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title: "At a certain temperature, the equilibrium constant \\(K_c\\) for the reaction represented below is \\(5.0\\).  \\[ 2\\,\\text{NO}_2\\text{(g)} \\rightleftharpoons \\text{N}_2\\text{O}_4\\text{(g)} \\]  A rigid \\(1.0\\text{ L}\\) vessel is charged with \\(0.50\\text{ mol}\\) of \\(\\text{NO}_2\\text{(g)}\\) and \\(2.0\\text{ mol}\\) of \\(\\text{N}_2\\text{O}_4\\text{(g)}\\). Which of the following correctly predicts the direction in which the reaction will proceed to reach equilibrium and provides the valid justification?"
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url: "https://nerd-notes.com/ubq/119712/"
date_modified: "2026-08-21T08:12:07+00:00"
---

# At a certain temperature, the equilibrium constant \(K_c\) for the reaction represented below is \(5.0\).

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

A rigid \(1.0\text{ L}\) vessel is charged with \(0.50\text{ mol}\) of \(\text{NO}_2\text{(g)}\) and \(2.0\text{ mol}\) of \(\text{N}_2\text{O}_4\text{(g)}\). Which of the following correctly predicts the direction in which the reaction will proceed to reach equilibrium and provides the valid justification?

At a certain temperature, the equilibrium constant \(K_c\) for the reaction represented below is \(5.0\).

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

A rigid \(1.0\text{ L}\) vessel is charged with \(0.50\text{ mol}\) of \(\text{NO}_2\text{(g)}\) and \(2.0\text{ mol}\) of \(\text{N}_2\text{O}_4\text{(g)}\). Which of the following correctly predicts the direction in which the reaction will proceed to reach equilibrium and provides the valid justification?

- **A.** The forward direction, because \(Q_c = 4.0\) and \(Q_c < K_c\).
- **B.** The forward direction, because \(Q_c = 8.0\) and \(Q_c > K_c\).
- **C.** The reverse direction, because \(Q_c = 0.13\) and \(Q_c < K_c\).
- **D.** The reverse direction, because \(Q_c = 8.0\) and \(Q_c > K_c\).

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/119712/*