At high temperatures bromine molecules can dissociate into bromine atoms. For the reaction Br2(g) = 2 Br (g) Kp = 2.48 x 10~³ at 1650 °C. A 5.00 L vessel at 1650 °C is filled with Br2(g) at an initial pressure of 7.00 atm and allowed to come to equilibrium. What will be the pressure (in atm) of Br·(g) at equilibrium?
At high temperatures bromine molecules can dissociate into bromine atoms. For the reaction Br2(g) = 2 Br (g) Kp = 2.48 x 10~³ at 1650 °C. A 5.00 L vessel at 1650 °C is filled with Br2(g) at an initial pressure of 7.00 atm and allowed to come to equilibrium. What will be the pressure (in atm) of Br·(g) at equilibrium?
Chemistry
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![At high temperatures, bromine molecules can dissociate into bromine atoms. For the reaction:
\[ \text{Br}_2(g) \rightleftharpoons 2 \text{Br}\cdot(g) \]
\[ K_p = 2.48 \times 10^{-3} \text{ at } 1650^\circ \text{C}. \]
A 5.00 L vessel at \( 1650^\circ \text{C} \) is filled with \(\text{Br}_2(g) \) at an initial pressure of 7.00 atm and allowed to come to equilibrium. What will be the pressure (in atm) of \(\text{Br}\cdot(g) \) at equilibrium?
### Explanation:
**Equation in Context:**
- The reaction describes the dissociation of bromine gas (\( \text{Br}_2 \)) into bromine atoms (\( \text{Br}\cdot \)).
- The equilibrium constant (\( K_p \)) is given as \( 2.48 \times 10^{-3} \) at a temperature of \( 1650^\circ \text{C} \).
**Calculation Steps:**
- To find the pressure of bromine atoms (\( \text{Br}\cdot \)) at equilibrium, use the equilibrium expression derived from \( K_p \).
**Graph or Diagram:**
- The image contains a virtual calculator on the right side, showing buttons for numerical input and pressure unit "atm" at the top.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcf594692-d96b-4ff4-ac1a-31ce1d3fad0a%2F30cbc5f5-fb15-4e11-a153-044ff65e3902%2Fx3i56y_processed.png&w=3840&q=75)
Transcribed Image Text:At high temperatures, bromine molecules can dissociate into bromine atoms. For the reaction:
\[ \text{Br}_2(g) \rightleftharpoons 2 \text{Br}\cdot(g) \]
\[ K_p = 2.48 \times 10^{-3} \text{ at } 1650^\circ \text{C}. \]
A 5.00 L vessel at \( 1650^\circ \text{C} \) is filled with \(\text{Br}_2(g) \) at an initial pressure of 7.00 atm and allowed to come to equilibrium. What will be the pressure (in atm) of \(\text{Br}\cdot(g) \) at equilibrium?
### Explanation:
**Equation in Context:**
- The reaction describes the dissociation of bromine gas (\( \text{Br}_2 \)) into bromine atoms (\( \text{Br}\cdot \)).
- The equilibrium constant (\( K_p \)) is given as \( 2.48 \times 10^{-3} \) at a temperature of \( 1650^\circ \text{C} \).
**Calculation Steps:**
- To find the pressure of bromine atoms (\( \text{Br}\cdot \)) at equilibrium, use the equilibrium expression derived from \( K_p \).
**Graph or Diagram:**
- The image contains a virtual calculator on the right side, showing buttons for numerical input and pressure unit "atm" at the top.
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