`DeltaH` and `DeltaS` for a reaction are `+30.558 kJ mol^(-1)` and `0.66 kJ mol^(-1)` at `1` atm pressure. The temperature at which free energy is equ

Question

`DeltaH` and `DeltaS` for a reaction are `+30.558 kJ mol^(-1)` and `0.66 kJ mol^(-1)` at `1` atm pressure. The temperature at which free energy is equ

`DeltaH` and `DeltaS` for a reaction are `+30.558 kJ mol^(-1)` and `0.66 kJ mol^(-1)` at `1` atm pressure. The temperature at which free energy is equal to zero and the nature of the reaction below this temperature are
A. `483 K`, spontaneous
B. `443 K`,non- spontaneous
C. `443 K`, spontaneous
D. `463 K`,non- spontaneous

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

Correct Answer - D
`DeltaG=DeltaH-TDeltaS=0`
`:. T=(DeltaH)/(DeltaS)=(+30.558kJmol^(-1))/(0.066 kJ K^(-1) mol^(-1))=463 K`
`DeltaG=DeltaH-TDeltaS`
`DeltaG=(30.558-Txx0.066) kJ mol^(-1)`
At `T lt 463 K`,
`0.066 T lt 463xx0.066`
`0.066 T lt 30.558`
`0 lt 30.558-0.066T`
`0 lt DeltaG`
Thus at `T lt 463 K`, `DeltaG gt ` i.e., process is non spontaneous