Consider the following reaction: 2NOBr(g) 2NO(g) + Br₂ (9) for this question. If 0.340 moles of NOBr(g), 0.327 moles of NO, and 0.499 moles of Br2 are at equilibrium in a 12.3 L container at 445 K, the value of the equilibrium constant, Ke, is

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**Equilibrium Constant Calculation**

**Consider the following reaction:**

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

**Given:**
- 0.340 moles of NOBr(g)
- 0.327 moles of NO
- 0.499 moles of \(\text{Br}_2\)

All substances are at equilibrium in a 12.3 L container at 445 K. 

**Find:**
- The value of the equilibrium constant, \( K_c \). 

(Note: There are no graphs or diagrams associated with this content.)
Transcribed Image Text:**Equilibrium Constant Calculation** **Consider the following reaction:** \[ 2 \text{NOBr}(g) \rightleftharpoons 2 \text{NO}(g) + \text{Br}_2(g) \] **Given:** - 0.340 moles of NOBr(g) - 0.327 moles of NO - 0.499 moles of \(\text{Br}_2\) All substances are at equilibrium in a 12.3 L container at 445 K. **Find:** - The value of the equilibrium constant, \( K_c \). (Note: There are no graphs or diagrams associated with this content.)
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