b)Predict the sign of entropy change in the following: i) I2 (g) →I2 (s) ii) 2SO3 (g) → 2SO2(g) + O2(g) 

Hi,

Reaction a) I₂​(g) ------> ​I₂​(s)

The degree of randomness will be more in gaseous phase and least will be in solid state. So, entropy decreases for above reaction as gaseous iodine is converted to solid state.

Entropy = -ve sign

Reaction : 2SO₃------> 2SO₂ (g) + O₂​(g)

As number of particles increases, randomness increases which reasults in increase in entropy.

2 molecules of SO₃ dissociate to give 3 particles i.e., 2 particles of SO₂ and  O₂.

Entropy = +ve sign

Regards