The free energy change for a reaction having △H=31400cal/mol,△S=32calK−1mol−1at727oC is:
A
- 600 cal
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B
-420 cal
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C
-850 cal
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D
600 cal
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Solution
The correct option is A - 600 cal Given: △H=31400cal/mol△S=32calK−1mol−1andT=727oC=727+273=1000K
The expression to calculate Gibbs free energy is given as △G=△H−T△S=31400−1000×32=−600cal
hence, the free energy for a reaction is −600cal