47)

T = boiling point = 337.42 k

           = 64.27 c

2)   N2 (g) + 3 H2 (g) → 2 NH3 (g)

   DH0rxn = (2*DH0NH3)-(dH0N2+3*DH0H2)

          = (2*-46.1)-(0+3*0) = -92.2 kj/mol

DS0rxn = (2*192.34)-(191.5+3*130.6) = -198.62 j/mol.k

DG0 > 0 (or) at least DG = 0 , for the process to be non-spontaneous.

0 = (-92.2*10^3)-(T*-198.62)

T = 464.2 K = 191.05 C

3) Entropy change(DS) = q/T

q = 40.67*(25/18) = 56.5 kj

T = 100 c = 373.15 k

DS = 56.5*10^3/373.15   = 151.41 j/mol.k