Problem 97 - Chemistry The Central Science

Question

(a) For each of the following reactions, predict the sign of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) without doing any calculations. (b) Based on your general chemical knowledge, predict which of these reactions will have \(K>1 .\) (c) In each case, indicate whether \(K\) should increase or decrease with increasing temperature. $$ \begin{array}{l}{\text { (i) } 2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{MgO}(s)} \\ {\text { (ii) } 2 \mathrm{KI}(s) \rightleftharpoons 2 \mathrm{K}(g)+\mathrm{I}_{2}(g)} \\ {\text { (iii) } \mathrm{Na}_{2}(g) \rightleftharpoons 2 \mathrm{Na}(g)} \\ {\text { (iv) } 2 \mathrm{V}_{2} \mathrm{O}_{5}(s) \rightleftharpoons 4 \mathrm{V}(s)+5 \mathrm{O}_{2}(g)}\end{array} $$

Step-by-Step Solution

Answer

The key findings of the analysis are: (i) Mg(s) + O₂(g) ↔ 2MgO(s): ΔH⁰ < 0, ΔS⁰ < 0, K > 1 at low temperatures, K decreases as temperature increases. (ii) 2KI(s) ↔ 2K(g) + I₂(g): ΔH⁰ > 0, ΔS⁰ > 0, K > 1 at high temperatures, K increases as temperature increases. (iii) Na₂(g) ↔ 2Na(g): ΔH⁰ > 0, ΔS⁰ > 0, K > 1 at high temperatures, K increases as temperature increases. (iv) 2V₂O₅(s) ↔ 4V(s) + 5O₂(g): ΔH⁰ > 0, ΔS⁰ > 0, K > 1 at high temperatures, K increases as temperature increases.

Step 1: (i) Reaction 1: Predicting ΔH⁰ and ΔS⁰

Mg(s) reacts with O₂(g) to form MgO(s). Since the product MgO is more stable than the reactants, the reaction is likely exothermic, meaning that ΔH⁰ will be negative. The reaction also decreases the number of gas molecules, implying a decrease in entropy (ΔS⁰ < 0).

Step 2: (i) Reaction 1: Predicting K > 1 and the Temperature Dependence of K

Since both ΔH⁰ and ΔS⁰ are negative for this reaction, its spontaneity is favored at lower temperatures (K > 1), and K decreases as the temperature increases.

Step 3: (ii) Reaction 2: Predicting ΔH⁰ and ΔS⁰

In this reaction, solid potassium iodide dissociates into gaseous potassium and iodine molecules. As this process requires breaking ionic bonds, it is endothermic (ΔH⁰ > 0). Moreover, the reaction results in an increase in the number of gas molecules, leading to an increase in entropy (ΔS⁰ > 0).

Step 4: (ii) Reaction 2: Predicting K > 1 and the Temperature Dependence of K

With both ΔH⁰ and ΔS⁰ being positive, the reaction is more spontaneous at higher temperatures, which means K > 1 at high temperatures. K increases with increasing temperature.

Step 5: (iii) Reaction 3: Predicting ΔH⁰ and ΔS⁰

In this reaction, Na₂(g) dissociates into 2Na(g). The reaction requires bond breaking, so it's endothermic (ΔH⁰ > 0). Additionally, since the reaction increases the number of gas molecules, there is an increase in entropy (ΔS⁰ > 0).

Step 6: (iii) Reaction 3: Predicting K > 1 and the Temperature Dependence of K

Both ΔH⁰ and ΔS⁰ are positive for this reaction, which means the reaction is more spontaneous at higher temperatures, and K increases with an increase in temperature.

Step 7: (iv) Reaction 4: Predicting ΔH⁰ and ΔS⁰

In this reaction, 2V₂O₅(s) decomposes into 4V(s) and 5O₂(g). The reaction requires bond breaking, so it's endothermic (ΔH⁰ > 0). Furthermore, the reaction leads to an increase in the number of gas molecules, suggesting an increase in entropy (ΔS⁰ > 0).

Step 8: (iv) Reaction 4: Predicting K > 1 and the Temperature Dependence of K

As both ΔH⁰ and ΔS⁰ are positive for this reaction, the reaction is more spontaneous at higher temperatures (K > 1), and K increases with increasing temperature.

Key Concepts

Enthalpy Change (ΔH⁰)

Enthalpy change, denoted as ΔH⁰, represents the heat change in a reaction at constant pressure. It's a vital concept in determining whether a reaction is exothermic or endothermic.

For exothermic reactions, ΔH⁰ is negative, meaning heat is released to the surroundings. These reactions feel warm to touch and typically involve bond formation. Conversely, endothermic reactions have a positive ΔH⁰, which indicates heat absorption, often involving bond breaking.

Understanding the enthalpy change of a reaction can help predict its feasibility and energy efficiency. For instance, in reaction (i) with Mg and O₂ forming MgO, ΔH⁰ is negative because forming ionic bonds releases energy. Remember, bond making releases energy, while bond breaking requires energy.

Entropy Change (ΔS⁰)

Entropy change, symbolized as ΔS⁰, quantifies the disorder or randomness in a system during a chemical reaction. Nature tends to favor more disordered states, hence an increase in entropy usually leads to more spontaneous processes.

When a reaction results in more gas molecules or breakdown into smaller particles, ΔS⁰ is positive. This means the system's entropy increases. An example is reaction (ii), where solid KI decomposes into gaseous K and I₂, increasing system randomness.

Conversely, if a reaction results in fewer gas molecules, ΔS⁰ is negative. This is seen in reaction (i) where MgO formation decreases the number of gas molecules, thus reducing entropy. Understanding ΔS⁰ helps determine reaction spontaneity.

Equilibrium Constant (K)

The equilibrium constant, denoted as K, provides insight into the position of equilibrium within a chemical reaction. It relates to the concentration of reactants and products at equilibrium.

For a given reaction at equilibrium, if K > 1, products are favored, meaning the reaction proceeds effectively in the forward direction. If K < 1, reactants are favored, indicating limited product formation.

In our example, reactions (ii), (iii), and (iv) have K > 1 at higher temperatures since they have both positive ΔH⁰ and ΔS⁰, making them more favorable thermodynamically at elevated temperatures. Understanding K and its dependence on temperature can predict how a reaction behaves under different conditions.

Temperature Dependence of Reactions

Temperature plays a crucial role in chemical reactions, influencing factors like reaction rate and equilibrium position. The interplay between enthalpy (ΔH⁰) and entropy (ΔS⁰) changes further dictates how temperature affects a reaction.

For reactions with both ΔH⁰ and ΔS⁰ positive, an increase in temperature leads to a larger equilibrium constant, making the reaction more spontaneous. This is why reactions such as (ii), (iii), and (iv) become more favorable at higher temperatures.

Alternatively, reactions with both ΔH⁰ and ΔS⁰ negative, like reaction (i), tend to have decreasing K with increasing temperature, reducing their spontaneity. Understanding these dynamics can guide predictions on how reactions shift with temperature changes.