Question

For the reaction
`2NO(g)+O_(2)(g)to2NO_(2)g`
Calculate `DeltaG` at 700K when enthalpy and entropy changes are `-113.0kJ" "mol^(-1) and `-145J" "K^(-1)mol^(-1)` respectively.

Answer 1

We know that `DeltaG=DeltaH-TDeltaS`
Given, `DeltaH=-113kJ" "mol^(-1)`
`=-113000J" "mol^(-1)`
`DeltaS=-145J" "K^(-1)" "mol^(-1)`
`T=700K`
Substituting these values in the above equation,
we get
`DeltaG=-113000-700xx(-145)`
`=-11500J" "mol^(-1)`
`=-11.5kJ" "mol^(-1)`