For a reaction at `25^(@)C` enthalpy change `(Delta H)` and entropy change `(Delta S)` are `-11.7xx10^(3)Jmol^(_1)` and `-105 J mol^(-1)K^(-1)` respectively. The reaction is :

Options

A. Spontaneous
B. Non spontaneous
C. At equilibrium
D. Can't say anything

Answer

B

Explanation

`Delta G = Delta H-T Delta S`
`=-11.7xx10^(3)-298xx(-105)`
`Delta G=+19590 J mol^(-1)`
`Delta G = +ve`, so it is non spontaneous process