Problem 42 - Chemistry An Atoms First Approach

Question

Predict the sign of \(\Delta S^{\circ}\) for each of the following changes. a. \(\mathrm{K}(s)+\frac{1}{2} \mathrm{Br}_{2}(g) \longrightarrow \mathrm{KBr}(s)\) b. \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) c. \(\mathrm{KBr}(s) \longrightarrow \mathrm{K}^{+}(a q)+\mathrm{Br}^{-}(a q)\) d. \(\mathrm{KBr}(s) \longrightarrow \mathrm{KBr}(l)\)

Step-by-Step Solution

Answer

a. \(\Delta S^{\circ} < 0\), since the reaction leads to a decrease in gaseous molecules. b. \(\Delta S^{\circ} < 0\), since the reaction results in a decrease in the number of gaseous molecules. c. \(\Delta S^{\circ} > 0\), as the solid KBr dissolves, leading to an increase in entropy. d. \(\Delta S^{\circ} > 0\), as the solid KBr melts to form a liquid, increasing entropy.

Step 1: Examine the physical states of reactants and products

Here, we have a solid potassium (K) reacting with gaseous bromine (Br2) to form a solid potassium bromide (KBr).

Step 2: Determine the change in entropy

The reactant side has one molecule of gas (Br2), while the product side has no gaseous molecules. The reaction results in a decrease in the number of gaseous molecules, and as gases have higher entropy than solids, this will result in a decrease in entropy. Therefore, the sign of \(\Delta S^{\circ}\) for this reaction will be negative (\(\Delta S^{\circ} < 0\)). #b. N2(g) + 3 H2(g) → 2 NH3(g)#

Step 1: Examine the physical states and number of molecules

In this reaction, gaseous nitrogen (N2) reacts with gaseous hydrogen (H2) to form gaseous ammonia (NH3).

Step 2: Determine the change in entropy

There are 4 gaseous molecules on the reactant side (1 N2 and 3 H2) and only 2 gaseous molecules on the product side (2 NH3). This reaction results in a decrease in the number of gaseous molecules going from reactants to products, which indicates a decrease in entropy. Thus, the sign of \(\Delta S^{\circ}\) for this reaction will be negative (\(\Delta S^{\circ} < 0\)). #c. KBr(s) → K+(aq) + Br-(aq)#

Step 1: Examine the physical states and number of particles

In this reaction, solid potassium bromide (KBr) is dissolving to form potassium ions (K+) and bromide ions (Br-) in an aqueous solution.

Step 2: Determine the change in entropy

Going from a solid to an aqueous solution leads to an increase in entropy as the ions become more randomly dispersed in the solution compared to their ordered arrangement in the solid. Therefore, the sign of \(\Delta S^{\circ}\) for this reaction will be positive (\(\Delta S^{\circ} > 0\)). #d. KBr(s) → KBr(l)

Step 1: Examine the physical states of reactants and products

In this reaction, solid potassium bromide (KBr) is melting to form liquid potassium bromide (KBr).

Step 2: Determine the change in entropy

The entropy of a substance increases as it transitions from solid to liquid. In this case, the reaction involves the melting of a solid, which results in an entropy increase due to the increased freedom of motion of particles in the liquid state compared to the solid state. Thus, the sign of \(\Delta S^{\circ}\) for this reaction will be positive (\(\Delta S^{\circ} > 0\)).