Question

Use Table 16.1 to calculate \(\Delta S^{\circ}\) for each of the following reactions: (a) \(4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\) (b) \(2 \mathrm{H}_{2} \mathrm{O}_{2}(l)+\mathrm{N}_{2} \mathrm{H}_{4}(l) \longrightarrow \mathrm{N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)\) (c) \(\mathrm{C}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)\) (d) \(\mathrm{CH}_{4}(g)+3 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CHCl}_{3}(l)+3 \mathrm{HCl}(g)\)

Short answer

For the given chemical reactions, we calculated the standard entropy change, \(\Delta S^{\circ}\), as follows: (a) \(\Delta S^{\circ} = 316.4 \, \mathrm{J/mol \cdot K}\) (b) \(\Delta S^{\circ} = 606.9 \, \mathrm{J/mol \cdot K}\) (c) \(\Delta S^{\circ} = -2.16 \, \mathrm{J/mol \cdot K}\) (d) \(\Delta S^{\circ} = -94.5 \, \mathrm{J/mol \cdot K}\)

Step-by-step solution

(a) Calculating \(\Delta S^{\circ}\) for reaction (a)

First, we will identify the species in the reaction and find their standard entropies in Table 16.1: - 4 NH3(g): \(S^{\circ} = 192.5 \, \mathrm{J/mol \cdot K}\) - 5 O2(g): \(S^{\circ} = 205.2 \, \mathrm{J/mol \cdot K}\) - 4 NO(g): \(S^{\circ} = 210.7 \, \mathrm{J/mol \cdot K}\) - 6 H2O(g): \(S^{\circ} = 188.7\, \mathrm{J/mol \cdot K}\) Now, we will calculate \(\Delta S^{\circ}\) for the reaction using the above entropies: \(\Delta S^{\circ} = (4 \times 210.7 + 6 \times 188.7) - (4 \times 192.5 + 5 \times 205.2)\) \(\Delta S^{\circ} = 2899.6 - 2583.2 = 316.4 \, \mathrm{J/mol \cdot K}\)

(b) Calculating \(\Delta S^{\circ}\) for reaction (b)

Similarly, we will identify the species in the reaction and find their standard entropies in Table 16.1: - 2 H2O2(l): \(S^{\circ} = 109.6 \, \mathrm{J/mol \cdot K}\) - N2H4(l): \(S^{\circ} = 121.2 \, \mathrm{J/mol \cdot K}\) - N2(g): \(S^{\circ} = 191.5 \, \mathrm{J/mol \cdot K}\) - 4 H2O(g): \(S^{\circ} = 188.7\, \mathrm{J/mol \cdot K}\) Now, calculate \(\Delta S^{\circ}\) for the reaction: \(\Delta S^{\circ} = (191.5 + 4 \times 188.7) - (2 \times 109.6 + 121.2)\) \(\Delta S^{\circ} = 946.3 - 339.4 = 606.9 \, \mathrm{J/mol \cdot K}\)

(c) Calculating \(\Delta S^{\circ}\) for reaction (c)

Identify the species in the reaction and find their standard entropies in Table 16.1: - C(s): \(S^{\circ} = 5.74 \, \mathrm{J/mol \cdot K}\) - O2(g): \(S^{\circ} = 205.2 \, \mathrm{J/mol \cdot K}\) - CO2(g): \(S^{\circ} = 213.7 \, \mathrm{J/mol \cdot K}\) Calculate \(\Delta S^{\circ}\) for the reaction: \(\Delta S^{\circ} = 213.7 - (5.74 + 205.2)\) \(\Delta S^{\circ} = -2.16 \, \mathrm{J/mol \cdot K}\)

(d) Calculating \(\Delta S^{\circ}\) for reaction (d)

