For a reaction `A to P`, the plots of [A] and [P] with time at temperatures `T_(1)` and `T_(2)` are given below.

If `T_(1) gt T_(2)`, the correct statements is (are):
(Assume `DeltaH^(Theta)` and `DeltaS^(Theta)` are independent of temperature and ratio of ln K at `T_(1)` to ln K at `T_(2)` is greater than `T_(2)/T_(1)`. Here H,S,G and K are enthalpy, entropy, Gibbs energy and equilibrium constant, respectively).

Options

A. `DeltaH^(Theta) lt 0, DeltaS^(Theta) lt 0`
B. `DeltaG^(Theta) lt 0, DeltaH^(Theta) gt 0`
C. `DeltaG^(Theta) lt 0, DeltaS^(Theta) lt 0`
D. `DeltaG^(Theta) lt 0, DeltaS^(Theta) gt 0`

Answer

ac

Explanation

As the concentration of product decreases by increasing temperature `DeltaH lt 0` as given `(ln (K_(1))/(ln (K_(2)) gt T_(2)/T_(1)))` Also, `[P]/[A]_(eq) gt 1`
`therefore K_(eq) gt 1`
As we know
`DeltaG^(@) = -RT ln K`
`(DeltaG^(@))_(1) = -RT_(1)ln K_(1)` `("as "K_(eq) gt 1, "T can't be" -ve)`
`(DeltaG^(@))_(2) = -RT_(2) ln K_(2) , therefore DeltaG^(@) lt 0`
`DeltaG_(1)^(@) = RT_(1) ln K_(1)`
`DeltaG_(2)^(@) = RT_(2) ln K_(2)`
`(DeltaG_(1)^(@))/(DeltaG_(1)^(@)) =T_(1)/T_(2) (ln K_(1))/(ln K_(2)) rArr (DeltaG_(1)^(@))/(DeltaG_(2)^(@)) = ((ln K_(1))/(ln K_(2)))/(T_(2)/T_(1))`
`(DeltaG_(1)^(@))/(DeltaG_(2)^(@)) gt 1 [as (ln K_(1))/(ln K_(2)) gt T_(2)/T_(1)]`
Since, `DeltaG lt 0`
So, `DeltaG_(1)^(@) lt DeltaG_(2)^(@)` because `DeltaG^(@)` increases on increase in temperature
As, `DeltaG^(@) uarr` on increases Temp.
`DeltaG = DeltaH - T DeltaS` and `DeltaH^(@) lt 0`
So, `DeltaS^(@) lt 0`