Problem 86 - Chemistry The Central Science

Question

The crystalline hydrate \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right) \cdot 4 \mathrm{H}_{2} \mathrm{O}(s)\) loses water when placed in a large, closed, dry vessel at room temperature: $$ \mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2} \cdot 4 \mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}(s)+4 \mathrm{H}_{2} \mathrm{O}(g) $$ This process is spontaneous and \(\Delta H^{\circ}\) is positive at room temperature. (a) What is the sign of \(\Delta S^{\circ}\) at room temperature? (b) If the hydrated compound is placed in a large, closed vessel that already contains a large amount of water vapor, does \(\Delta S^{\circ}\) change for this reaction at room temperature?

Step-by-Step Solution

Answer

(a) The sign of ΔS° at room temperature is positive. (b) The presence of a large amount of water vapor in the closed vessel does not change ΔS° for this reaction at room temperature.

Step 1: Examine the reaction

First, let's examine the given reaction: $$ \mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2} \cdot 4 \mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}(s)+4 \mathrm{H}_{2} \mathrm{O}(g) $$ We can observe that a solid substance (crystalline hydrate) is losing water and converting into its anhydrous form along with water molecules in the gaseous state. The process is spontaneous and has a positive enthalpy change (ΔH°).

Step 2: Determine the sign of ΔS°

We know that for a spontaneous reaction, the Gibbs free energy change (ΔG°) is negative. Taking the relationship between Gibbs free energy change, enthalpy change, and entropy change into account: $$ \Delta G^{\circ} = \Delta H^{\circ} - T \times \Delta S^{\circ} $$ Given that ΔH° is positive and ΔG° is negative for the spontaneous reaction, we can infer that ΔS° must be positive to make the right side of the equation negative: $$ \Delta G^{\circ} = (+) - T \times (+) $$ Thus, the sign of ΔS° at room temperature is positive.

Step 3: Determine if ΔS° changes in the presence of a large amount of water vapor

In the given scenario, a large amount of water vapor is already present in the closed vessel. The entropy change (ΔS°) is mainly dependent on the difference in the number of moles of gaseous products and reactants and the arrangement of particles in each state. Since the water is in the gaseous state in both the reaction and the vessel, the presence of excess water vapor would not change the nature of the reaction or the arrangement of the particles. Therefore, ΔS° remains the same in this case, and it is still positive at room temperature. In summary: (a) The sign of ΔS° at room temperature is positive. (b) The presence of a large amount of water vapor in the closed vessel does not change ΔS° for this reaction at room temperature.

Key Concepts

Gibbs free energy

Gibbs free energy is a crucial concept in thermodynamics. It helps determine whether a chemical reaction occurs spontaneously. The equation for Gibbs free energy is given by:\[\Delta G = \Delta H - T \Delta S\]where:

• \(\Delta G\) is the change in Gibbs free energy
• \(\Delta H\) is the change in enthalpy
• \(T\) is the temperature in Kelvin
• \(\Delta S\) is the change in entropy

In simple terms, a spontaneous reaction is one that occurs without additional energy input.
For a reaction to be spontaneous, \(\Delta G\) must be negative. This happens when the entropy increase (\(\Delta S\)) compensates for a positive enthalpy change (\(\Delta H\)).
In the case of the given reaction, even though \(\Delta H\) is positive, the spontaneous nature suggests that \(\Delta S\) must also be positive and large enough to make \(\Delta G\) negative. This means that the process increases the disorder or randomness of the system.

enthalpy change

Enthalpy change, symbolized by \(\Delta H\), represents the heat absorbed or released during a reaction at constant pressure.
It is an important factor in determining how much energy is necessary for a reaction. In thermodynamic terms, it involves the making and breaking of chemical bonds.For our specific reaction, the enthalpy change is noted as positive.
A positive \(\Delta H\) means the system absorbs heat from the surroundings, a characteristic of endothermic reactions. Endothermic reactions feel cool to the touch because they pull heat in.
This reaction, while absorbing heat, also increases the system's energy but does not stop it from being spontaneous due to the increase in entropy.In conclusion, while a positive \(\Delta H\) might seem counterintuitive for a spontaneous reaction, in such cases the role of entropy becomes significant.
Here, the creation of gaseous water molecules from solid increases the disorder, driving the spontaneity despite the energy absorption.

entropy change

Entropy change, denoted as \(\Delta S\), indicates the degree of disorder or randomness in a system.
Increasing entropy often drives reactions to occur spontaneously. Simply put, systems naturally progress toward higher states of disorder.In our reaction, solid hydrate changes into solid cadmium nitrate and gaseous water (\(\mathrm{H}_{2}\mathrm{O}(g)\)).
This transition from a crystalline solid to a mixture with gaseous components means an increase in the randomness and distribution of particles.Key points about \(\Delta S\):

• A positive \(\Delta S\) signifies increased disorder.
• Gaseous substances usually have higher entropy than solids or liquids.

The increase in entropy (positive \(\Delta S\)) is a significant driving force behind the spontaneity of the reaction.
Even if the system absorbs heat, the increased disorder from forming gaseous water ensures that the overall process is energetically favorable.
Thus, a positive \(\Delta S\) can make a reaction proceed spontaneously, even when \(\Delta H\) is positive as well.