---
title: "The decomposition of nitrogen monoxide gas is represented by the equation below.  \\[ 2\\,\\text{NO(g)} \\xrightarrow{\\text{Pt(s)}} \\text{N}_2\\text{(g)} + \\text{O}_2\\text{(g)} \\]  In the absence of a catalyst, the reaction has a very high activation energy and proceeds extremely slowly at \\(500\\text{ K}\\). In the presence of a solid platinum catalyst, \\(\\text{Pt(s)}\\), the reaction proceeds rapidly at the same temperature. The diagram illustrates a proposed particulate model for the process occurring on the platinum surface.  Which of the following best explains how the \\(\\text{Pt}\\) surface increases the rate of the reaction?"
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url: "https://nerd-notes.com/ubq/123743/"
date_modified: "2026-09-28T12:02:24+00:00"
---

# The decomposition of nitrogen monoxide gas is represented by the equation below.

\[ 2\,\text{NO(g)} \xrightarrow{\text{Pt(s)}} \text{N}_2\text{(g)} + \text{O}_2\text{(g)} \]

In the absence of a catalyst, the reaction has a very high activation energy and proceeds extremely slowly at \(500\text{ K}\). In the presence of a solid platinum catalyst, \(\text{Pt(s)}\), the reaction proceeds rapidly at the same temperature. The diagram illustrates a proposed particulate model for the process occurring on the platinum surface.

Which of the following best explains how the \(\text{Pt}\) surface increases the rate of the reaction?

The decomposition of nitrogen monoxide gas is represented by the equation below.

\[ 2\,\text{NO(g)} \xrightarrow{\text{Pt(s)}} \text{N}_2\text{(g)} + \text{O}_2\text{(g)} \]

In the absence of a catalyst, the reaction has a very high activation energy and proceeds extremely slowly at \(500\text{ K}\). In the presence of a solid platinum catalyst, \(\text{Pt(s)}\), the reaction proceeds rapidly at the same temperature. The diagram illustrates a proposed particulate model for the process occurring on the platinum surface.

Which of the following best explains how the \(\text{Pt}\) surface increases the rate of the reaction?

![A grayscale particulate diagram displays three sequential rectangular frames labeled Stage 1, Stage 2, and Stage 3 from left to right. A legend maps: large gray circle with crosshatching = Pt atom, small black circle = N atom, small white circle = O atom. In each frame, a single horizontal row of six identical large gray crosshatched circles forms the solid bottom surface. In Stage 1, two NO molecules (each consisting of one black circle bonded to one white circle) are positioned at the surface, with dashed vertical lines extending between the N atoms and the top of the Pt atoms. In Stage 2, the intramolecular bonds are broken; two black circles are bonded together as an N2 molecule and two white circles are bonded together as an O2 molecule directly above the surface. In Stage 3, the N2 molecule and O2 molecule are positioned near the top of the frame away from the intact row of Pt atoms. No other particles, labels, text, or annotations appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1790596944-zxyEmg.jpg)

- **A.** Adsorption of \(\text{NO}\) molecules onto the \(\text{Pt}\) surface weakens the \(\text{N}-\text{O}\) bonds, providing an alternative reaction pathway with a lower activation energy.
- **B.** Adsorption of \(\text{NO}\) molecules onto the \(\text{Pt}\) surface increases the average kinetic energy of the reactant molecules, increasing the frequency of successful collisions.
- **C.** Interaction with the \(\text{Pt}\) surface increases the thermodynamic stability of the products, making \(\Delta H^\circ\) for the overall reaction significantly more negative.
- **D.** The \(\text{Pt}\) surface reacts chemically with \(\text{NO}\) to form a permanent intermediate compound, shifting the position of chemical equilibrium toward the products.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/123743/*