Question

(a) For each of the following reactions, predict the sign of \(\Delta H^{*}\) and \(\Delta S^{\circ}\) without doing any calculations. (b) Based on your general chemical knowledge, predict which of these reactions will have \(K>1\) at \(25^{\circ} \mathrm{C} .(\mathbf{c})\) In each case, indicate whether \(K\) should increase or decrease with increasing temperature. (i) \(2 \mathrm{Fe}(s)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{FeO}(s)\) (ii) \(\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{Cl}(g)\) (iii) \(\mathrm{NH}_{4} \mathrm{Cl}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{HCl}(g)\) (iv) \(\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{CaO}(s) \rightleftharpoons \mathrm{CaCO}_{3}(s)\)

Short answer

Reaction 1: ΔH* is negative (exothermic), ΔS° is negative, K > 1 at 25°C, and K decreases with increasing temperature. Reaction 2: ΔH* is positive (endothermic), ΔS° is positive, K > 1 at 25°C is uncertain, and K increases with increasing temperature. Reaction 3: ΔH* is positive (endothermic), ΔS° is positive, K > 1 at 25°C is uncertain, and K increases with increasing temperature. Reaction 4: ΔH* is negative (exothermic), ΔS° is negative, K > 1 at 25°C is uncertain, and K decreases with increasing temperature.

Step-by-step solution

Since this is a formation reaction where we are producing a compound from its elements, the enthalpy change is likely to be negative, meaning the reaction is exothermic. ##Step 2: Determine sign of ΔS°##

In this reaction, we have one mole of a gaseous reactant forming two moles of solid product. This means a decrease in entropy as gas has more entropy than solids. So, ΔS° would be negative. ##Step 3: Predict K > 1 at 25°C##

Given that ΔH* is negative and ΔS° is negative, the behavior of the reaction depends on the relative magnitudes of the two values. However, typically, formation reactions tend to proceed spontaneously, so we can say that K > 1 at 25°C. ##Step 4: Dependence of K on temperature##

According to Van't Hoff's equation(\( \frac{d \ln K }{ dT } = \frac{ \Delta H }{ RT^2 } \)), a reaction with negative ΔH* and negative ΔS° will have K decrease with increasing temperature. #Reaction 2: Cl₂(g) ⇌ 2 Cl(g)# ##Step 1: Determine sign of ΔH*##

This reaction involves breaking the Cl-Cl bond in Cl₂, which requires energy input. Therefore, the enthalpy change will be positive, meaning the reaction is endothermic. ##Step 2: Determine sign of ΔS°##

The reaction involves one mole of reactant gas forming two moles of product gas. As a result, there is an increase in entropy in the system. Therefore, ΔS° is positive. ##Step 3: Predict K > 1 at 25°C##

As both ΔH* and ΔS° are positive, the reaction will tend to be more spontaneous at higher temperatures. At 25°C, we cannot predict with certainty whether or not K > 1 without more information. ##Step 4: Dependence of K on temperature##

A reaction with positive ΔH* and positive ΔS° will have K increase with increasing temperature according to Van't Hoff's equation. #Reaction 3: NH₄Cl(s) ⇌ NH₃(g) + HCl(g)## ##Step 1: Determine sign of ΔH*##

This reaction involves breaking the ionic bond in NH₄Cl and forming two moles of gas, so energy input is required. The enthalpy change is positive, meaning the reaction is endothermic. ##Step 2: Determine sign of ΔS°##

Going from one mole of a solid substance to two moles of gases results in an increase in entropy. Therefore, ΔS° is positive. ##Step 3: Predict K > 1 at 25°C##

With positive ΔH* and ΔS°, this reaction will be favored at higher temperatures, making it difficult to predict whether K > 1 at 25°C. ##Step 4: Dependence of K on temperature##

A reaction with positive ΔH* and positive ΔS° will have K increase with increasing temperature according to Van't Hoff's equation. #Reaction 4: CO₂(g) + CaO(s) ⇌ CaCO₃(s)## ##Step 1: Determine sign of ΔH*##

This is a formation reaction, so the enthalpy change is likely to be negative, meaning the reaction is exothermic. ##Step 2: Determine sign of ΔS°##

One mole of gas and one mole of solid are converted into one mole of solid product, leading to a decrease in entropy as a gas has more entropy than solids. Therefore, ΔS° is negative. ##Step 3: Predict K > 1 at 25°C##

Generally, in formation reactions, K tends to be greater than 1. However, we cannot predict with certainty without more information on the relative magnitudes of ΔH* and ΔS°. ##Step 4: Dependence of K on temperature##

A reaction with negative ΔH* and negative ΔS° will have K decrease with increasing temperature according to Van't Hoff's equation.

Key concepts

Enthalpy Change (ΔH)

In chemical reactions, understanding the concept of enthalpy change (ΔH) is crucial. It refers to the heat energy absorbed or released during a reaction at constant pressure. If a reaction releases heat, it is considered exothermic, and the enthalpy change (ΔH) will be negative. Conversely, if a reaction absorbs heat from its surroundings, it is endothermic, and the enthalpy change will be positive.

For example, the reaction between iron and oxygen to form iron oxide is exothermic. Here, a compound is produced from its elements, releasing energy, and thus ΔH is negative. On the other hand, the dissociation of chlorine gas into two chlorine atoms requires energy input to break the Cl-Cl bond, making ΔH positive for this reaction.

Entropy Change (ΔS)

Entropy change (ΔS) provides insight into the disorder or randomness of a system. In a chemical process, if the disorder increases, ΔS is positive; if the disorder decreases, ΔS is negative. This can generally be predicted by looking at the states of reactants and products.

Consider a reaction with one mole of gaseous reactants forming two moles of gaseous products. The system's entropy increases because gases have higher entropy than solids or liquids, leading to a positive ΔS. Conversely, if a gas reacts to form a solid, like in the production of calcium carbonate from carbon dioxide and calcium oxide, entropy decreases, resulting in a negative ΔS.

Equilibrium Constant (K)

The equilibrium constant (K) is a valuable indicator of the position of equilibrium in a chemical reaction. It mathematically describes the ratio of concentrations of products to reactants at equilibrium. If K > 1, the reaction favors the formation of products. If K < 1, reactants are favored.

For reactions like the formation of iron oxide, where both ΔH and ΔS are negative, the tendency is for K > 1, especially at lower temperatures. In contrast, reactions such as the dissociation of chlorine, with both ΔH and ΔS positive, may require higher temperatures for K to be greater than 1. The interplay of these variables can be analyzed further using equations such as Van't Hoff's equation to understand how changing temperatures might affect the equilibrium.

Exothermic and Endothermic Reactions

Exothermic and endothermic reactions are classifications based on the net release or absorption of heat energy. In exothermic reactions, energy is released to the surroundings, and the system loses heat, making ΔH negative. Combustion reactions are common examples of exothermic processes.

Endothermic reactions absorb energy from their surroundings and require added energy in the form of heat. Here, ΔH is positive, as observed in reactions like ammonium chloride dissolving into its constituent gases.

• Exothermic: Product formation is heat-releasing, ΔH is negative.
• Endothermic: Energy is needed for reactants to turn into products, ΔH is positive.

These definitions help predict and understand the energetics and potential spontaneity of chemical processes.