Problem 91 - Chemistry: The Central Science

Question

(a) For each of the following reactions, predict the sign of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) and discuss briefly how these factors determine the magnitude of \(K .\) (b) Based on your general chemical knowledge, predict which of these reactions will have \(K>0 .\) (c) In each case indicate whether \(\underline{K}\) should increase or decrease with increasing temperature. (i) \(2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{MgO}(s)\) (ii) \(2 \mathrm{KI}(s) \rightleftharpoons 2 \mathrm{~K}(g)+\mathrm{I}_{2}(g)\) (iii) \(\mathrm{Na}_{2}(g) \rightleftharpoons 2 \mathrm{Na}(g)\) (iv) \(2 \mathrm{~V}_{2} \mathrm{O}_{5}(s) \rightleftharpoons 4 \mathrm{~V}(s)+5 \mathrm{O}_{2}(g)\)

Step-by-Step Solution

Answer

(i) \(\Delta H^{\circ}<0\), \(\Delta S^{\circ}<0\), \(K>0\) at low temperatures and possibly \(K<0\) at high temperatures, and \(K\) decreases with increasing temperature. (ii) \(\Delta H^{\circ}>0\), \(\Delta S^{\circ}>0\), \(K<0\) at low temperatures and possibly \(K>0\) at high temperatures, and \(K\) increases with increasing temperature. (iii) \(\Delta H^{\circ}>0\), \(\Delta S^{\circ}>0\), \(K<0\) at low temperatures and possibly \(K>0\) at high temperatures, and \(K\) increases with increasing temperature. (iv) \(\Delta H^{\circ}>0\), \(\Delta S^{\circ}>0\), \(K<0\) at low temperatures and possibly \(K>0\) at high temperatures, and \(K\) increases with increasing temperature.

Step 1: Determining \(\Delta H^{\circ}\)

In this reaction, solid magnesium combines with gaseous oxygen to form solid magnesium oxide. The formation of a compound from its constituent elements is an exothermic process, releasing heat; thus, the enthalpy change \(\Delta H^{\circ}<0\).

Step 2: Determining \(\Delta S^{\circ}\)

Since the reaction involves a decrease in the number of gas molecules (from one to zero), there is a reduction in entropy. Therefore, \(\Delta S^{\circ}<0\).

Step 3: Magnitude of \(K\)

Based on Gibbs' free energy formula and \(ΔG=-RT\ln K\), an exothermic reaction with a decrease in entropy will have a negative \(ΔG\) at low temperatures but might become positive at high temperatures, leading to a value of \(K>0\) at low temperatures and possibly \(K<0\) at high temperatures.

Step 4: Temperature Dependence of \(K\)

As temperature increases, the effect of the negative entropy change becomes more significant, making \(ΔG\) more positive. Therefore, for this reaction, \(K\) should decrease with increasing temperature. **(ii) \(2 \mathrm{KI}(s) \rightleftharpoons 2 \mathrm{~K}(g)+\mathrm{I}_{2}(g)\)**

Step 5: Determining \(\Delta H^{\circ}\)

In this reaction, solid potassium iodide decomposes into gaseous potassium and gaseous iodine. Decomposition reactions are usually endothermic, as energy is required to break the bonds. Thus, \(\Delta H^{\circ}>0\).

Step 6: Determining \(\Delta S^{\circ}\)

As two moles of gas form from a single mole of solid, there is an increase in entropy, so \(\Delta S^{\circ}>0\).

Step 7: Magnitude of \(K\)

Based on Gibbs' free energy formula, an endothermic reaction with an increase in entropy will have a positive \(ΔG\) at low temperatures but might become negative at high temperatures, leading to \(K<0\) at low temperatures and possibly \(K>0\) at high temperatures.

Step 8: Temperature Dependence of \(K\)

As temperature increases, the effect of the positive entropy change becomes more significant, making \(ΔG\) more negative. Therefore, for this reaction, \(K\) should increase with increasing temperature. **(iii) \(\mathrm{Na}_{2}(g) \rightleftharpoons 2 \mathrm{Na}(g)\)**

Step 9: Determining \(\Delta H^{\circ}\)

This reaction is dissociation of diatomic sodium gas into separate sodium gas atoms. This process requires energy to break the bond, so the enthalpy change is positive: \(\Delta H^{\circ}>0\).

