Charles Recitation 7 ?-> ::: Apps ? O fi ?Online Banking-ur https://learn.vccs.e

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Charles Recitation 7 ?-> ::: Apps ? O fi ?Online Banking-ur https://learn.vccs.edu/bbcswebdav/pid-114991363-dt-content-rid-243578636-2/courses/GC297.CHM.112.WOLSU18/GC297.CHM.112.WOLSU18 = D Credit Card Accoun LG HOME-myCCC LG HOME-myNVCC * Late Nite Labs Course selected. Pic VCU-Central Authe?FOREVER CHORDS b 4. Thermodynamic data for Cgraphite) and Cidiamend) at 298 K is given in the table below. (KJ/mol) graphite) 0.0 Cdiamond)1.895 5.740 2.377 a) Calculate AH° and AS° and ?6" for the transformation of one mole of graphite to diamond at 298 K b) Is there a temperature at which this transformation will occur spontaneously at atmospheric pressure? Justify your answer.

Explanation / Answer

a) Given data:

?Hfo (graphite) = 0.0 kJ/mol

?Hfo (diamond) = 1.895 kJ/mol

?Sfo (diamond) = 2.377 J/mol K

?Sfo (graphite) = 5.740 J/mol K

For ?Ho calculation:-
?Ho = ?Hfo (diamond) - ?Hfo (graphite)
        = 1.895 -0
        = 1.895 kJ/mol
        = 1895 J/mol

For ?So calculation:-

?So = ?Sfo (diamond) - ?Sfo (graphite)
        = 2.377 -5.740
        = -3.363 J/mol K

For ?Go calculation:-

?Go = ?Ho - T?So
        = 1995 - 298(-3.363)
        = 2997 J/mol
        = 2.997 kJ/mol

b)
?Go = ?Ho - T?So
The reaction is said to be spontaneous if ?Go < 0
Therefore,
         ?Ho - T?So < 0
1895 - T(-3.363) < 0
1895 + 3.363 T < 0

For the reaction to be spontaneous the temperature must be negative, which is not possible.

Hence, there is no temperature at whih this transform will occur spontaneously.

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