For a particular reaction, `DeltaH=-38.3 kJ and DeltaS^(@)=-113JxxK^(-1).` This reaction is:

Question

For a particular reaction, `DeltaH=-38.3 kJ and DeltaS^(@)=-113JxxK^(-1).` This reaction is:

For a particular reaction, `DeltaH=-38.3 kJ and DeltaS^(@)=-113JxxK^(-1).` This reaction is:
A. spontaneous at all temperature
B. non-spontaneous at all temperature
C. spontaneous at temperature below `66^(@)C`
D. spontaneous at temperature above `66^(@)C`

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - D
`DeltaG=DeltaH-TDeltaS`
For spontaneous process, `DeltaG lt 0`
`(DeltaH)/(DeltaS) lt T`
`(-38.3xx1000)/(-113) lt T`
i.e., `T gt 338.93K`, i.e., `66^(@)C`

4 views Answered 2023-02-05 15:29