For a particular reversible reaciton at temperature `T, DeltaH` and `DeltaS` were found to be both `+ve`. If `T_e` is the temperature at equilibrium, the reaciton would be spontaneous when :

Options

A. `T=T_(e)`
B. `T_(e) gt T`
C. `T_(e) lt T`
D. `T_(e)` is `5` times `T`.

Answer

C

Explanation

`DektaG=DeltaH-TDeltaS`
At equilibrium , `DeltaG=0`
For a reaction to be spontaneous `DeltaG` should be negative, therefore, `T` should be greater than `T_(e)`.