Question

Calculate \(\Delta S_{\text {surr }}\) for the following reactions at \(25^{\circ} \mathrm{C}\) and 1 atm. a. \(\mathrm{C}_{3} \mathrm{H}_{5}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 3 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(t) \Delta H^{\circ}=-2221 \mathrm{~kJ}\) b. \(2 \mathrm{NO}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)\) \(\Delta H^{\circ}=112 \mathrm{~kJ}\)

Short answer

For the given reactions at 25°C and 1 atm, the change in the entropy of the surroundings (\(\Delta S_{\text{surr}}\)) are: a) \(\Delta S_{\text{surr}} = 7.45\ \frac{kJ}{K}\) b) \(\Delta S_{\text{surr}} = -0.38\ \frac{kJ}{K}\)

Step-by-step solution

Convert temperature to Kelvin

Since the given temperature is 25°C, convert it to Kelvin: \(T = 25 + 273.15 = 298.15\ K\)

Calculate the heat flow

Using the given standard enthalpy change, \(\Delta H^{\circ} = -2221\ kJ\), calculate the heat flow as: \(q = \Delta H^{\circ} = -2221\ kJ\)

Calculate ΔSsurr

By using the formula for the change in the entropy of the surroundings, we get: \(\Delta S_{\text{surr}} = -\frac{q}{T} = -\frac{-2221}{298.15} = 7.45\ \frac{kJ}{K}\) **Reaction b:** Calculating ΔSsurr for \(2NO_2(g) \to 2NO(g) + O_2(g)\) Calculating the heat flow:

Step 1:Convert temperature to Kelvin

Since the given temperature is 25°C, convert it to Kelvin: \(T = 25 + 273.15 = 298.15\ K\)

Calculate the heat flow

Using the given standard enthalpy change, \(\Delta H^{\circ} = 112\ kJ\), calculate the heat flow as: \(q = \Delta H^{\circ} = 112\ kJ\)

Calculate ΔSsurr

By using the formula for the change in the entropy of the surroundings, we get: \(\Delta S_{\text{surr}} = -\frac{q}{T} = -\frac{112}{298.15} = -0.38\ \frac{kJ}{K}\) So, the change in the entropy of the surroundings (\(\Delta S_{\text{surr}}\)) for the reactions a and b are 7.45 kJ/K and -0.38 kJ/K respectively.

Key concepts

Standard Enthalpy Change

Understanding the Standard Enthalpy Change is crucial for students studying thermodynamics or chemistry. It refers to the heat change associated with a chemical reaction at standard conditions, which commonly includes a temperature of 298 K (25°C) and a pressure of 1 atmosphere. To grasp this concept, think of it as the energy change happening when substances interact chemically under controlled conditions.

Standard enthalpy change, represented as \( \Delta H^\circ \) in scientific equations, is measured in either kilojoules per mole (kJ/mol) or just kilojoules (kJ). Whether it's exothermic (releasing heat) or endothermic (absorbing heat) helps us understand the inherent energy dynamics of a reaction.

Thermodynamic Reactions

Delving into Thermodynamic Reactions, these are the bread and butter of chemical processes where energy is either given off or absorbed. The key factor is the relationship between heat (enthalpy) and disorder (entropy).

There are a few main types of thermodynamic reactions:

• Exothermic reactions: These release heat to the surroundings, often making them warmer. They typically have negative \( \Delta H^\circ \) values, signifying energy release.
• Endothermic reactions: In contrast, these absorb heat, which can make their surroundings cooler. Positive \( \Delta H^\circ \) values are indicative of heat absorption.

Thermodynamic reactions propel the progress of chemical reactions and determine the direction and feasibility of processes.

Gibbs Free Energy

Moving on to Gibbs Free Energy, symbolized as \( G \), this concept is a cornerstone of chemical thermodynamics. It is a measure of the maximum reversible work that a thermodynamic system can perform at constant temperature and pressure, and it's instrumental in predicting the direction of chemical reactions.

The equation that relates Gibbs Free Energy to enthalpy and entropy changes is \( G = H - TS \), where \( H \) is the enthalpy, \( T \) the temperature in Kelvin, and \( S \) the entropy. A negative \( \Delta G \) indicates a spontaneous process, which will proceed without any input of energy. This is an essential criterion for spontaneous reactions and understanding the balance between enthalpy, entropy, and free energy changes is crucial for mastering thermodynamics.

Chemical Equilibrium

Lastly, Chemical Equilibrium is a state where the rate of the forward reaction equals the rate of the reverse reaction. This means that the concentrations of the reactants and products remain unchanged over time - the system is dynamically balanced. The principle of Le Chatelier's contributes to understanding how changes in conditions (concentration, pressure, temperature) can shift the equilibrium position.

A fundamental equation associated with chemical equilibrium is the equilibrium constant expression \( K \) which relates the concentration of the products to the reactants at equilibrium. The Gibbs Free Energy is also related to the equilibrium constant through the relation \( \Delta G = -RT \ln K \) where \( R \) is the gas constant, and \( T \) the temperature. Comprehending chemical equilibrium is fundamental in predicting how reactions will behave in different conditions and is tightly knit with the other core concepts discussed.