Thermodynamics Test - 11

MgO is the oxide of weak base and we know that heat of neutralisation of 1 equivalent of strong acid with strong base is - 57.33 kJ/mol.
⇒ With weak base, some heat is absorbed in dissociation of weak base.
⇒ The amount of heat released during the neutralisation of weak base with strong acid will be less than 57.33 kJ/mol.

As the energy available for muscular work on consumption of 180 g of glucose is 2880×25/100 = 720 kJ/mol
So, the energy available for muscular work after consuming 120 g of glucose = 720 × 120/180 = 480 kJ/mol
As 100 kJ is used for walking 1 km,
So, 480 kJ is sufficient for walking = 1 × 480/100 = 4.8 km

The standard molar enthalpy of formation of element in its standard state is taken as zero.
Since the standard state of Cl₂ is gaseous, its ∆_fH^o = 0.
However, Br₂ is also an element, but its standard state is liquid.

A process is spontaneous if ∆G (Gibbs free energy) calculated by the Gibbs Helmholtz equation is negative.

∆G = ∆H - T∆S

The conditions favourable for spontaneity of a process are: ∆H is -ve (process is exothermic) and ∆S is +ve (there is an increase in the disorder).
A process for which both the conditions are favourable will always be spontaneous.

Both (i) and (ii) are spontaneous.
(i) H₂O(l) → H₂O(g)
At 25^oC, evaporation of water will take place spontaneously as the relative humidity is low.
The reaction involves an increase in entropy.
(ii) H₂O(s) → H₂O(l)
Melting of ice is spontaneous at 0^oC or any temperature above that.

S₂ (s) + 2O₂ (g) → 2SO₂ (g); ∆G = - 544 kJ …….. (i)
2Zn (s) + S₂ (s) → 2ZnS (s); ∆G = - 293 kJ ………(ii)
2Zn (s) + O₂ (g) → 2ZnO (s); ∆G = - 480 kJ ……...(iii)
Add (i) and (iii) and subtract (ii) from it, we get
∆G = - 731 kJ