---
title: "Two gas-phase equilibria and their equilibrium constants at \\(298\\text{ K}\\) are shown below.  \\[2\\text{NO}(g) + \\text{O}_2(g) \\rightleftharpoons 2\\text{NO}_2(g) \\quad K_1 = 1.6 \\times 10^5\\]  \\[\\text{N}_2\\text{O}_4(g) \\rightleftharpoons 2\\text{NO}_2(g) \\quad K_2 = 4.0 \\times 10^{-4}\\]  What is the value of the equilibrium constant, \\(K\\), for the reaction below at \\(298\\text{ K}\\)?  \\[\\text{NO}(g) + \\dfrac{1}{2}\\text{O}_2(g) \\rightleftharpoons \\dfrac{1}{2}\\text{N}_2\\text{O}_4(g)\\]"
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date_modified: "2026-08-23T05:04:54+00:00"
---

# Two gas-phase equilibria and their equilibrium constants at \(298\text{ K}\) are shown below.

\[2\text{NO}(g) + \text{O}_2(g) \rightleftharpoons 2\text{NO}_2(g) \quad K_1 = 1.6 \times 10^5\]

\[\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g) \quad K_2 = 4.0 \times 10^{-4}\]

What is the value of the equilibrium constant, \(K\), for the reaction below at \(298\text{ K}\)?

\[\text{NO}(g) + \dfrac{1}{2}\text{O}_2(g) \rightleftharpoons \dfrac{1}{2}\text{N}_2\text{O}_4(g)\]

Two gas-phase equilibria and their equilibrium constants at \(298\text{ K}\) are shown below.

\[2\text{NO}(g) + \text{O}_2(g) \rightleftharpoons 2\text{NO}_2(g) \quad K_1 = 1.6 \times 10^5\]

\[\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g) \quad K_2 = 4.0 \times 10^{-4}\]

What is the value of the equilibrium constant, \(K\), for the reaction below at \(298\text{ K}\)?

\[\text{NO}(g) + \dfrac{1}{2}\text{O}_2(g) \rightleftharpoons \dfrac{1}{2}\text{N}_2\text{O}_4(g)\]

- **A.** \(2.0 \times 10^4\)
- **B.** \(1.0 \times 10^6\)
- **C.** \(8.0 \times 10^6\)
- **D.** \(4.0 \times 10^8\)

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/121507/*