Problem 99 - Chemistry: The Central Science

Question

(a) For each of the following reactions, predict the sign of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) 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. (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) \(\mathrm{Na}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{Na}(\mathrm{g})\) (iv) \(2 \mathrm{~V}_{2} \mathrm{O}_{5}(s) \rightleftharpoons 4 \mathrm{~V}(\mathrm{~s})+5 \mathrm{O}_{2}(\mathrm{~g})\)

Step-by-Step Solution

Answer

(i) ΔH° < 0, ΔS° < 0, K > 1, K decreases with increasing temperature. (ii) ΔH° > 0, ΔS° > 0, K < 1, K increases with increasing temperature. (iii) ΔH° > 0, ΔS° > 0, K < 1, K increases with increasing temperature. (iv) ΔH° > 0, ΔS° > 0, K < 1, K increases with increasing temperature.

Step 1: Enthalpy Change Prediction

In this reaction, oxygen gas reacts with magnesium solid to form magnesium oxide solid. It is an exothermic reaction because it forms a strong ionic bond. Therefore, we expect ΔH° to be negative.

Step 2: Entropy Change Prediction

Since a gas is consumed and no gases are formed, the reaction will exhibit a decrease in entropy. So we predict ΔS° to be negative.

Step 3: K Prediction

Considering the exothermic nature of the reaction and the decrease in entropy, we can predict that K > 1 because the forward reaction is favored.

Step 4: Temperature's Effect on K

Since the reaction is exothermic (ΔH° < 0), increasing the temperature would favor the reverse reaction and thus decrease K. (ii) \(2 \mathrm{KI}(s) \rightleftharpoons 2 \mathrm{K}(g)+\mathrm{I}_{2}(g)\)

Step 5: Enthalpy Change Prediction

In this reaction, solid potassium iodide dissociates into potassium gas and iodine gas. This reaction requires the input of energy to break ionic bonds, making it an endothermic reaction. Thus, ΔH° will be positive.

Step 6: Entropy Change Prediction

Two gas molecules are formed from one solid, causing an increase in entropy. Therefore, ΔS° is positive.

Step 7: K Prediction

With positive ΔH° and ΔS° values, K < 1 indicating that the reverse reaction is favored at low temperature and the forward reaction will be favored at high temperature.

Step 8: Temperature's Effect on K

For endothermic reactions (ΔH° > 0), increasing the temperature favors the forward reaction, leading to an increase in K. (iii) \(\mathrm{Na}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{Na}(\mathrm{g})\)

Step 9: Enthalpy Change Prediction

This reaction involves the dissociation of diatomic sodium gas molecules into individual sodium atoms. Breaking the bond requires energy input, making it an endothermic reaction. Therefore, ΔH° should be positive.

Step 10: Entropy Change Prediction

As one gas molecule dissociates into two gas molecules, the entropy increases. Thus, ΔS° is positive.

Step 11: K Prediction

With positive ΔH° and ΔS° values, K < 1 indicating that the reverse reaction is favored at low temperature and forward reaction will be favored at high temperature.

Step 12: Temperature's Effect on K

As this is an endothermic reaction (ΔH° > 0), increasing the temperature will favor the forward reaction, leading to an increase in K. (iv) $2 \mathrm{~V}_{2} \mathrm{O}_{5}(s) \rightleftharpoons 4 \mathrm{~V}(\mathrm{~s})+5 \mathrm{O}_{2}(\mathrm{~g})$

Step 13: Enthalpy Change Prediction

In this reaction, solid vanadium pentoxide decomposes into solid vanadium and oxygen gas. Breaking these bonds requires energy input, making it an endothermic reaction. Thus, ΔH° is positive.

Step 14: Entropy Change Prediction

Five gas molecules are formed from two solid molecules, increasing the entropy. Therefore, ΔS° is positive.

Step 15: K Prediction

With positive ΔH° and ΔS° values, K < 1, indicating that the reverse reaction is favored at low temperature and the forward reaction will be favored at high temperature.

Step 16: Temperature's Effect on K

Since this is an endothermic reaction (ΔH° > 0), increasing the temperature will favor the forward reaction, leading to an increase in K.

