Question

#1)Suppose a small decorative piece of ice (system) at a temperature of 0°C and weighing 538 g is placed on a granite tabletop (surroundings) at 32 °C. Determine ∆S_sys, ∆S_surr and ∆S_univ (in J/K) associated for the reversible melting of this piece of ice.
Assume the temperature of the tabletop does not change and that the final temperature of the water is 0 °C. To solve this problem you are given the molar heat of fusion of ice is 6.01 kJ mol^-1.
∆S_sys =
∆S_surr =
∆S_univ =

#2)Determine ∆_rG° (in kJ mol^-1) for the following reaction at 298 K,

SO₃(g) + H₂O(l) ----> H₂SO₄(aq)

Given that ∆_fG° for the species are,
SO₃(g) = -371.06 kJ mol^-1
H₂O(l) = -237.13 kJ mol^-1
H₂SO₄(aq) = -744.53 kJ mol^-1

Answer 1

A thermodynamically irreversible process is always accompanied by an increase of entropy of the system and its surrounding taken together while in a thermodynamically reversible process, the entropy of the system and its surrounding taken together remains unaltered. Thus we can write

∆Ssys + ∆Ssurr = 0