---
title: "A student studies the equilibrium decomposition of nitrosyl chloride in a closed, rigid vessel at a certain temperature.  \\(2\\text{NOCl(g)} \\rightleftharpoons 2\\text{NO(g)}+\\text{Cl}_2\\text{(g)}\\)  At this temperature, \\(K_c=4.0\\times10^{-4}\\). The concentrations immediately after the gases are mixed, before appreciable reaction occurs, are shown below.  | Species | Initial concentration | |———|———————–| | \\(\\text{NOCl(g)}\\) | \\(0.10\\text{ M}\\) | | \\(\\text{NO(g)}\\) | \\(0.020\\text{ M}\\) | | \\(\\text{Cl}_2\\text{(g)}\\) | \\(0.10\\text{ M}\\) |  Which prediction and justification concerning the change in \\([\\text{NO}]\\) as the system approaches equilibrium are correct?"
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date_modified: "2026-08-19T12:40:30+00:00"
---

# A student studies the equilibrium decomposition of nitrosyl chloride in a closed, rigid vessel at a certain temperature.

\(2\text{NOCl(g)} \rightleftharpoons 2\text{NO(g)}+\text{Cl}_2\text{(g)}\)

At this temperature, \(K_c=4.0\times10^{-4}\). The concentrations immediately after the gases are mixed, before appreciable reaction occurs, are shown below.

| Species | Initial concentration |
|———|———————–|
| \(\text{NOCl(g)}\) | \(0.10\text{ M}\) |
| \(\text{NO(g)}\) | \(0.020\text{ M}\) |
| \(\text{Cl}_2\text{(g)}\) | \(0.10\text{ M}\) |

Which prediction and justification concerning the change in \([\text{NO}]\) as the system approaches equilibrium are correct?

A student studies the equilibrium decomposition of nitrosyl chloride in a closed, rigid vessel at a certain temperature.

\(2\text{NOCl(g)} \rightleftharpoons 2\text{NO(g)}+\text{Cl}_2\text{(g)}\)

At this temperature, \(K_c=4.0\times10^{-4}\). The concentrations immediately after the gases are mixed, before appreciable reaction occurs, are shown below.

| Species | Initial concentration |
|---------|-----------------------|
| \(\text{NOCl(g)}\) | \(0.10\text{ M}\) |
| \(\text{NO(g)}\) | \(0.020\text{ M}\) |
| \(\text{Cl}_2\text{(g)}\) | \(0.10\text{ M}\) |

Which prediction and justification concerning the change in \([\text{NO}]\) as the system approaches equilibrium are correct?

- **A.** No net shift occurs because \(Q_c=\dfrac{(0.020)^2(0.10)}{0.10}=4.0\times10^{-4}=K_c\).
- **B.** The system shifts right and \([\text{NO}]\) increases because \(Q_c=4.0\times10^{-3}>K_c\), so the forward reaction is favored.
- **C.** The system shifts left and \([\text{NO}]\) decreases because preparing a product-rich mixture decreases the value of \(K_c\).
- **D.** The system shifts left and \([\text{NO}]\) decreases because \(Q_c=4.0\times10^{-3}>K_c\), so the net reverse reaction consumes products.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/119458/*