Animated Solution for Chemistry - Chemical Thermodynamics: The process with negative entropy change is

Welcome to a fascinating journey into the world of thermodynamics! Today, we are tackling a classic question about entropy. But before we dive into the options, let's take a moment to truly understand what entropy is.

In simple terms, entropy (denoted by $S$) is a measure of the randomness or disorder within a system. Imagine a perfectly organized library versus a room where books are scattered all over the floor. The scattered room has much higher entropy. In chemistry, the physical state of a substance is the biggest clue to its entropy. Solids are highly ordered (low entropy), liquids are somewhat disordered, and gases are wildly chaotic (high entropy).

When dealing with chemical reactions, especially those involving gases, we have a powerful shortcut to predict the sign of the entropy change ($\Delta S$). We simply look at the change in the number of gaseous moles, denoted as $\Delta n_g$.

If a reaction produces more gas molecules than it consumes ($\Delta n_g>0$), the system is becoming more chaotic, and the entropy increases ($\Delta S>0$). Conversely, if the number of gas molecules decreases ($\Delta n_g<0$), the system is becoming more ordered, and the entropy decreases ($\Delta S<0$).

Let's put our shortcut to the test by evaluating the given options:

Option (a): Synthesis of Ammonia The reaction is: $N_2(g)+3H_2(g)\to 2N{H}_3(g)$ Here, we have $1+3=4$ moles of gas on the reactant side, and only $2$ moles of gas on the product side. Calculating the change: $\Delta n_g=2-4=-2$. Since the number of gas molecules is decreasing, the randomness is decreasing. Therefore, the entropy change is negative.

Option (b): Dissociation of Calcium Sulphate The reaction is: $CaS{O}_4(s)\to CaO(s)+S{O}_3(g)$ We start with zero gas moles (only a solid) and end up with $1$ mole of gas. Calculating the change: $\Delta n_g=1-0=+1$. Creating a gas from a solid is a massive increase in disorder, so the entropy change is positive.

Option (c): Dissolution of Iodine The process is: $I_2(s)\to I_2(aq)$ When solid iodine dissolves in water, its rigid crystal lattice breaks apart. The iodine molecules become free to move around in the aqueous solution. This transition from a locked solid state to a mobile dissolved state means the randomness increases, making the entropy change positive.

Option (d): Sublimation of Dry Ice The process is: $C{O}_2(s)\to C{O}_2(g)$ Sublimation is the direct conversion of a solid to a gas. As we discussed earlier, gases are the kings of chaos. Going from a highly ordered solid directly to a highly disordered gas results in a huge positive entropy change.

After carefully analyzing all the options, it is clear that the only process where the system becomes more ordered is the synthesis of ammonia. The decrease in gaseous moles directly leads to a decrease in randomness.

Therefore, the correct answer is (a).