For the reaction:
`X_(2)O_(4)(l)rarr2XO_(2)(g)`
`DeltaU=2.1 cal , DeltaS =20 "cal" K^(-1) "at" 300 K`
Hence `DeltaG` is

Options

A. 2.7 kcal
B. `-2.7` kcal
C. 9.3 kcal
D. `-9.3` kcal

Answer

B

Explanation

The change in Gibbs free energy is given by
`DeltaG=DeltaH-TDeltaS`
where, `DeltaH`= change enthalpy of the reaction
`DeltaS` =change entropy of the reaction
Thus, in order to determine `DeltaG`, the values of `DeltaH` must be known. The value of `DeltaH` can be calculated by using equation
`DeltaH=DeltaU+Deltan_(g)RT" "...(i)`
where, `DeltaU`= change in internal energy using
`Deltan_(g)` =number of moles of gaseous products-number of moles of gaesous reactants
`=2-0=2`
R=gas constant =2 kcal
given, `DeltaU=2.1 Kcal`
`=2.1 xx10^(3)` cal [`because` 1 kcal `=10^(3)` cal ]
But putting the values in eq. (i) we get,
`therefore DeltaH=(2.1xx10^(3))+(2xx2xx300)`
=3300 cal
Hence, `DeltaG=DeltaH-TDeltaS`
`implies DeltaG=(3300)-(300 xx 20)`
`DeltaG=-2700` cal
`therefore DeltaG=-2.7` kcal