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

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## 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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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:

\[ K_p = \frac{(\text{P}_{\text{NOBr}})^2}{(\text{P}_{\text{NO}})^2 \times \text{P}_{\text{Br}_2}} \]
Transcribed Image Text:## 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 buttons 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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