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title: "The elementary gas-phase reaction between carbon monoxide and nitrogen dioxide is represented by the following equation: \\[ \\text{CO}(g) + \\text{NO}_2(g) \\rightarrow \\text{CO}_2(g) + \\text{NO}(g) \\] The diagram represents two different collision pathways between a \\(\\text{CO}\\) molecule and an \\(\\text{NO}_2\\) molecule in a reaction vessel where all colliding molecules have kinetic energy exceeding the activation energy, \\(E_a\\). Which of the following statements best explains whether both collision pathways will result in the formation of products?"
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url: "https://nerd-notes.com/ubq/119791/"
date_modified: "2026-08-21T08:12:21+00:00"
---

# The elementary gas-phase reaction between carbon monoxide and nitrogen dioxide is represented by the following equation: \[ \text{CO}(g) + \text{NO}_2(g) \rightarrow \text{CO}_2(g) + \text{NO}(g) \] The diagram represents two different collision pathways between a \(\text{CO}\) molecule and an \(\text{NO}_2\) molecule in a reaction vessel where all colliding molecules have kinetic energy exceeding the activation energy, \(E_a\). Which of the following statements best explains whether both collision pathways will result in the formation of products?

The elementary gas-phase reaction between carbon monoxide and nitrogen dioxide is represented by the following equation: \[ \text{CO}(g) + \text{NO}_2(g) \rightarrow \text{CO}_2(g) + \text{NO}(g) \] The diagram represents two different collision pathways between a \(\text{CO}\) molecule and an \(\text{NO}_2\) molecule in a reaction vessel where all colliding molecules have kinetic energy exceeding the activation energy, \(E_a\). Which of the following statements best explains whether both collision pathways will result in the formation of products?

![A grayscale particulate diagram showing two collision models in two separate adjacent rectangular panels labeled Collision Pathway 1 and Collision Pathway 2. A legend at the top specifies: solid black circle = C atom, open white circle = O atom, hatched circle = N atom. In Collision Pathway 1 on the left: one CO molecule consisting of one solid black circle bonded to one open white circle moves to the right, with the solid black C atom facing right; one NO2 molecule consisting of one central hatched N atom bonded to two open white O atoms moves to the left, with one open white O atom directly facing the approaching solid black C atom. In Collision Pathway 2 on the right: one CO molecule moves to the right with its open white O atom facing right; one NO2 molecule moves to the left with its central hatched N atom facing left toward the approaching open white O atom. Straight dashed arrows show the trajectory of each molecule toward the central collision point. No other particles, labels, text, or annotations appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1787299941-OqoAOL.jpg)

- **A.** No, because only Collision Pathway 1 has the proper steric orientation to form the new \(\text{C}-\text{O}\) bond necessary to produce the activated complex.
- **B.** No, because only Collision Pathway 2 possesses sufficient kinetic energy to overcome electrostatic repulsions between the approaching nuclei.
- **C.** Yes, because in both pathways the colliding molecules possess kinetic energy exceeding \(E_a\), which is the sole requirement for an effective collision.
- **D.** Yes, because the high collision energy causes the electron clouds to temporarily distort and bond regardless of the angle of molecular approach.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/119791/*