Problem 99 - 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 \(K\) 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 \(K\) decreases with increasing temperature. (ii) \(\Delta H^{\circ} > 0\), \(\Delta S^{\circ} > 0\), \(K>0\) at high temperatures, and \(K\) increases with increasing temperature. (iii) \(\Delta H^{\circ} > 0\), \(\Delta S^{\circ} > 0\), \(K>0\) at high temperatures, and \(K\) increases with increasing temperature. (iv) \(\Delta H^{\circ} > 0\), \(\Delta S^{\circ} > 0\), \(K>0\) at high temperatures, and \(K\) increases with increasing temperature.

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

For this reaction, the formation of a solid product (MgO) occurs from the solid reactants (Mg) and gaseous reactant (O2). Since energy is released when compounds are formed from their elements, the enthalpy change, \(\Delta H^{\circ}\), should be negative.

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

The reaction is a combination of solid and gaseous reactants forming a solid product. There will be a decrease in the number of gas molecules in the process, resulting in a decrease in entropy. Therefore, \(\Delta S^{\circ}\) is negative.

Step 3: Determine \(K\) range and temperature dependence

Since both \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are negative, this reaction would be favorable at lower temperatures, where the exothermic aspect dominates. Hence, \(K > 0\) at lower temperatures. However, for higher temperatures, the positive contribution from \(\Delta S^{\circ}T\) would make \(K\) decrease with increasing temperature. (ii) \(2 \mathrm{KI}(s) \rightleftharpoons 2 \mathrm{~K}(g)+\mathrm{I}_{2}(g)\)

Step 4: Determine \(\Delta H^{\circ}\) sign

When solid KI decomposes into its gaseous elements, energy is needed to break the ionic bonds. Therefore, \(\Delta H^{\circ}\) is positive.

Step 5: Determine \(\Delta S^{\circ}\) sign

In this reaction, solid KI is converted into gaseous K and I2, resulting in an increase in entropy since the number of gas molecules increases. Therefore, \(\Delta S^{\circ}\) is positive.

Step 6: Determine \(K\) range and temperature dependence

Since both \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are positive, this reaction is favored at higher temperatures where the contribution from \(\Delta S^{\circ}T\) is significant. Therefore, \(K>0\) at high temperatures, and \(K\) will increase with increasing temperature. (iii) \(\mathrm{Na}_{2}(g) \rightleftharpoons 2 \mathrm{Na}(g)\)

Step 7: Determine \(\Delta H^{\circ}\) sign

In this reaction, energy is needed to break the bond between Na atoms. Therefore, \(\Delta H^{\circ}\) is positive.

Step 8: Determine \(\Delta S^{\circ}\) sign

Since the number of gas molecules increases from one to two, the entropy increases. Therefore, \(\Delta S^{\circ}\) is positive.

Step 9: Determine \(K\) range and temperature dependence

With both \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) being positive, this reaction is favored at higher temperatures. Therefore, \(K>0\) at high temperatures, and \(K\) will increase with increasing temperature. (iv) \(2 \mathrm{~V}_{2} \mathrm{O}_{5}(s) \rightleftharpoons 4 \mathrm{~V}(s)+5 \mathrm{O}_{2}(g)\)

Step 10: Determine \(\Delta H^{\circ}\) sign

The decomposition of V2O5 into its elements requires the input of energy to break the chemical bonds. Therefore, \(\Delta H^{\circ}\) is positive.

Step 11: Determine \(\Delta S^{\circ}\) sign

The reaction involves the formation of gaseous O2 from solid V2O5, leading to an increase in the number of gas molecules and an increase in entropy. Therefore, \(\Delta S^{\circ}\) is positive.

Step 12: Determine \(K\) range and temperature dependence

Since both \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are positive, this reaction is favored at higher temperatures. Therefore, \(K>0\) at high temperatures and \(K\) will increase with increasing temperature.

Key Concepts

Enthalpy Change (ΔH)

The concept of enthalpy change ( ΔH ) in thermodynamics refers to the amount of heat absorbed or released during a chemical reaction at constant pressure. In simpler terms, it tells us if a reaction gives off heat (exothermic) or absorbs heat (endothermic).

• **Exothermic Reactions ( ΔH < 0):** These reactions release heat to the surroundings. For instance, when magnesium (Mg) reacts with oxygen (O2) to form magnesium oxide (MgO), the reaction is exothermic. The reaction releases energy, making ΔH negative.


• **Endothermic Reactions ( ΔH > 0):** These reactions require heat input. A classic example is the decomposition of potassium iodide (KI) into potassium (K) and iodine gas (I2). Breaking ionic bonds requires energy, resulting in a positive ΔH .

Understanding whether a reaction is exothermic or endothermic helps predict how temperature changes could affect the reaction outcome, notably the equilibrium constant (K). Exothermic reactions tend to favor equilibrium with lower temperatures, while endothermic reactions are more favorable at higher temperatures.

Entropy Change (ΔS)

Entropy ( ΔS ) is the measure of disorder or randomness in a system. A positive entropy change ( ΔS > 0) indicates that the disorder increases, while a negative value ( ΔS < 0) means that the system becomes more ordered.

• **Entropy Decreases ( ΔS < 0):** This often occurs in reactions where gases are converted to solids or liquids. For example, when solid magnesium reacts with gaseous oxygen to form solid MgO, the orderliness increases and ΔS is negative.


• **Entropy Increases ( ΔS > 0):** This is common in reactions where solid or liquid reactants turn into gases, as gases have more disorder. In the decomposition of KI (s) to K (g) and I2 (g), ΔS is positive since gases are formed, increasing disorder.

Entropy contributes to determining the spontaneity of a reaction along with enthalpy. Reactions where both ΔH and ΔS are negative favor lower temperatures, while both positive ΔH and ΔS favor higher temperatures, boosting the equilibrium constant (K) at these temperatures.

Equilibrium Constant (K)

The equilibrium constant ( K ) is a value that indicates the ratio of product concentrations to reactant concentrations at equilibrium for a particular reaction. It gives us insight into which direction a reaction is favored.

• **K > 1:** The reaction favors product formation, meaning at equilibrium, there are more products than reactants. This is common in exothermic reactions at lower temperatures where ΔH is negative.


• **K < 1:** The reaction favors the reactants, indicating more reactants remain at equilibrium. Endothermic reactions at low temperatures often display a lower K value as ΔH is positive.

The relationship between temperature and equilibrium constant is determined by the Gibbs Free Energy equation: ΔG = ΔH - TΔS . When both ΔH and ΔS are positive, increasing the temperature generally increases K , making the reaction more product-favored as disorder contributes positively. Conversely, when both are negative, K tends to decrease with increasing temperature. Understanding these concepts of K allows accurate predictions about how reaction conditions will affect chemical equilibrium.