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

Delta H is positive and Delta S is negative

Step-by-step explanation:

The expression for the standard change in free energy is:

`\Delta G=\Delta H-T* \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* \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

Answer 2

ΔH is positive and ΔG is always positive

Step-by-step 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 .