Question

Calculate $\mathrm{\Delta }{S}_{\text{ surr }}$ for the following reactions at ${25}^{\circ }\mathrm{C}$ and 1 $\mathrm{atm}$ . a. ${\mathrm{C}}_3{\mathrm{H}}_8(g)+5{\mathrm{O}}_2(g)⟶3{\mathrm{CO}}_2(g)+4{\mathrm{H}}_2\mathrm{O}(l)\mathrm{\Delta }{H}^{\circ }=-2221\mathrm{kJ}$ b. $2{\mathrm{NO}}_2(g)⟶2\mathrm{NO}(g)+{\mathrm{O}}_2(g)\mathrm{\Delta }{H}^{\rho }=112\mathrm{kJ}$

Short answer

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

Step-by-step solution

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 $\mathrm{\Delta }{S}_{\text{surr}}$ for Reaction a#

Apply the formula for reaction a

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

Apply the formula for reaction b

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

Key concepts

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 $\mathrm{\Delta }{S}_{\text{surr}}$, can be calculated using the formula: $$\mathrm{\Delta }{S}_{\text{surr}}=-\frac{\mathrm{\Delta }{H}^{\circ }}{T}$$ where $\mathrm{\Delta }{H}^{\circ }$ 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 $\mathrm{\Delta }{H}^{\circ }$ is used to identify whether a reaction gives off or absorbs heat. For example, if a reaction has a negative $\mathrm{\Delta }{H}^{\circ }$, it suggests exothermic processes, where energy is released. Conversely, a positive $\mathrm{\Delta }{H}^{\circ }$ indicates endothermic processes.

By calculating the quotient $-\frac{\mathrm{\Delta }{H}^{\circ }}{T}$, 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: $$T_K=25+273.15=298.15K$$ 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.