Short Answer Type
Long Answer Type(a) If ∆G < 0 i.e. negative, the reaction will be spontaneous. ∆G can be negative if:
(i) Both energy and the entropy factors are favourable and may have any magnitude i.e. ∆H is –ve and T∆S are +ve.
(ii) When energy factor favours (∆H= –ve), but entropy factor opposes (T∆S = –ve), but ∆H ( –ve) > T∆S (–ve).
(iii) When energy factor is not favouring (∆H = +ve) but entropy factor favours (T∆S = +ve). But T∆S (+ve) > ∆H (+ve).
(b) ∆G = 0, the reaction is in an equilibrium state and thus there is no net change in either direction.
This happens when one of the factors is favourable and the other is opposite but they are equal in magnitude.
(c) ∆G > 0 i.e. +ve, the reaction will not be spontaneous, ∆G can be positive if:
(i) both the factors oppose i.e. ∆H is +ve and T∆S is –ve,
(ii) both the factors i.e. ∆H and T∆S have –ve sign and T∆S > ∆H,
(iii) both the factors i.e. ∆H and T∆S have a +ve sign and ∆H > T∆S.
Short Answer TypeLong Answer TypeWhat is free energy change? Show that the change in free energy is equal to useful work done.
or
prove that –∆G = w(useful work)
Short Answer TypePredict the enthalpy change, free energy change and entropy change when ammonium chloride is dissolved in water and the solution becomes colder.
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, ice and water are in equilibrium and 
for the process 
Calculate 
for the conversion of ice to liquid water.
For the reaction
calculate 
at 700K when enthalpy and entropy changes are -113 kJ mol-1 and -145 JK-1 mol-1 respectively.
Long Answer TypeFrom the following values of ∆H and ∆S, decide whether or not these reactions will be spontaneous at 298 K:
Reaction A:
∆H = – 10.5 X 103 J mol–1
∆S = + 31 JK–1 mol–1
Reaction B:
∆H = – 11.7 X 103 J mol–1 ;
∆S = –105 jK–1 mol–1.
Switch
Calculate its melting point.
