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Calculate \(\Delta S_{\text { surr }}\) for the following reactions at \(25^{\circ} \mathrm{C}\) and 1 \(\mathrm{atm}\) . a. \(\mathrm{C}_{3} \mathrm{H}_{8}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 3 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)\Delta H^{\circ}=-2221 \mathrm{kJ}\) b. \(2 \mathrm{NO}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \qquad \Delta H^{\rho}=112 \mathrm{kJ}\)

Short Answer

Expert verified
For the given reactions, the change in entropy of the surroundings at \(25^{\circ}\mathrm{C}\) and 1 \(\mathrm{atm}\) are: a. \(\Delta S_{\text{surr}}^\text{(a)} = 7454\,\text{J}\,\mathrm{K^{-1}\,mol^{-1}}\) b. \(\Delta S_{\text{surr}}^\text{(b)} = -376\,\text{J}\,\mathrm{K^{-1}\,mol^{-1}}\)

Step by step solution

01

Convert Celsius to Kelvin

To convert the given temperature from Celsius to Kelvin, we add 273.15 to it. \(T_\text{K} = 25^\circ\text{C} + 273.15 = 298.15\,\text{K}\) #Step 2: Calculate \(\Delta S_{\text{surr}}\) for Reaction a#
02

Apply the formula for reaction a

Using the formula for \(\Delta S_{\text{surr}}\) and the provided values, we can calculate the change in entropy for reaction a: \(\Delta S_{\text{surr}}^\text{(a)} = -\frac{\Delta H^{\circ}_\text{(a)}}{T} = -\frac{-2221\,\text{kJ}\,\mathrm{mol^{-1}}}{298.15\,\text{K}}\) First, we need to convert \(\Delta H^{\circ}\) from kJ/mol to J/mol: \(\Delta H^{\circ}_\text{(a)} \times 1000 = -2221000\,\text{J}\,\mathrm{mol^{-1}}\) Now, we can plug in the values: \(\Delta S_{\text{surr}}^\text{(a)} = -\frac{-2221000\,\text{J}\,\mathrm{mol^{-1}}}{298.15\,\text{K}} = 7454\,\text{J}\,\mathrm{K^{-1}\,mol^{-1}}\) Therefore, the change in entropy of the surroundings for reaction a is: \(\Delta S_{\text{surr}}^\text{(a)} = 7454\,\text{J}\,\mathrm{K^{-1}\,mol^{-1}}\) #Step 3: Calculate \(\Delta S_{\text{surr}}\) for Reaction b#
03

Apply the formula for reaction b

Using the formula for \(\Delta S_{\text{surr}}\) and the provided values, we can calculate the change in entropy for reaction b: \(\Delta S_{\text{surr}}^\text{(b)} = -\frac{\Delta H^{\circ}_\text{(b)}}{T} = -\frac{112\,\text{kJ}\,\mathrm{mol^{-1}}}{298.15\,\text{K}}\) First, we need to convert \(\Delta H^{\circ}\) from kJ/mol to J/mol: \(\Delta H^{\circ}_\text{(b)} \times 1000 = 112000\,\text{J}\,\mathrm{mol^{-1}}\) Now, we can plug in the values: \(\Delta S_{\text{surr}}^\text{(b)} = -\frac{112000\,\text{J}\,\mathrm{mol^{-1}}}{298.15\,\text{K}} = -376\,\text{J}\,\mathrm{K^{-1}\,mol^{-1}}\) Therefore, the change in entropy of the surroundings for reaction b is: \(\Delta S_{\text{surr}}^\text{(b)} = -376\,\text{J}\,\mathrm{K^{-1}\,mol^{-1}}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Thermodynamics and Entropy Change
Thermodynamics is the study of energy, heat, and work, and how they interrelate. It addresses the transfer and transformation of energy in physical and chemical processes. An essential concept within thermodynamics is entropy, which is a measure of disorder or randomness in a system.

In thermodynamics, there is a principle known as the second law, which states that the total entropy of an isolated system always increases over time. When considering chemical reactions, it is crucial to evaluate not only the system itself but also its surroundings. Entropy change in the surroundings, denoted by \( \Delta S_{\text{surr}} \), can be calculated using the formula: \[ \Delta S_{\text{surr}} = -\frac{\Delta H^{\circ}}{T} \] where \( \Delta H^{\circ} \) stands for the standard change in enthalpy and \( T \) is the temperature in Kelvin.

