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Thermodynamics

Question
CBSEENCH11006295

(i) What are the limitations of criterion for randomness?
(ii) Calculate the standard free energy of formation of straight H subscript 2 straight O subscript 2.The free change for the reaction.
straight H subscript 2 straight O subscript 2 space space rightwards arrow space space space straight H subscript 2 straight O space plus space 1 half straight O subscript 2 space space is
space space space space space space space space increment straight G to the power of 0 space equals space minus 125.10 space kJ space and
increment straight G subscript straight f superscript 0 left parenthesis straight H subscript 2 straight O right parenthesis space equals space minus 228.4 space kJ space mol to the power of negative 1 end exponent

Solution
(i) Processes like solidification of a liquid and liquefaction of a gas are accompanied by a decrease in randomness (entropy). If randomness factor were the only criterion of the spontaneity of a reaction, then these processes would not have been feasible. In other words, the tendency to acquire maximum randomness (entropy) is a factor which determines the spontaneity of a reaction, but it is not the sole criterion.
(ii) We know that,
increment subscript straight f straight G degree space equals space sum from blank to blank of triangle straight G subscript Products superscript 0 space space minus space straight G subscript reactants superscript 0
equals space open square brackets increment subscript straight f straight G subscript left parenthesis straight H subscript 2 straight O right parenthesis end subscript superscript 0 space plus space 1 half increment straight G subscript fO subscript 2 end subscript superscript 0 space straight O subscript 2 superscript 0 close square brackets space minus space increment subscript straight f straight G to the power of 0 left parenthesis straight H subscript 2 straight O subscript 2 right parenthesis
Substituting the values, we have,
   negative 125.10 space equals space open square brackets negative 228.4 space plus space 1 half cross times space zero close square brackets space minus space increment subscript straight f straight G subscript left parenthesis straight H subscript 2 straight O subscript 2 right parenthesis end subscript superscript 0
or space space space space space increment subscript straight f straight G subscript left parenthesis straight H subscript 2 straight O subscript 2 right parenthesis end subscript superscript 0 space equals space minus 228.4 space plus space 125.10 space equals space minus 103.30 space kJ space mol to the power of negative 1 end exponent.