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. `-9.3kcal`

Answer

b

Explanation

`DeltaH=DeltaU+Deltan_(g) RT`
Given `: DeltaU =2.1 kcal ,Deltan_(f) = 2-0=2`,
`R = 2 xx 10^(-3) kcal, T= 300K`
`:. DeltaH =2.1+ 2 xx 2xx 10^(-3) xx 300 = 3.3 kcal`
Again, `DeltaG= DeltaH - T DeltaS`
Given `: DeltaS =20 xx 10^(-3) kcal K^(-1)`
`DeltaH= 3.3 kcal` ( calculateabove)
`:. DeltaG = 3.3 - 300 xx (20 xx 10^(-3))`
`=3.3-6 = - 2.7 kcal`