Identify the species in reaction (d) and find their standard entropies in Table 16.1: - CH4(g): \(S^{\circ} = 186.2 \, \mathrm{J/mol \cdot K}\) - 3 Cl2(g): \(S^{\circ} = 223.0 \, \mathrm{J/mol \cdot K}\) - CHCl3(l): \(S^{\circ} = 199.0 \, \mathrm{J/mol \cdot K}\) - 3 HCl(g): \(S^{\circ} = 186.9 \, \mathrm{J/mol \cdot K}\) Calculate \(\Delta S^{\circ}\) for the reaction: \(\Delta S^{\circ} = (199.0 + 3 \times 186.9) - (186.2 + 3 \times 223.0)\) \(\Delta S^{\circ} = 760.7 - 855.2 = -94.5 \, \mathrm{J/mol \cdot K}\)

Key concepts

Thermodynamics

Thermodynamics is a branch of physics that deals with the relationships between heat and other forms of energy. In the context of chemistry, it plays a crucial role in understanding how substances interact with one another during chemical reactions. The first law of thermodynamics, also known as the law of energy conservation, states that energy cannot be created or destroyed, only transformed from one form to another. In chemical reactions, this means the total energy of the reactants must be equal to the total energy of the products plus or minus any energy transferred to or from the surroundings.

Entropy is an essential concept in thermodynamics, representing the degree of disorder or randomness in a system. The second law of thermodynamics states that the total entropy of an isolated system can never decrease over time, and as such, natural processes prefer states with higher entropy. When we calculate the standard entropy change \( \Delta S^{\circ} \) for a chemical reaction, we are determining how much the entropy of the system changes from the reactants to the products under standard conditions. This gives us insight into the spontaneity of the reaction; typically, reactions with positive entropy changes are more likely to occur spontaneously.

Chemical Reactions

Chemical reactions involve the transformation of one set of chemical substances to another through the breaking and forming of chemical bonds. Reactants are transformed into products, and this process involves changes in energy, specifically in the form of enthalpy and entropy. During these transformations, we observe conservation of mass, where the amount of each element remains the same before and after the reaction, a principle known as the law of conservation of mass.

To analyze the thermodynamic properties of chemical reactions, scientists use the standard state, which refers to a specified set of conditions including a pressure of 1 bar and a concentration of 1 molar, generally at a temperature of 25°C (298 K). Under these conditions, we can calculate the standard change in enthalpy \( \Delta H^{\circ} \) and the standard entropy change \( \Delta S^{\circ} \) for a reaction. These values can be looked up in tables like Table 16.1 mentioned in the original exercise. By using these standard values, we can predict the behavior of a chemical reaction and determine whether it will release energy (exothermic) or absorb energy (endothermic), as well as whether it will lead to an increase or decrease in disorder.

Enthalpy and Entropy

Enthalpy (\( H \)) and entropy (\( S \)) are two fundamental thermodynamic quantities that often dictate the direction and extent of chemical reactions. Enthalpy is a measure of the total heat content of a system, while entropy measures the distribution of energy within it and its overall disorder. The change in enthalpy (\( \Delta H \)) during a reaction indicates whether the process is exothermic (releasing heat) or endothermic (absorbing heat).

The standard change in enthalpy (\( \Delta H^{\circ} \)) and standard entropy change (\( \Delta S^{\circ} \)) of a reaction can be calculated from tabulated values for individual substances under standard conditions. When calculating the standard entropy change for a reaction, we take the sum of the standard entropies of the products and subtract the sum of the standard entropies of the reactants. Mathematically, this is expressed as \( \Delta S^{\circ} = \sum S^{\circ}_{products} - \sum S^{\circ}_{reactants} \). It's important to understand that a positive entropy change suggests an increase in disorder during the reaction, often favoring the reaction to proceed, whereas a negative entropy change suggests a decrease in disorder. Coupled with the enthalpy change, entropy helps to predict the feasibility of reactions through the Gibbs free energy equation \( \Delta G = \Delta H - T\Delta S \), identifying whether reactions are favorable under certain conditions.