Calculate the standard free energy change for the following reaction at `27^@C`.
`H_2(g) + I_2(g) to 2HI(g) , DeltaH^@ = + 51.9 kJ`
[Given : `DeltaS_(H_2)^@ = 130.6 JK^(-1) mol^(-1)`
`DeltaS_(I_2)^@ = 116.7 JK^(-1) mol^(-1)`
`DeltaS_(HI)^@ = 206.3 JK^(-1) mol^(-1)]`.
Predict whether the reaction is feasible at 27°C or not.

Explanation

The given reaction is `H_(2)(g) + I_(2)(g) rarr 2HI (g), Delta H^(@)= +51.9kJ`
The change in entropy for this reaction is: `DletaS^(@)= SigmaDeltaS_(("products"))^(@)- SgimaDeltaS_(("reactants"))^(@)`
`={2 xx DeltaS^(@) [HI (g)]} - {DeltaS^(@)[ H_(2) (g)] + DeltaS^(@) [I_(2) (g)]}`
`=(2 xx 206.3) - (130.6+ 116.7)`
`=165.3 JK^(-1)`
The standard free energy for the reaction is given by
`DeltaG^(@)= DletaH^(@)+ T DeltaS^(@)`
`=+51.9 xx 1000-(273 + 27) xx 165.3`
`=51900-49590`
= +2310J
As `DeltaG^(@)` is positive at `27^(@)C`, the reaction is not feasible at this temperature.