Question

A reaction is non-spontaneous at any temperature whenDelta H is negative and Delta S is positiveDelta H is positive and Delta S is negativeDelta H is negative and Delta S is negativeDelta H is positive and Delta S is positive

Answer 1

Answer:

ΔH is positive and  ΔG is always positive

Explanation:

According the equation of Gibb's free energy -

∆G = ∆H -T∆S

∆G = is the change in gibb's free energy

∆H = is the change in enthalpy

T = temperature

∆S = is the change in entropy .

And , the sign of the  ΔG , determines whether the reaction is Spontaneous or non Spontaneous or at equilibrium ,

i.e. ,

if

• ΔG < 0 , the reaction is Spontaneous

• ΔG > 0 , the reaction is non Spontaneous

• ΔG = 0 , the reaction is at equilibrium

From ,

∆G = ∆H -T∆S

ΔG > 0 , the reaction is non Spontaneous

Hence ,

To make the above equation , positive ,

The value of ∆H is also positive .

Answer 2

Answer:

Delta H is positive and Delta S is negative

Explanation:

The expression for the standard change in free energy is:

$\Delta G=\Delta H-T\times \Delta S$

Where,  

$\Delta G$ is the change in the Gibbs free energy.

T is the absolute temperature. (T in kelvins)

$\Delta H$ is the enthalpy change of the reaction.

$\Delta S$ is the change in entropy.

For a reaction to be non-spontaneous, $\Delta G>0$

So,

$\Delta H-T\times \Delta S>0$

To be the above equation satisfied, the conditions are:-

$\Delta H>0$ and $\Delta S<0$

Thus, answer is:- Delta H is positive and Delta S is negative