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For a reaction, A hArr P, the plots of [...

For a reaction, `A hArr P`, the plots of [A[ and [P] with time at temperature `T_(1) and T_2` are given ahead:
.
If `T_(1) gt T_(1)`, the correct statement(s) is (are):
(Assume `DeltaH^(@)` and `DeltaS^(@)` are independent of temperature and ratio of ln K at `T_(1)` to ln K at `T_(2)` as greater than `(T_(2))/(T_(1))`. Here, H,S,G and K are enthalpy, entropy, Gibbs energy and equilibrium constant, respectively.

A

`DeltaH^(@) lt 0, DeltaS^(@) lt0`

B

`DeltaG^(@) lt 0, DeltaH^(@) gt0`

C

`DeltaG^(@) lt 0, DeltaS^(@) lt 0`

D

`DeltaG^(@) lt 0, DeltaS^(@) gt 0`

Text Solution

Verified by Experts

The correct Answer is:
A, C
Given that:
`("ln "K_(1))/("ln "K_(2)) gt (T_(2))/(T_(1))" "(Here" "T_(2)gtT_(1))`
`T_(1)" ln "K_(1) gt T_(2)lnK_(2)`
`RT_(1)lnK_(1)gtRT_(2)lnK_(2)` (It shows that `DeltaG^(@)lt0)`
`(-DeltaH^(@)-T_(1)DeltaS^(@))gt - (DeltaG^(@)-T_(2)DeltaS^(@))`
`-DeltaH^(@)+T_(1)DeltaS^(@) gt -DeltaH^(@)+T_(2)DeltaS^(@)`
Above expression will be correct ony when `(DeltaH^(@) lt 0)`
`T_(1)DeltaS^(@) gt T_(2)DeltaS^(@)`
Given that `T_(2) gt T_(1)`
`thereforeDeltaS^(@) lt0`
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