Key Concepts

enthalpy change (ΔH°)

In the study of thermodynamics in chemistry, enthalpy change, often denoted as \(\Delta H^{\circ}\), refers to the heat content change during a reaction. This concept helps determine whether a reaction is endothermic or exothermic:

• Exothermic reactions: These release heat to the surroundings, indicated by a negative \(\Delta H^{\circ}\). This is characteristic of reactions that form stronger bonds, such as magnesium reacting with oxygen to form magnesium oxide (\(2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{MgO}(s)\)).
• Endothermic reactions: These absorb heat, reflected by a positive \(\Delta H^{\circ}\). Such reactions require energy to break bonds, as seen in the dissociation of ionic solids like potassium iodide (\(2 \mathrm{KI}(s) \rightarrow 2 \mathrm{K}(g)+\mathrm{I}_{2}(g)\)).

The sign of \(\Delta H^{\circ}\) gives insights into the energy requirements of the reaction, which is critical when predicting how a reaction might respond to temperature changes.

entropy change (ΔS°)

Entropy change, represented as \(\Delta S^{\circ}\), reflects the disorder or randomness change in the molecules involved in a reaction. This concept is vital for predicting reaction spontaneity:

• Decrease in entropy: For reactions where gases are converted to solids, such as in \(2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{MgO}(s)\), a decrease in disorder occurs, resulting in a negative \(\Delta S^{\circ}\).
• Increase in entropy: Reactions converting solids to gases, like \(2 \mathrm{KI}(s) \rightarrow 2 \mathrm{K}(g)+\mathrm{I}_{2}(g)\), lead to an increase in disorder, yielding a positive \(\Delta S^{\circ}\).

Understanding \(\Delta S^{\circ}\) helps in predicting whether a reaction is thermodynamically favorable, as increased entropy typically favors reaction spontaneity.

Le Chatelier's Principle

Le Chatelier's Principle is a critical aspect of equilibrium that helps chemists understand how a reaction will adjust when conditions such as temperature, pressure, or concentration change. This principle posits that if a dynamic equilibrium is upset, the system will adjust to counteract the disturbance.

• Temperature Changes: For exothermic reactions (negative \(\Delta H^{\circ}\)), increasing temperature shifts the equilibrium towards the reactants. In contrast, if \(\Delta H^{\circ}\) is positive (endothermic reactions), an increase in temperature favors the forward reaction, producing more products.

Applying Le Chatelier's principle allows for the prediction and manipulation of reaction conditions to achieve desired equilibrium states, which is crucial for chemical manufacturing and synthesis.

equilibrium constant (K)

The equilibrium constant \(K\) quantifies the ratio of product concentrations to reactant concentrations at equilibrium. It provides critical insight into the extent to which a reaction will proceed. Here are some important aspects:

• K > 1: Implies products are favored, as observed in exothermic reactions with strong heat release, like \(2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{MgO}(s)\).
• K < 1: Suggests reactants are favored, typically seen in endothermic reactions where energy input is necessary, such as \(2 \mathrm{KI}(s) \rightarrow 2 \mathrm{K}(g)+\mathrm{I}_{2}(g)\).

Understanding \(K\) helps chemists predict reaction behavior under various conditions and design processes accordingly.

temperature dependence on equilibrium

The temperature of a system affects the position of equilibrium, as described by the Van 't Hoff equation, which relates changes in temperature to changes in the equilibrium constant \(K\):

• Exothermic Reactions (negative \(\Delta H^{\circ}\)): Increasing the temperature decreases \(K\), shifting equilibrium towards the reactants. This behavior is observed when heat is released, such as in \(2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{MgO}(s)\).
• Endothermic Reactions (positive \(\Delta H^{\circ}\)): A rise in temperature increases \(K\), shifting equilibrium towards the products. This is seen in reactions like \(2 \mathrm{KI}(s) \rightarrow 2 \mathrm{K}(g)+\mathrm{I}_{2}(g)\), where heat absorption is necessary.

Understanding the temperature dependence of \(K\) helps in optimizing reaction conditions for desired outcomes, crucial in industrial applications.