calculate the change in gibbs free energy in kJ for the reaction below at 7l0.4°C when there is a pressure Of. O548atm Hz, o235atm Clz and 7.48atm Hl. For Hulg), AO$= -95.3K7/md.
calculate the change in gibbs free energy in kJ for the reaction below at 7l0.4°C when there is a pressure Of. O548atm Hz, o235atm Clz and 7.48atm Hl. For Hulg), AO$= -95.3K7/md.
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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![**Topic: Calculating Gibbs Free Energy Change**
**Objective:**
Learn how to calculate the change in Gibbs free energy for a chemical reaction using given conditions of temperature and pressure.
**Problem Statement:**
Calculate the change in Gibbs free energy \( \Delta G \) in kJ for the reaction below at 715°C when there is a pressure of 1.235 atm \( O_2 \) and 0.9548 atm \( H_2 \):
\[ 2H_2 (g) + O_2 (g) \leftrightarrow 2H_2O (g) \]
Given data:
- Standard Gibbs free energy change \( \Delta G^0 \) for \( H_2O(g) \) = -9.53 kJ/mol
**Explanation:**
This problem requires the application of thermodynamic principles to determine the Gibbs free energy change.
**Step-by-step Solution:**
1. **Write down the reaction:**
\[ 2H_2 (g) + O_2 (g) \leftrightarrow 2H_2O (g) \]
2. **Use the given standard Gibbs free energy change for water in its gaseous form:**
\[ \Delta G^0_{H2O(g)} = -9.53 \text{ kJ/mol} \]
3. **Convert the temperature from Celsius to Kelvin:**
\[ T(K) = 715°C + 273.15 = 988.15 \text{ K} \]
4. **Determine the reaction quotient \( Q \):**
\[ Q = \frac{(\text{Pressure of } H_2O)^2}{(\text{Pressure of } H_2)^2 \times (\text{Pressure of } O_2)} = \frac{(P_{H_2O})^2}{(0.9548 \text{ atm})^2 \times (1.235 \text{ atm})} \]
Since pressures of water are not given, proceed assuming \( P_{H_2O} \) can be derived or set under standard conditions if unavailable.
5. **Use the Gibbs free energy equation that relates \( \Delta G \) to \( \Delta G^0 \) and \( Q \):**
\[ \Delta G = \Delta G^0 + RT \ln Q](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F08c67838-b49a-475f-ad93-59e7305e5b28%2F2e115878-c14f-4f5a-a69f-5c8788fbb854%2F9fmc7xu_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Topic: Calculating Gibbs Free Energy Change**
**Objective:**
Learn how to calculate the change in Gibbs free energy for a chemical reaction using given conditions of temperature and pressure.
**Problem Statement:**
Calculate the change in Gibbs free energy \( \Delta G \) in kJ for the reaction below at 715°C when there is a pressure of 1.235 atm \( O_2 \) and 0.9548 atm \( H_2 \):
\[ 2H_2 (g) + O_2 (g) \leftrightarrow 2H_2O (g) \]
Given data:
- Standard Gibbs free energy change \( \Delta G^0 \) for \( H_2O(g) \) = -9.53 kJ/mol
**Explanation:**
This problem requires the application of thermodynamic principles to determine the Gibbs free energy change.
**Step-by-step Solution:**
1. **Write down the reaction:**
\[ 2H_2 (g) + O_2 (g) \leftrightarrow 2H_2O (g) \]
2. **Use the given standard Gibbs free energy change for water in its gaseous form:**
\[ \Delta G^0_{H2O(g)} = -9.53 \text{ kJ/mol} \]
3. **Convert the temperature from Celsius to Kelvin:**
\[ T(K) = 715°C + 273.15 = 988.15 \text{ K} \]
4. **Determine the reaction quotient \( Q \):**
\[ Q = \frac{(\text{Pressure of } H_2O)^2}{(\text{Pressure of } H_2)^2 \times (\text{Pressure of } O_2)} = \frac{(P_{H_2O})^2}{(0.9548 \text{ atm})^2 \times (1.235 \text{ atm})} \]
Since pressures of water are not given, proceed assuming \( P_{H_2O} \) can be derived or set under standard conditions if unavailable.
5. **Use the Gibbs free energy equation that relates \( \Delta G \) to \( \Delta G^0 \) and \( Q \):**
\[ \Delta G = \Delta G^0 + RT \ln Q
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