Question

Which of the following describes a system that CANNOT be thermodynamically favored?

A.) ΔS is negative and ΔH is positive
B.) ΔS is positive and ΔH is positive
C.) ΔS is negative and ΔH is negative
D.) ΔS is positive and ΔH is negative

Answer 1

Answer: d) ΔS is positive and ΔH is negative

Step-by-step explanation:

According to Gibb's equation:

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

`\Delta G` = Gibbs free energy

`\Delta H` = enthalpy change

`\Delta S` = entropy change

T = temperature in Kelvin

`\Delta G`= +ve, reaction is non spontaneous

`\Delta G`= -ve, reaction is spontaneous

`\Delta G`= 0, reaction is in equilibrium

a) ΔS is negative and ΔH is positive

`\Delta G=(-ve)-T(+ve)`

`\Delta G=(-ve)(-ve)=-ve`

Reaction is spontaneous at all temperatures.

b) ΔS is positive and ΔH is positive

`\Delta G=(+ve)-T(+ve)`

`\Delta G=(+ve)(-ve)=-ve`

Reaction is spontaneous at high temperatures.

c) ΔS is negative and ΔH is negative

`\Delta G=(-ve)-T(-ve)`

`\Delta G=(-ve)(+ve)=-ve`

Reaction is spontaneous at low temperatures.

d) ΔS is positive and ΔH is negative

`\Delta G=(+ve)-T(-ve)`

`\Delta G=(+ve)(+ve)=+ve`

Reaction is non spontaneous or thermodynamically unfavored at all temperatures.