## Equilibrium Reaction Problem**The Reaction:**\[ 2 \text{NO}(g) + \text{Br}_2(g) \rightleftharpoons 2 \text{NOBr}(g) \]This reaction has an equilibrium constant, \( K_p = 109 \) at 25°C. **Given:**- Equilibrium partial pressure of \(\text{Br}_2\) = 0.0282 atm- Equilibrium partial pressure of \(\text{NOBr}\) = 0.0680 atm**Task:**Calculate the partial pressure of \(\text{NO}\) at equilibrium.**Equation:** Partial pressure = \(\_\_\_\_\_\) atm**Interactive Components:**- Submit Answer button- Try Another Version button**Navigation:**- Show Hint button- Previous and Next buttonsThis problem requires the application of the equilibrium constant expression for gaseous reactions to find the unknown partial pressure of \(\text{NO}\). The equilibrium expression for this reaction based on partial pressures is:\[ K_p = \frac{(\text{P}_{\text{NOBr}})^2}{(\text{P}_{\text{NO}})^2 \times \text{P}_{\text{Br}_2}} \]

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The reaction 2 NO(g) + Br2 (g) = 2 NOBr(g) has Kp = 109 at 25°C. If the equilibrium partial pressure of Br2 is 0.0282 atm and the equilibrium partial pressure of NOBr is 0.0680 atm, calculate the partial pressure of NO at equilibrium. Partial pressure = atm

The reaction 2 NO(g) + Br2 (g) = 2 NOBr(g) has Kp = 109 at 25°C. If the equilibrium partial pressure of Br2 is 0.0282 atm and the equilibrium partial pressure of NOBr is 0.0680 atm, calculate the partial pressure of NO at equilibrium. Partial pressure = atm

Chemistry

10th Edition

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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Question

## Equilibrium Reaction Problem

**The Reaction:**

\[ 2 \text{NO}(g) + \text{Br}_2(g) \rightleftharpoons 2 \text{NOBr}(g) \]

This reaction has an equilibrium constant, \( K_p = 109 \) at 25°C.

**Given:**

- Equilibrium partial pressure of \(\text{Br}_2\) = 0.0282 atm
- Equilibrium partial pressure of \(\text{NOBr}\) = 0.0680 atm

**Task:**

Calculate the partial pressure of \(\text{NO}\) at equilibrium.

**Equation:**

Partial pressure = \(\_\_\_\_\_\) atm

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**Navigation:**

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This problem requires the application of the equilibrium constant expression for gaseous reactions to find the unknown partial pressure of \(\text{NO}\). The equilibrium expression for this reaction based on partial pressures is:

\[ K_p = \frac{(\text{P}_{\text{NOBr}})^2}{(\text{P}_{\text{NO}})^2 \times \text{P}_{\text{Br}_2}} \]

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