Question

The signs of \[\Delta H\] , \[\Delta S\] and \[\Delta G\] for a non-spontaneous reaction at all temperatures would be:
A. \[ + ,{\text{ }} + ,{\text{ }} - \]
B. \[ + ,{\text{ }} - ,{\text{ }} + \]
C. \[ - ,{\text{ }} - ,{\text{ }} - \]
D. \[ + ,{\text{ }} + ,{\text{ }} + \]

Answer 1

Hint: A non-spontaneous reaction is the reaction which is not feasible or continuous and needs an external energy to proceed. The change in enthalpy will be negative as the energy is absorbed in the reaction, the change in entropy is negative and the change in Gibbs free energy will be positive.

Complete step by step answer:
As we know that, for a non-spontaneous reaction, the change in Gibbs free energy is positive and for a spontaneous reaction, it is negative. The mathematical expression from which we can predict the signs of these three thermodynamic terms is:
\[\Delta G = \Delta H - T\Delta S\]
The key point for a reaction to be non-spontaneous is that it will have a positive change in Gibbs free energy (\[\Delta G\]) what so ever be the change in enthalpy (\[\Delta H\]) or the change in entropy (\[\Delta S\]) may be. But, from the above equation we can deduce that if the values of change in enthalpy and change in entropy are positive and negative respectively, then the change in Gibbs free energy will remain positive.
$
  \Delta G = + \Delta H - T( - \Delta S) \\
  \Delta G = + ve \\
\ $
Thus, the correct option is B. \[ + ,{\text{ }} - ,{\text{ }} + \]

Note:
The spontaneous process is also known as a non-reversible process and proceeds in the forward direction because of a negative value of change in Gibbs free energy. It is not dependent on the values of the change in the enthalpy and the change in entropy solely.