We need the process with negative entropy change

Entropy, $S$, generally measures randomness/disorder.

• If disorder increases, then $\Delta S > 0$.
• If disorder decreases, then $\Delta S < 0$.

Check each option

Option A

Dissociation of $CaSO_4(s)$ to $CaO(s)$ and $SO_3(g)$ $CaSO_4(s) \rightarrow CaO(s) + SO_3(g)$

Here, a gas is produced from solids. Formation of a gaseous product greatly increases disorder.

Therefore, $\Delta S > 0$

So, A is not correct.

Option B

Dissolution of iodine in water

On dissolving, the solid iodine particles disperse in the solvent, which generally increases randomness.

Therefore, entropy usually increases: $\Delta S > 0$

So, B is not correct.

Option C

Synthesis of ammonia from $N_2$ and $H_2$ $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$

Count gaseous moles:

• Reactants: $1 + 3 = 4$ moles of gas
• Products: $2$ moles of gas

Since the number of gas molecules decreases from $4$ to $2$, randomness decreases.

Therefore, $\Delta S < 0$

So, C is correct.

Option D

Sublimation of dry ice $CO_2(s) \rightarrow CO_2(g)$

Solid changes directly to gas, causing a large increase in disorder.

Therefore, $\Delta S > 0$

So, D is not correct.

Final conclusion

The only process with negative entropy change is: $\boxed{\text{C}}$

Comparison with stored correct answer

Stored correct answer: C

My derived answer: C

Hence, they agree.