Thermodynamics Test - 7

2 x [C (s) + O₂ → CO₂ (g); ∆H = x kJ] …………..(i)
3 x [H₂ (g) + ½ O₂ (g) → H₂O (l); ∆H = y kJ] ……..(ii)
C₂H₆ (g) + 7/2 O₂ (g) → 2CO₂ (g); ∆H = z kJ ……………….(iii)
Add (i) and (ii) and substracting from (iii), we get
p = 2x + 3y - z

The standard molar entropy of H₂O (s) or ice is less than H₂O (l) since H₂O molecules in ice are more ordered than those in water.

∆G = 0 because the reaction is at equilibrium.
∆G^o = - 2.303 RT logK;
∆G^o = - 2.303 x 8.314 JK^-1 mol^-1 x 300K x log 10² = - 11.48 kJ

The essential condition for feasibility of a reaction is that the reaction is to be accompanied with a decrease in free energy.

Note: Although both the conditions of decrease in enthalpy (the reaction being exothermic) as well as the increase in entropy are favourable for spontaneity, but the free energy is calculated taking both the factors into account.

Pressure, volume and density are intensive properties as these are independent of the given mass of the of the substance.
Internal energy of a substance is the sum of the kinetic and potential energies of a system and depends on the quantity of matter under consideration. It is an extensive property. On dividing the substance into two equal parts, the internal energy of each part would reduce to half.

Enthalpy H is a state function and is expressed by the relation:
H = U + pV
During an adiabatic change, ∆H = 0 and ∆U = w
Hence, w = nCv∆T
Gibbs energy (G) is the free energy of a system and is a state function. It is expressed as: G = H − TS.

During the melting of a cube of ice at 273 K and atmospheric pressure, the heat is transferred to the system (is positive).
Since the volume decreases, positive work is done on the ice-water system by the atmosphere.
Therefore, change in internal energy, ∆U = q + w