---
title: "The reaction between \\(\\text{NO(g)}\\) and \\(\\text{Cl}_2\\text{(g)}\\) proceeds according to the equation below.  \\[ 2\\text{NO(g)} + \\text{Cl}_2\\text{(g)} \\rightarrow 2\\text{NOCl(g)} \\]  Based on experimental data, the reaction is first order with respect to \\(\\text{Cl}_2\\text{(g)}\\) and second order with respect to \\(\\text{NO(g)}\\). In a second trial conducted at the same temperature, the initial concentration of \\(\\text{Cl}_2\\text{(g)}\\) is doubled and the initial concentration of \\(\\text{NO(g)}\\) is tripled. By what factor does the initial rate of the reaction increase?"
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url: "https://nerd-notes.com/ubq/119686/"
date_modified: "2026-08-21T08:12:03+00:00"
---

# The reaction between \(\text{NO(g)}\) and \(\text{Cl}_2\text{(g)}\) proceeds according to the equation below.

\[ 2\text{NO(g)} + \text{Cl}_2\text{(g)} \rightarrow 2\text{NOCl(g)} \]

Based on experimental data, the reaction is first order with respect to \(\text{Cl}_2\text{(g)}\) and second order with respect to \(\text{NO(g)}\). In a second trial conducted at the same temperature, the initial concentration of \(\text{Cl}_2\text{(g)}\) is doubled and the initial concentration of \(\text{NO(g)}\) is tripled. By what factor does the initial rate of the reaction increase?

The reaction between \(\text{NO(g)}\) and \(\text{Cl}_2\text{(g)}\) proceeds according to the equation below.

\[ 2\text{NO(g)} + \text{Cl}_2\text{(g)} \rightarrow 2\text{NOCl(g)} \]

Based on experimental data, the reaction is first order with respect to \(\text{Cl}_2\text{(g)}\) and second order with respect to \(\text{NO(g)}\). In a second trial conducted at the same temperature, the initial concentration of \(\text{Cl}_2\text{(g)}\) is doubled and the initial concentration of \(\text{NO(g)}\) is tripled. By what factor does the initial rate of the reaction increase?

- **A.** It increases by a factor of \(6\).
- **B.** It increases by a factor of \(9\).
- **C.** It increases by a factor of \(12\).
- **D.** It increases by a factor of \(18\).

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/119686/*