We know that the reverse of the reaction:H2O(g) ↔ H2(g) + 1⁄2O2(g) is rather explosive, yet there is still a finite equilibrium constant of 8.581×10-41 at 298 K. It’s hard to imagine how an equilibrium could exist! How could there be any hydrogen or oxygen gas left over the bang?a. To understand what is going on, first fill out the remaining entries of the table of mole fractions below like in the example from question 1. Let’s assume the pressure is maintained at a constant 1 bar, as such we don’t need to worry about partial pressures etc.b. Using the data above please write the expression for ?? = [(XH2) (Xo2 )^1/2 ] / XH2o using The table provides information on the amounts and mole fractions of different substances at equilibrium. It focuses on the substances H₂O, O₂, and H₂.**Table Overview:**- **Columns:** - **H₂O**: Water - **O₂**: Oxygen - **H₂**: Hydrogen- **Rows:** 1. **Amount at equilibrium:** - **H₂O:** \( n(1 - \alpha) \) - **O₂:** (No value provided) - **H₂:** \( n\alpha \) 2. **Mole fractions:** (Values are not provided for any of the substances)This table likely serves as a representation of a chemical equilibrium scenario involving the decomposition or formation of water, where \( n \) is the initial mole of a substance, and \( \alpha \) is the degree of dissociation or conversion.

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Answered: We know that the reverse of the…

Chemistry

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ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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We know that the reverse of the reaction:
H2O(g) ↔ H2(g) + 1⁄2O2(g)

is rather explosive, yet there is still a finite equilibrium constant of 8.581×10-41 at 298 K. It’s hard to imagine how an equilibrium could exist! How could there be any hydrogen or oxygen gas left over the bang?
a. To understand what is going on, first fill out the remaining entries of the table of mole fractions below like in the example from question 1. Let’s assume the pressure is maintained at a constant 1 bar, as such we don’t need to worry about partial pressures etc.

b. Using the data above please write the expression for ?? = [(X_H2) (X_o2 )^1/2 ] / X_H2ousing

The table provides information on the amounts and mole fractions of different substances at equilibrium. It focuses on the substances H₂O, O₂, and H₂.

**Table Overview:**

- **Columns:**
- **H₂O**: Water
- **O₂**: Oxygen
- **H₂**: Hydrogen

- **Rows:**
1. **Amount at equilibrium:**
- **H₂O:** \( n(1 - \alpha) \)
- **O₂:** (No value provided)
- **H₂:** \( n\alpha \)

2. **Mole fractions:** (Values are not provided for any of the substances)

This table likely serves as a representation of a chemical equilibrium scenario involving the decomposition or formation of water, where \( n \) is the initial mole of a substance, and \( \alpha \) is the degree of dissociation or conversion.

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