QUESTION 1 CH 3 OH (l) CH 3 OH (g) Although the H o and S o values do change sli

Question

QUESTION 1

CH3OH (l)    CH3OH (g)

Although the Ho and So values do change slightly with temperature, assumethat the change is not large enough to significantly affect the outcome of the following calculation.

Use the Ho and So values to determine the temperature (in oC) at which this phase change occurs (find the boiling point).

QUESTION 2

N2 (g) + 3 H2 (g)    2 NH3 (g)

Although the Ho and So 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 Ho and So 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 100oC.

Hvaporization = 40.67 kJ/mole

Explanation / Answer

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

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