Step 10: Determining \(\Delta S^{\circ}\)

This reaction has no change in the number of gas molecules and is symmetrical, so there will be a small increase in entropy due to the increased freedom of movement for the individual atoms. Therefore, \(\Delta S^{\circ}>0\).

Step 11: Magnitude of \(K\)

Similar to (ii), the endothermic reaction with an increase in entropy leads to \(ΔG>0\) at low temperatures and possibly \(ΔG<0\) at high temperatures. Therefore, \(K<0\) at low temperatures and possibly \(K>0\) at high temperatures.

Step 12: Temperature Dependence of \(K\)

As temperature increases, the positive entropy change becomes more significant, making \(ΔG\) more negative. Therefore, for this reaction, \(K\) should increase with increasing temperature. **(iv) \(2 \mathrm{~V}_{2} \mathrm{O}_{5}(s) \rightleftharpoons 4 \mathrm{~V}(s)+5 \mathrm{O}_{2}(g)\)**

Step 13: Determining \(\Delta H^{\circ}\)

In this reaction, solid vanadium(V) oxide decomposes into solid vanadium and gaseous oxygen. As a decomposition reaction, it requires energy to break the bonds. Thus, \(\Delta H^{\circ}>0\).

Step 14: Determining \(\Delta S^{\circ}\)

Five moles of gas form from two moles of solid, leading to an increase in entropy. Therefore, \(\Delta S^{\circ}>0\).

Step 15: Magnitude of \(K\)

Similar to (ii), endothermic reaction with an increase in entropy leads to \(ΔG>0\) at low temperatures and possibly \(ΔG<0\) at high temperatures. Therefore, \(K<0\) at low temperatures and possibly \(K>0\) at high temperatures.

Step 16: Temperature Dependence of \(K\)

As temperature increases, the positive entropy change becomes more significant, making \(ΔG\) more negative. Therefore, for this reaction, \(K\) should increase with increasing temperature.

Key Concepts

Enthalpy Change

Enthalpy change, often denoted as \( \Delta H \), describes the amount of heat absorbed or released during a chemical reaction at constant pressure. Knowing whether a reaction is exothermic (

• \( \Delta H < 0 \): Releases heat, often making\ \( \Delta G \) negative at low temperatures
• \( \Delta H > 0 \): Absorbs heat, requires energy to proceed

The sign of \( \Delta H \) provides insight into the heat flow within the reaction. Exothermic reactions, like the formation of magnesium oxide from magnesium and oxygen, tend to drive the reaction forward by releasing energy to the surroundings. Conversely, endothermic reactions, such as the decomposition of potassium iodide into gaseous products, necessitate an input of energy, which can affect the feasibility of the reaction at different temperatures.

Entropy Change

Entropy change, expressed as \( \Delta S \), measures the disorder or randomness in a system. This concept helps determine the spontaneity of a reaction and its effect on the equilibrium constant. Key points include:

• \( \Delta S > 0 \): An increase in entropy, indicating greater disorder, usually occurs when solids or liquids change to gases or when a single solid decomposes into multiple gases.
• \( \Delta S < 0 \): A decrease in entropy, meaning reduced disorder, often results from gas formation into a liquid or solid.

In the given reactions, an increase in gaseous products generally led to an increase in entropy, as seen in the decomposition of potassium iodide. This change directly influences the Gibbs free energy and the equilibrium constant of the reaction, impacting its direction and favorability under different conditions.

Equilibrium Constant

The equilibrium constant, \( K \), describes the balance reached in a reversible reaction and how it shifts according to temperature changes and other factors. It is related to the Gibbs free energy through the formula: \[ \Delta G = -RT\ln K \]Where:

• \( \Delta G < 0 \): \( K > 1 \), indicating a reaction that favors products at equilibrium
• \( \Delta G > 0 \): \( K < 1 \), indicating a reaction that favors reactants

The temperature dependency of \( K \) arises from the interplay of \( \Delta H \) and \( \Delta S \). As temperature increases, an exothermic reaction with negative entropy change may see a decrease in \( K \), while endothermic reactions with positive entropy may exhibit an increase in \( K \). This is particularly significant in reactions involving phase changes or bond breakages, which impact the Gibbs free energy significantly, altering the position of equilibrium.