For a given reaction, `DeltaH=35.5 KJ "mol"^(-1)` and `DeltaS=83.6 JK^(-1) "mol"^(-1)`. The reaction is spontaneous at: (Assume that `DeltaH and delta

Question

For a given reaction, `DeltaH=35.5 KJ "mol"^(-1)` and `DeltaS=83.6 JK^(-1) "mol"^(-1)`. The reaction is spontaneous at: (Assume that `DeltaH and delta

For a given reaction, `DeltaH=35.5 KJ "mol"^(-1)` and `DeltaS=83.6 JK^(-1) "mol"^(-1)`. The reaction is spontaneous at: (Assume that `DeltaH and deltaS` so not vary with temperature)
A. `T lt 425 K`
B. `T gt 425 K`
C. all temperature
D. `T gt 298 K`

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

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

Correct Answer - B
Accoding to Gibbs-Helmholtz equation,
Gibbs energy `(DeltaG)=DeltaH-TDeltaS`
Where , `DeltaH`=Enthalpy change
`DeltaS` =Entropy change
T=Temperature
For a reaction to be spontaneous
`DeltaG lt 0`
`therefore` Gibbs -Helmholtz equation becomes,
`DeltaG=DeltaH-TDeltaS lt 0`
or , `DeltaH lt TDeltaS`
or, `T gt (DeltaH)/(DeltaS)=(35.5 KJ "mol"^(-1))/(83.6 JK^(-1) "mol"^(-1))=(355xx1000)/(83.6)`
`=425 K`
`T gt 425 K`