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For a given reaction, DeltaH=35.5 KJ "mo...

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`

Text Solution

Verified by Experts

The correct Answer is:
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`
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Knowledge Check

  • For a given reaction, DeltaH = 35.5 kJ "mol"^(-1) and DeltaS = 83.6 J K^(-1) "mol"^(-1) . The reaction is spontaneous at (assume that DeltaH and DeltaS do not vary with temperature.

    A
    `T gt 425K`
    B
    all temperatures
    C
    `T lt 298K`
    D
    `T lt 425K`

  • 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 do not vary with temperature)

    A
    `T gt 298 K`
    B
    `T lt 425 K`
    C
    `T gt 425 K`
    D
    All temperatures

  • 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 do not very with temperature)

    A
    `T gt 425 K`
    B
    All temperatures
    C
    `T gt 298 K`
    D
    `T lt 425 K`

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