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Chapter 12: Q9E (page 702)

Go to the PhET Reactions & Rates interactive. Use the Single Collision tab to represent how the collision between monatomic oxygen (O) and carbon monoxide (CO) results in the breaking of one bond and the formation of another. Pull back on the red plunger to release the atom and observe the results. Then, click on “Reload Launcher” and change to “Angled shot” to see the difference.

  1. What happens when the angle of the collision is changed?
  2. Explain how this is relevant to rate of reaction.

Short Answer

Expert verified
  1. When angle of collision is changed it affects the formation of bonds and breaking of bonds between monoatomic oxygen and carbon monoxide.
  2. Angle of collision increases or decreases the rate of reaction.

Step by step solution

01

Angle of collision

In order for a reaction to occur, the particles of the reactant must have enough energy, and must collide at the correct angle.

Collision between monoatomic oxygen and carbon monoxide results in breaking of one bond and the formation of another bond. When angle of the collision is changed the collisions between monoatomic oxygen and carbon monoxide effect in way that the collisions may be increase or decrease resulting in the formation and breaking of bonds. It depends on the angle of collision or proper orientation.

02

Relevant to the rate of reaction

Molecules of reactants must come into contact with each other before reaction. If the angle of collision is changed resulting in the collisions between monoatomic oxygen and carbon monoxide increase, the rate of reaction or formation and breaking of bonds increases.

If angle of collision is changed resulting in the collisions between monoatomic oxygen and carbon monoxide decrease, the rate of reaction or formation and breaking of bonds decreases.

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Most popular questions from this chapter

The rate constant for the first-order decomposition at 45 °C of dinitrogen pentoxide, N2O5, dissolved in chloroform, \({\bf{CHC}}{{\bf{l}}_3}\), is 6.2 × 10−4 min−1.

\({\bf{2}}{{\bf{N}}_{\bf{2}}}{{\bf{O}}_{\bf{5}}} \to {\bf{4N}}{{\bf{O}}_{\bf{2}}}{\bf{ + }}{{\bf{O}}_{\bf{2}}}\)

What is the rate of the reaction when \({{\bf{N}}_{\bf{2}}}{{\bf{O}}_{\bf{5}}}{\bf{\; = 0}}{\bf{.40 M}}\)

What is the half-life for the decomposition of \({{\bf{O}}_{\bf{3}}}\) when the concentration of \({{\bf{O}}_{\bf{3}}}\)is \({\bf{2}}{\bf{.35 \times 1}}{{\bf{0}}^{{\bf{ - 6}}}}\)M? The rate constant for this second-order reaction is 50.4 L/mol/h.

In the PhET Reactions & Rates (http://openstaxcollege.org/l/16PHETreaction) interactive, on the Many Collisions tab, set up a simulation with 15 molecules of A and 10 molecules of BC. Select “Show Bonds” under Options.

  1. Leave the Initial Temperature at the default setting. Observe the reaction. Is the rate of reaction fast or slow?
  2. Click “Pause” and then “Reset All,” and then enter 15 molecules of A and 10 molecules of BC once again. Select “Show Bonds” under Options. This time, increase the initial temperature until, on the graph, the total average energy line is completely above the potential energy curve. Describe what happens to the reaction

Explain why an egg cooks more slowly in boiling water in Denver than in New York City. (Hint: Consider the effect of temperature on reaction rate and the effect of pressure on boiling point.)

Acetaldehyde decomposes when heated to yield methane and carbon monoxide according to the equation: \({\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{CHO}}\)(g) ⟶\({\bf{C}}{{\bf{H}}_{\bf{4}}}\)(g) +\({\bf{CO}}\)(g)

Determine the rate law and the rate constant for the reaction from the following experimental data:

Trial

(\({\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{CHO}}\)) (mol/L)

\(\frac{{ - \Delta \left( {{\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{CHO}}} \right)}}{{\Delta t}}\)(mol )(Ls−1)

1.

1.75 × 10−3

2.06 × 10−11

2.

3.50 × 10−3

8.24 × 10−11

3.

7.00 × 10−3

3.30 × 10−10
See all solutions

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