Problem 99 - Chemistry Matter and Change

Question

Calculate \(\Delta G_{\text { system for each process, and state whether }}\) the process is spontaneous or nonspontaneous. a) \(\Delta H_{\text { system }}=145 \mathrm{kJ}, T=293 \mathrm{K}, \Delta S_{\mathrm{system}}=195 \mathrm{JK}\) b) \(\Delta H_{\text { system }}=-232 \mathrm{kJ}, T=273 \mathrm{K}, \Delta S_{\text { system }}=138 \mathrm{J} / \mathrm{K}\) c) \(\Delta H_{\text { system }}=-15.9 \mathrm{kJ}, T=373 \mathrm{K}, \Delta S_{\text { syyttem }}=-268 \mathrm{J} / \mathrm{K}\)

Step-by-Step Solution

Answer

a) \(\Delta G = 87.865 \: \text{kJ}\), the process is non-spontaneous. b) \(\Delta G = -269.674 \: \text{kJ}\), the process is spontaneous. c) \(\Delta G = 84.024 \: \text{kJ}\), the process is non-spontaneous.

Step 1: Identify the given values for process A

∆H = 145 kJ T = 293 K ∆S = 195 J/K

Step 2: Convert the units from J/K to kJ/K

The ∆S is given in J/K, but ∆H is in kJ. To have consistent units, convert ∆S to kJ/K: ∆S = 195 J/K * (1 kJ/1000 J) = 0.195 kJ/K

Step 3: Calculate ∆G for process A

Using the Gibbs free energy formula: ∆G = ∆H - T∆S ∆G = 145 kJ - (293 K)(0.195 kJ/K) ∆G = 145 kJ - 57.135 kJ ∆G = 87.865 kJ Since ∆G is positive, the process is non-spontaneous. #Process B#

Step 1: Identify the given values for process B

∆H = -232 kJ T = 273 K ∆S = 138 J/K

Step 2: Convert the units from J/K to kJ/K

∆S = 138 J/K * (1 kJ/1000 J) = 0.138 kJ/K

Step 3: Calculate ∆G for process B

∆G = ∆H - T∆S ∆G = -232 kJ - (273 K)(0.138 kJ/K) ∆G = -232 kJ - 37.674 kJ ∆G = -269.674 kJ Since ∆G is negative, the process is spontaneous. #Process C#

Step 1: Identify the given values for process C

∆H = -15.9 kJ T = 373 K ∆S = -268 J/K

Step 2: Convert the units from J/K to kJ/K

∆S = -268 J/K * (1 kJ/1000 J) = -0.268 kJ/K

Step 3: Calculate ∆G for process C

∆G = ∆H - T∆S ∆G = -15.9 kJ - (373 K)(-0.268 kJ/K) ∆G = -15.9 kJ + 99.924 kJ ∆G = 84.024 kJ Since ∆G is positive, the process is non-spontaneous.

Key Concepts

Spontaneity in Chemical Reactions

Understanding spontaneity in chemical reactions is key to predicting if a process will occur naturally without any outside intervention. This concept revolves around Gibbs Free Energy, represented by the symbol \( \Delta G \). A negative \( \Delta G \) suggests that a reaction is spontaneous, meaning it can occur on its own. Conversely, a positive \( \Delta G \) indicates a nonspontaneous process, which requires energy input to proceed.

Spontaneity doesn't mean that a reaction happens quickly. It just means it can happen without external energy. The relationship for this is given by:

• \( \Delta G = \Delta H - T\Delta S \)

\( \Delta H \) is the change in enthalpy, while \( T\Delta S \) incorporates temperature and change in entropy. These elements work together to determine the spontaneity of a process.

By examining the values of \( \Delta G \) calculated for different exercises, we can clarify whether a process is spontaneous or not, helping students see how theoretical concepts apply to real scenarios.

Enthalpy and Entropy

Enthalpy (\( \Delta H \)) and entropy (\( \Delta S \)) are fundamental components in understanding thermodynamics. Enthalpy change refers to the heat exchanged during a process at constant pressure. When \( \Delta H \) is negative, the process releases heat, termed exothermic. A positive \( \Delta H \) means it's endothermic, requiring heat.

Entropy measures disorder or randomness in a system. A positive \( \Delta S \) indicates more disorder, while a negative value suggests greater order. The universe favors systems moving toward higher entropy, aligning with the second law of thermodynamics.

• Enthalpy changes (\( \Delta H \)) account for whether energy is absorbed or released.
• Entropy changes (\( \Delta S \)) reflect the degree of randomness or order in the system.

Together, these parameters shape the spontaneity of a reaction. Both contribute to the Gibbs Free Energy equation, which dictates whether a reaction is likely to occur naturally. In our examples, we've seen various values of \( \Delta H \) and \( \Delta S \) guide the spontaneity analysis.

Thermodynamics Calculations

Thermodynamic calculations involving Gibbs Free Energy allow us to quantitatively assess and predict chemical reaction tendencies. The equation \( \Delta G = \Delta H - T\Delta S \) is vital for calculating whether a process is spontaneous or not.

First, consistent units are essential. If \( \Delta H \) is in kilojoules and \( \Delta S \) in joules per Kelvin, converting \( \Delta S \) to kilojoules per Kelvin is crucial for an accurate \( \Delta G \) calculation.

• Convert \( \Delta S \) from J/K to kJ/K by dividing by 1000.
• Apply the Gibbs equation to find \( \Delta G \).

For each scenario in the exercises, substituting the values into the equation gives insights into reaction spontaneity.

In understanding thermodynamic calculations, practice with different values ensures mastery. Analyzing examples like those presented helps students familiarize themselves with practical applications. By engaging with these calculations, one gains confidence in predicting chemical behavior in different scenarios.