Calculate the equilibrium constant for the following reaction at 60°C using the data at 298°K H₂O(g) + CH3OH(g) ↔ CO₂(g) + 3H₂(g) CH3OH(g) - 161.96 - 200.66 Ag 1,298 (kJ/mol) Ah 298 (kJ/mol) H₂O(g) - 228.57 241.82 CO₂(g) - 394.36 393.51 H₂(g) 0
Calculate the equilibrium constant for the following reaction at 60°C using the data at 298°K H₂O(g) + CH3OH(g) ↔ CO₂(g) + 3H₂(g) CH3OH(g) - 161.96 - 200.66 Ag 1,298 (kJ/mol) Ah 298 (kJ/mol) H₂O(g) - 228.57 241.82 CO₂(g) - 394.36 393.51 H₂(g) 0
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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![**Equilibrium Constant Calculation for the Reaction at 60°C**
**Reaction:**
\[ \text{H}_2\text{O(g)} + \text{CH}_3\text{OH(g)} \leftrightarrow \text{CO}_2\text{(g)} + 3\text{H}_2\text{(g)} \]
**Thermodynamic Data at 298K:**
| | \(\text{H}_2\text{O(g)}\) | \(\text{CH}_3\text{OH(g)}\) | \(\text{CO}_2\text{(g)}\) | \(\text{H}_2\text{(g)}\) |
|---------------|---------------------------|-----------------------------|---------------------------|-------------------------|
| \(\Delta G^\circ_{f,298}\) (kJ/mol) | -228.57 | -161.96 | -394.36 | 0 |
| \(\Delta H^\circ_{f,298}\) (kJ/mol) | -241.82 | -200.66 | -393.51 | 0 |
**Explanation:**
- **\(\Delta G^\circ_{f,298}\):** This is the standard Gibbs free energy of formation for each substance at 298 K.
- **\(\Delta H^\circ_{f,298}\):** This is the standard enthalpy of formation for each substance at 298 K.
These values are essential for calculating the equilibrium constant at a different temperature (60°C or 333 K) using the van 't Hoff equation. This involves determining the standard reaction Gibbs free energy change and using it to find \( K_{eq} \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7447eb07-fceb-4d7b-a8f5-73349ee04b18%2F3c7ae246-5979-421d-9d25-5f84dda2837e%2F23zh2i_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Equilibrium Constant Calculation for the Reaction at 60°C**
**Reaction:**
\[ \text{H}_2\text{O(g)} + \text{CH}_3\text{OH(g)} \leftrightarrow \text{CO}_2\text{(g)} + 3\text{H}_2\text{(g)} \]
**Thermodynamic Data at 298K:**
| | \(\text{H}_2\text{O(g)}\) | \(\text{CH}_3\text{OH(g)}\) | \(\text{CO}_2\text{(g)}\) | \(\text{H}_2\text{(g)}\) |
|---------------|---------------------------|-----------------------------|---------------------------|-------------------------|
| \(\Delta G^\circ_{f,298}\) (kJ/mol) | -228.57 | -161.96 | -394.36 | 0 |
| \(\Delta H^\circ_{f,298}\) (kJ/mol) | -241.82 | -200.66 | -393.51 | 0 |
**Explanation:**
- **\(\Delta G^\circ_{f,298}\):** This is the standard Gibbs free energy of formation for each substance at 298 K.
- **\(\Delta H^\circ_{f,298}\):** This is the standard enthalpy of formation for each substance at 298 K.
These values are essential for calculating the equilibrium constant at a different temperature (60°C or 333 K) using the van 't Hoff equation. This involves determining the standard reaction Gibbs free energy change and using it to find \( K_{eq} \).
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