`Delta H^(0)` & `Delta S^(0)` for certain reaction are `-10kJ` & `-44J/K`, then `Delta G^(0)` is
`+3112J`, non -spontaneous
`-3112J`, spontaneous
`+31.12J`, non -spontaneous
`-31.12J`, spontaneous

Suppose that a reaction has DeltaH=-40kJ and DeltaS =- 50J//K . At what temperature range will it change from spontaneous to non-spontaneous?

If Delta G gt 0 , then the nature of the process will be (spontaneous/non-spontaneous)?

If Delta G gt 0, then the nature of the process will be (spontaneous/non-spontaneous)?

For a given reaction Delta G obtained was having positive sign convention. State whether the reaction was spontaneous or non-spontaneous.

For a given reaction Delta G obtained was having positive sign convention. State whether the reaction was spontaneous or non-spontaneous.

For a reaction R_(1), Delta G = x KJ mol^(-1) . For a reaction R_(2), Delta G = y KJ mol^(-1) . Reaction R_(1) is non-spontaneous but along with R_(2) it is spontaneous. This means that

A particular reaction at 27^(@)C for which DeltaH gt 0 and DeltaS gt 0 is found to be non-spontaneous. The reaction may proceed spontaneously if

If Delta H gt 0 and Delta S gt 0 , the reaction proceeds spontaneously when :-

A change in the free energy of a system at constant temperature and pressure will be: Delta_(sys)G = Delta_(sys)H -T Delta_(sys)S At constant temperature and pressure Delta_(sys) G lt 0 (spontaneous) Delta_(sys)G = 0 (equilibrium) Delta_(sys)G gt 0 (non-spontaneous) For a spontaneous reaction DeltaG , equilibrium K and E_(cell)^(Theta) will be, respectively

A process is non-spontaneous when DeltaS_("total") lt 0 .