At a particular temperature a 2.00-L flask at equilibrium contains 5.10 x 10-4 moles of N₂, 2.50 x 10-5 moles of 02 and 2.00 x 10-2 moles of N₂O. Calculate K at this temperature for the reaction 2N₂ (g) + O₂(g) ⇒ 2N₂O(g) I K= If [N₂] = 2.00 × 10-4 M, [N₂O] = 0.200 M, and [0₂] = 0.00813 M, does this represent a system at equilibrium? Submit Answer 6 item attempts remaining Try Another Version

Chemistry
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**Equilibrium Calculation Exercise**

At a particular temperature, a 2.00-L flask at equilibrium contains:
- \(5.10 \times 10^{-4}\) moles of \(N_2\)
- \(2.50 \times 10^{-5}\) moles of \(O_2\)
- \(2.00 \times 10^{-2}\) moles of \(N_2O\)

Calculate the equilibrium constant, \(K\), at this temperature for the reaction:

\[ 2N_2(g) + O_2(g) \rightleftharpoons 2N_2O(g) \]

**Given:**

- \([N_2] = 2.00 \times 10^{-4} \, M\)
- \([N_2O] = 0.200 \, M\)
- \([O_2] = 0.00813 \, M\)

**Question:**

Does this represent a system at equilibrium?

[Submit Answer] [Try Another Version]

*6 item attempts remaining*
Transcribed Image Text:**Equilibrium Calculation Exercise** At a particular temperature, a 2.00-L flask at equilibrium contains: - \(5.10 \times 10^{-4}\) moles of \(N_2\) - \(2.50 \times 10^{-5}\) moles of \(O_2\) - \(2.00 \times 10^{-2}\) moles of \(N_2O\) Calculate the equilibrium constant, \(K\), at this temperature for the reaction: \[ 2N_2(g) + O_2(g) \rightleftharpoons 2N_2O(g) \] **Given:** - \([N_2] = 2.00 \times 10^{-4} \, M\) - \([N_2O] = 0.200 \, M\) - \([O_2] = 0.00813 \, M\) **Question:** Does this represent a system at equilibrium? [Submit Answer] [Try Another Version] *6 item attempts remaining*
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