Given: for ethylene oxide `DeltaH^@_f` and `DeltaS^@` are `-51kJmol^(-1), 243Jmol^(-1)K^(-1)` respectively and acetaldehyde `DeltaH^@_f` and `DeltaS^@` are `-166kJmol^(-1)` and `266 J mol^(-1) K^(-1)` respectively, then select the correct statements:

Options

A. `DeltaH^(@)` for the reaction is `-115 kJ mol^(-1)`
B. `DeltaS^(@)=0.0236 k J mol^(-1) K^(-1)`
C. Reaction is thermodynamically favourable
D. High temperature can favour formation of more product.

Answer

A, B, C, D

Explanation

`DeltaH^(@)=DeltaH_(f("Product"))^(@)-DeltaH_(f("Reactants"))^(@)`
`=-166-(-51)=-115 kJ mol^(-1)`
`DeltaS^(@)=266-243=23 J`
`DeltaH^(@)=-ve,DeltaS^(@)=+ve` and `DeltaG^(@)=DeltaH^(@)-TDeltaS^(@)` must be `-ve`. Therefore, the reaction will be thermodynamically favourable. for reaction having `-ve DeltaH^(@)` and `+ve`
`DeltaS^(@)`, all temperature conditions will be favourable.