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Thermodynamics

Question
CBSEENCH11006274

For the melting of ice at 25 degree straight C comma
               straight H subscript 2 straight O left parenthesis straight s right parenthesis space space space rightwards harpoon over leftwards harpoon space space straight H subscript 2 straight O left parenthesis l right parenthesis comma
the enthalpy of fusion is 6.97 space kJ space mol to the power of negative 1 end exponent and entropy of fusion is 25.4 JK to the power of negative 1 end exponent mol to the power of negative 1 end exponent. Calculate the free energy change and predict whether the melting of ice is spontaneous or not at this temperature. 

Solution

According to Gibb's Helmoltz equation
   increment straight G space equals space increment straight H space minus space straight T increment straight S space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis
 Here space space space space increment straight H space equals space 6.97 space kJ space mol to the power of negative 1 end exponent space equals space 6970 space straight J space mol to the power of negative 1 end exponent
space space space space space space space space space space space increment straight S space equals space 25.4 space JK to the power of negative 1 end exponent mol to the power of negative 1 end exponent
space space space space space space space space space space space space space space straight T space equals space 25 plus 273 space equals space 298 space straight K
Substituting the values in (1), we have,
increment straight G space equals space 6970 space straight J space mol to the power of negative 1 end exponent
space space space space space space space equals space minus left parenthesis 298 space straight K right parenthesis space cross times space left parenthesis 25.4 space JK to the power of negative 1 end exponent space mol to the power of negative 1 end exponent right parenthesis
space space space space space space space equals space 69700 space mol to the power of negative 1 end exponent space minus space 7569.2 space straight J space mol to the power of negative 1 end exponent
space space space space space space space equals space minus 599.2 space straight J space mol to the power of negative 1 end exponent space