For a particular reaction, AH = 81.95 kJ/mol and AS = 27.0 J / (mol-K) Calculate AG for this reaction at 298 K. AG = kJ/mol What can be said about the spontaneity of the reaction at 298 K? The system is spontaneous in the reverse direction O The system is at equilibrium O The system is spontaneous as written O O

The system is spontaneous in the reverse direction

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 the question,

∆H = 81.95 kJ/mol

(since, 1 KJ = 1000 J)

∆H = 81950 J/mol

∆S = 27.0 J / (mol-K)

The ∆G is calculated from the above formula -

∆G = ∆H - T∆S

∆G = (81950 J/mol) - [ (298 K) x (27.0 J / (mol·K)) ]

∆G = (81950 J/mol) - (8046 J/mol)

∆G = 73904 J/mol

∆G = 73.904 kJ/mol

Since ΔG > 0, the system is non spontaneous in the forward direction and hence it will be spontaneous in the reverse direction.