Test: Free Energy Change & Spontaneity - NEET MCQ

Since the reaction is not reversible, equilibrium constant is zero.

∆H_Vap = 30kJ/mol and ∆S = 75 J/K
∆G = ∆H - T∆S
At eqm, ∆G = 0
Therefore, ∆H = T∆S
Or T = 30×10³/75 = 400 K

∆G = ∆H - T∆S
For opt (c), ∆G = 48000 - 400(135)
= 48000 - 54000
= -6000
∆G is -ve
Therefore reaction is spontaneous.

1/2X₂  + 3/2Y₂ ⟶XY₃,
ΔH= −30 kJ
ΔS_reaction = ∑ΔS_product−∑ΔS_reactant 
X₂ + 3Y₂ → 2XY₃
​ΔH=−60 kJ
ΔS_reaction = 2×50−3×40−1×60 =100−120−60=−80 JK⁻¹mol⁻¹
 ΔG=ΔH−TΔS=0
ΔH=TΔS
1000×(−60)=−80×T
T=750 K

∆G° = ∆H° - T∆S°
       = -54070 - 298 10
       = -57050
∆G° = -2.303 RT log₁₀k
-57050 = -2.303 8.314 298 log₁₀k
57050 = 5705 log₁₀k
log₁₀k = 10

ΔG=ΔH−nFT(dE/dT)_P
​and  ΔG=ΔH−TΔS
∴ΔS/nF=(dE/dT)_P
or  
96.5/2×96500=(dE/dT)_P
​∴(dE_cell/dT)P
​=1×10⁻³ / 2
=5×10⁻⁴VK⁻¹

Statement I: Every endothermic reaction is spontaneous if TΔS > ΔH.

Endothermic reactions have a positive ΔH (heat absorbed). For spontaneity, we use the Gibbs free energy equation:
ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta SΔG=ΔH−TΔS For a process to be spontaneous, ΔG<0\Delta G < 0ΔG<0. If TΔS>ΔHT\Delta S > \Delta HTΔS>ΔH, then ΔG\Delta GΔG becomes negative, indicating spontaneity. Hence, Statement I is correct.

Statement II: Sign of ΔG is the true criterion for deciding spontaneity of a reaction.

This is a fundamental thermodynamic principle. The sign of ΔG\Delta GΔG indeed determines whether a reaction is spontaneous (ΔG<0\Delta G < 0ΔG<0) or non-spontaneous (ΔG>0\Delta G > 0ΔG>0). Hence, Statement II is also correct.

Correct Answer: Since both statements are correct and the second statement explains the first, the correct option is:

A: Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I.

The entropy of a substance increases on going from the liquid to the vapour state at any temperature. It is true. As randomness increases, entropy also increases.
∆G = ∆H - T∆S. So for spontaneous reaction, all 3(T, ∆H and ∆S) are needed.
An exothermic reaction will always be spontaneous. THis is false
Reactions with a positive (ΔH° and ΔS°) can never be product favoured. False, for a larger value of T, ∆G might be negative. 
If ΔG° for a reaction is negative, the reaction will have an equilibrium constant greater than one. True, for ΔG° less than 0, reaction is spontaneous and so, ith=s eqm constant will have value greater than 1.

According to me, correct options are b and d.
For a reversible reaction, ∆S°_(Universe) will be zero. ∆S°_(Universe) is greater than zero for an irreversible reaction. SInce the randomness decreases, ∆S°_(system) becomes less than zero and ∆S°(surrounding) becomes more than zero. BOth becomes equal with sign opposite and thus for reversible reaction, ∆S°_(universe) = ∆S°_(System) + ∆S°_{(surrounding)} = 0

We have ∆G = ∆H - T∆S
For spontaneous reaction ∆G<0 and vice versa,
∆G at 80° C:-
3213 - (80+273)8.73 = 141.9 kJ mol-1
As it comes positive, the conversion of rhombic and monoclinic is non- spontaneous at 80° C.In other words, rhombic is more stable than monoclinic at this temperature.
∆G at 110° C:-
3213 - (110+273)8.73 = -119.1 kJ mol-1
As it comes positive, the conversion of rhombic and monoclinic is spontaneous at 110° C.In other words, rhombic is less stable than monoclinic at this temperature.

We have ∆G = ∆H - T∆S
At eqm. ∆G=0,
T = ∆H/∆S
    = 3.213 kJ mol-1 / 8.71 JK-1 mol-1
    = 368.88 K

Thus, the correct alternate would indicate that the oxidation of iron is spontaneous at 298 K.

We have ΔG = ΔH – TΔS; by substituting ΔH = 19.07 kcal and ΔS = 90 cal/K, we get ΔG = 19.07 Kcal – 293(90 cal/K) = 19.07 Kcal – 26.37 Kcal = -7300 cal = -7.3 Kcal. The Gibbs free energy change is -7.3 Kcal.