This formula gives insight into how heat exchange affects entropy. If heat is released to the surroundings (exothermic process), entropy increases, whereas if heat is absorbed (endothermic process), entropy decreases. These calculations help us understand the energetic feasibility of reactions and processes.
The Role of Enthalpy in Thermodynamics
Enthalpy, represented as \( H \), is a thermodynamic property that reflects the total heat content of a system. It is especially significant when studying reactions at constant pressure. Entropy and enthalpy are closely related in thermodynamics because both affect energy distribution and balance in reactions.

In the context of entropy change for surroundings, enthalpy change \( \Delta H^{\circ} \) is used to identify whether a reaction gives off or absorbs heat. For example, if a reaction has a negative \( \Delta H^{\circ} \), it suggests exothermic processes, where energy is released. Conversely, a positive \( \Delta H^{\circ} \) indicates endothermic processes.

By calculating the quotient \(-\frac{\Delta H^{\circ}}{T}\), we quantify how much the enthalpy change influences entropy change, allowing us to deduce a reaction's impact on surroundings. Understanding the enthalpy change helps predict reaction feasibility and guides interpretations of energy transformations.
Kelvin Conversion in Thermodynamics Calculations
Converting temperatures to Kelvin is a foundational step in thermodynamic calculations. The Kelvin scale is absolute, starting at zero, which corresponds to absolute zero—a point where all thermal motion ceases. This is critical because many thermodynamic equations, including those for entropy change, require absolute temperatures.

To convert Celsius to Kelvin, you add 273.15 to the Celsius temperature. For instance, a temperature of \( 25^{\circ} C \) converts to Kelvin as: \[ T_{K} = 25 + 273.15 = 298.15\, K \] Using the Kelvin scale assures that our calculations for reaction spontaneity, energy distribution, and entropy are accurate.

Kelvin conversion is not just a mathematical step, but rather an integral part of applying thermodynamic principles at a universal scale. Ensuring all temperatures are in Kelvin is crucial for maintaining consistency and accuracy in scientific analysis across different conditions and contexts.

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Most popular questions from this chapter

Monochloroethane \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\right)\) can be produced by the direct reaction of ethane gas \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) with chlorine gas or by the reaction of ethylene gas \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\) with hydrogen chloride gas. The second reaction gives almost a 100\(\%\) yield of pure \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) at a rapid rate without catalysis. The first method requires light as an energy source or the reaction would not occur. Yet \(\Delta G^{\circ}\) for the first reaction is considerably more negative than \(\Delta G^{\circ}\) for the second reaction. Explain how this can be so.

At what temperatures will the following processes be spontaneous? a. \(\Delta H=-18 \mathrm{kJ}\) and \(\Delta S=-60 . \mathrm{J} / \mathrm{K}\) b. \(\Delta H=+18 \mathrm{kJ}\) and \(\Delta S=+60 . \mathrm{JK}\) c. \(\Delta H=+18 \mathrm{kJ}\) and \(\Delta S=-60 . \mathrm{J} / \mathrm{K}\) d. \(\Delta H=-18 \mathrm{kJ}\) and \(\Delta S=+60 . \mathrm{J} / \mathrm{K}\)

Liquid water at \(25^{\circ} \mathrm{C}\) is introduced into an evacuated, insulated vessel. Identify the signs of the following thermodynamic functions for the process that occurs: \(\Delta H, \Delta S, \Delta T_{\text { water }} \Delta S_{\text { surr }}\) \(\Delta S_{\text { univ }}\)

The synthesis of glucose directly from \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) and the synthesis of proteins directly from amino acids are both nonspontaneous processes under standard conditions. Yet it is necessary for these to occur for life to exist. In light of the second law of thermodynamics, how can life exist?

Which of the following reactions (or processes) are expected to have a negative value for \(\Delta S^{\circ} ?\) a. \(\mathrm{SiF}_{6}(a q)+\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{HF}(g)+\mathrm{SiF}_{4}(g)\) b. \(4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Al}_{2} \mathrm{O}_{3}(s)\) c. \(\mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{COCl}_{2}(g)\) d. \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)\) e. \(\mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\)

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