Question

The crystalline hydrate \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2} \cdot 4 \mathrm{H}_{2} \mathrm{O}(s)\) loses water when placed in a large, closed, dry vessel: \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2} \cdot 4 \mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}(s)+4 \mathrm{H}_{2} \mathrm{O}(g)\) This process occurs even though it is endothermic; that is, \(\Delta H\) is positive. Is this process an exception to Bertholet's generalization? Explain.

Short answer

Although the given process is endothermic with a positive ΔH, we cannot conclude that it is an exception to Bertholet's generalization without more specific information about the reaction's entropy change and temperature. As the reaction involves an increase in entropy (ΔS > 0), it is possible that the term -TΔS can offset the positive ΔH value and make ΔG negative, resulting in a spontaneous process. This example highlights the importance of considering both changes in enthalpy and entropy when determining a reaction's spontaneity.

Step-by-step solution

Understand Bertholet's generalization

Bertholet's generalization states that for a reaction to be spontaneous, it should preferably release energy as heat (exothermic reaction), and not absorb energy (endothermic reaction). In simple terms, the reaction with a negative ΔH value (exothermic reaction) has a higher probability of being spontaneous, whereas a reaction with a positive ΔH value is unlikely to be spontaneous.

Recall the relationship between Gibbs free energy, enthalpy, and entropy

To determine the spontaneity of a reaction, we need to consider the Gibbs free energy change, ΔG, not only the change in enthalpy, ΔH. The relationship between ΔG, ΔH, and ΔS (change in entropy) is given by the following equation: $$ \Delta G = \Delta H - T\Delta S $$ Where T represents the temperature of the process. For a spontaneous process, the change in Gibbs free energy, ΔG, must be negative.

Analyze the given reaction in terms of ΔG

The given reaction is: $$ \mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2} \cdot 4 \mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}(s)+4 \mathrm{H}_{2} \mathrm{O}(g) $$ And the information provided is that this process is endothermic. Therefore, ΔH is positive. Unfortunately, we are not given the value of ΔS (change in entropy) or the reaction's temperature, so we cannot directly calculate ΔG. However, since the reaction involves the conversion of water molecules from solid to gaseous phase, we can expect a significant increase in entropy (ΔS > 0). Due to the increase in entropy, the term -TΔS will be negative and quite large, especially at high temperatures. This means that if the increase in entropy is significant enough, it might offset the positive ΔH value and make ΔG negative, making the reaction spontaneous.

Conclude if this process is an exception to Bertholet's generalization

The given process is an endothermic reaction as ΔH is positive. However, we can reasonably assume that the reaction will lead to an increase in entropy, making ΔS positive. If the increase in entropy is large enough, it can offset the positive ΔH value and make ΔG negative, resulting in a spontaneous process. Therefore, we cannot conclude that this process is an exception to Bertholet's generalization without more specific information about the reaction's entropy change and temperature. Rather, this example highlights the importance of considering both the changes in enthalpy and entropy when determining a reaction's spontaneity.

Key concepts

Bertholet's Generalization

Bertholet's generalization provides a simple rule to predict the spontaneity of chemical reactions. It suggests that reactions are more likely to occur spontaneously if they release energy as heat, known as exothermic reactions. In these cases, the change in enthalpy,

• \( \Delta H \),

is negative. This means that the system loses energy to its surroundings.
On the other hand, if a reaction absorbs heat (endothermic), resulting in a positive \( \Delta H \), it is generally thought to be non-spontaneous.
However, this rule doesn't always hold true. While an exothermic reaction is often spontaneous, not all endothermic reactions are non-spontaneous.
This is because spontaneity is influenced by other factors like entropy and temperature, not just enthalpy alone. Bertholet's generalization is a useful guideline, but it's essential to consider the whole picture.

Entropy

Entropy, denoted by \( \Delta S \), is a measure of disorder or randomness in a system. In chemical reactions, an increase in entropy typically favors spontaneity.
When a solid turns into a gas, as in the given reaction of

• \( \mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2} \cdot 4 \mathrm{H}_{2} \mathrm{O}(s) \to \mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}(s)+4 \mathrm{H}_{2} \mathrm{O}(g) \),

entropy increases significantly. This is because gas particles are far more disordered compared to the organized solid state.
In the equation for Gibbs free energy, \( \Delta G = \Delta H - T\Delta S \), entropy plays a key role along with temperature. If \( \Delta S \) is positive and large, it can drive a reaction to be spontaneous, even if it is endothermic.
Therefore, understanding entropy helps in assessing the overall energy changes and predictions about whether a reaction will occur spontaneously.

Spontaneity of Reactions

The spontaneity of a reaction is determined by the change in Gibbs free energy, \( \Delta G \). A reaction is spontaneous if \( \Delta G \) is negative.
The equation \( \Delta G = \Delta H - T\Delta S \) integrates enthalpy and entropy changes to assess the true nature of a reaction's spontaneity.

• If a reaction has positive \( \Delta H \) but also has a strong positive \( \Delta S \) at high temperatures, the term \( -T\Delta S \) becomes significant enough to make \( \Delta G \) negative.

That's why some endothermic reactions can occur spontaneously under the right conditions. Practically, this means that temperature and entropy changes can overpower the enthalpy effect.
In the example given, the large increase in entropy is likely making the reaction spontaneous even though it's absorbing heat. This illustrates the importance of considering both enthalpy and entropy when evaluating reaction spontaneity.