Question

Predict the sign of \(\Delta S_{\text {sys }}\) for each of the following processes: (a) Gaseous Ar is liquefied at \(80 \mathrm{~K}\). (b) Gaseous \(\mathrm{N}_{2} \mathrm{O}_{4}\) dissociates to form gaseous \(\mathrm{NO}_{2}\). (c) Solid potassium reacts with gaseous \(\mathrm{O}_{2}\) to form solid potassium superoxide, \(\mathrm{KO}_{2}\). (d) Lead bromide precipitates upon mixing \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q)\) and \(\mathrm{KBr}(a q)\)

Short answer

The sign of $\Delta S_{sys}$ for each process is as follows: (a) Negative, due to liquefaction of gaseous Ar into a more ordered liquid state. (b) Positive, due to the increase in the number of gaseous molecules during the dissociation of N2O4 into NO2. (c) Negative, due to the decrease in gaseous molecules during the reaction of solid potassium and gaseous O2 to form solid KO2. (d) Negative, as the overall process results in a decrease of randomness in the system during the precipitation of lead bromide.

Step-by-step solution

Process (a): Liquefaction of gaseous Ar

In this process, gaseous Ar is liquefied at 80 K. Since the gaseous state is more disordered than the liquid state, the entropy of the system decreases during this process. Therefore, the change in entropy (ΔS) will have a negative sign.

Process (b): Dissociation of N2O4 to form NO2

In this process, gaseous N2O4 dissociates to form gaseous NO2. The balanced chemical equation for this reaction is: \[ N_{2}O_{4}(g) \rightarrow 2NO_{2}(g) \] Since the number of gaseous molecules increases from 1 to 2 during this reaction, the system becomes more disordered. Therefore, the change in entropy (ΔS) will have a positive sign.

Process (c): Reaction of solid potassium and gaseous O2 to form solid KO2

In this process, solid potassium reacts with gaseous O2 to form solid potassium superoxide (KO2). The balanced chemical equation for this reaction is: \[ 2K(s) + O_{2}(g) \rightarrow 2KO_{2}(s) \] Since the number of gaseous molecules decreases during this reaction, the system becomes less disordered. Therefore, the change in entropy (ΔS) will have a negative sign.

Process (d): Precipitation of lead bromide

In this process, lead bromide precipitates upon mixing Pb(NO3)2(aq) and KBr(aq). The balanced chemical equation for this reaction is: \[ Pb(NO_{3})_{2}(aq) + 2KBr(aq) \rightarrow PbBr_{2}(s) + 2KNO_{3}(aq) \] In this reaction, the lead bromide ions in the aqueous solution combine to form solid lead bromide, which is less disordered than the aqueous ions. Although there is still one aqueous product (KNO3) in the mixture, the overall process results in a decrease of randomness in the system. Therefore, the change in entropy (ΔS) will have a negative sign.