Question

Consider the reaction $$\begin{array}{l}{2 \mathrm{O}(g) \longrightarrow \mathrm{O}_{2}(g)} \\\ {\text { sof } \Delta H \text { and } \Delta S}\end{array}$$ a. Predict the signs of \(\Delta H\) and \(\Delta S .\) b. Would the reaction be more spontaneous at high or low temperatures?

Short answer

a. The signs of \(\Delta H\) and \(\Delta S\) are both negative, with \(\Delta H < 0\) and \(\Delta S < 0\). b. The reaction would be more spontaneous at low temperatures.

Step-by-step solution

Predict the sign of ∆H

To predict the sign of the change in enthalpy (∆H), we need to consider the bond energies of the reactants and products. In the given reaction, two individual oxygen molecules (O) are combining to form an O2 molecule (oxygen gas). Oxygen has a double bond, which is relatively strong. Since bond formation is an exothermic process (energy is released), we can expect that the change in enthalpy for this reaction will be negative (∆H < 0).

Predict the sign of ∆S

Entropy (∆S) is a measure of the disorder of a system. In this reaction, two oxygen molecules are combining to form one molecule of oxygen gas. The reaction results in a decrease in the number of particles in the system. So, the order of the system increases, and we can expect a decrease in entropy (∆S < 0).

Determine the spontaneity of the reaction considering temperature

According to the Gibbs free energy equation, a reaction is spontaneous if ΔG < 0, where: ΔG = ΔH - TΔS We have predicted the signs of ∆H and ∆S to be negative. As the temperature (T) increases, the term TΔS becomes more positive. Since the signs of ∆H and ΔS are both negative, and TΔS becomes more positive with increasing temperature, the value of ΔG will be less negative (or even positive) at higher temperatures. This means that the reaction is more favorable (spontaneous) at lower temperatures. In summary: a. ∆H < 0 (negative) and ∆S < 0 (negative) b. The reaction would be more spontaneous at low temperatures.

Key concepts

enthalpy sign prediction

In a chemical reaction, predicting the sign of the change in enthalpy (\( \Delta H \)) helps us understand whether the process absorbs or releases energy. The given reaction involves the transformation from individual oxygen atoms to a gaseous \( \mathrm{O}_2 \) molecule, a process that usually involves forming new bonds. Since forming bonds generally releases energy, this is an exothermic reaction. Exothermic reactions typically result in a negative \( \Delta H \), indicating that energy is being given off as heat. It's important to note that a negative value of \( \Delta H \) usually suggests that the products are energetically more favorable than the reactants. Thus, by looking at how bonds are formed or broken, we can accurately predict the sign of enthalpy change.

• Bond Formation: Releases energy, making \( \Delta H \) negative.
• Exothermic Reaction: Indicated by \( \Delta H < 0 \)

Recognizing these patterns in reactions makes it easier to predict whether \( \Delta H \) will be positive or negative.

entropy change

Understanding entropy change (\( \Delta S \)) involves assessing the disorder or randomness in a system. In our example reaction, two separate oxygen atoms combine to create one oxygen molecule. This transition moves from a state of higher disorder—more individual particles—to lower disorder—less individual particles—which implies that the entropy has decreased. Entropy is often associated with the number of ways the particles can be arranged. Greater particle variety typically means higher entropy. Since \( \Delta S \) is negative in this case, the reaction results in higher order or lower disorder in the final product.

• Decrease in Particle Number: Reduction of disorder, hence a negative \( \Delta S \).
• Higher order: Fewer ways to arrange the particles lead to decreased entropy.

Recognizing the change from disorder to order helps in understanding how \( \Delta S \) will affect reaction spontaneity.

reaction spontaneity

The spontaneity of a reaction is a key factor that determines whether or not it occurs naturally. Spontaneity is assessed using the Gibbs free energy equation:\[\Delta G = \Delta H - T\Delta S\]A reaction is spontaneous if \( \Delta G < 0 \).In our reaction, both \( \Delta H \) and \( \Delta S \) are negative. At lower temperatures, the \( T\Delta S \) term has a smaller impact, meaning \( \Delta G \) is more likely to remain negative, making the reaction more spontaneous. As temperature increases, the negative \( T\Delta S \) becomes larger, which could make \( \Delta G \) positive, thus reducing spontaneity.

• Low Temperatures: Favor spontaneous reactions since \( \Delta G \) stays negative more easily.
• Higher Temperatures: Increase the influence of the \( T\Delta S \) term, making spontaneity less likely if both \( \Delta H \) and \( \Delta S \) are negative.

Recognizing the temperature's role helps predict when a particular chemical reaction will proceed on its own.