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Chapter 12: Q11E (page 702)

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 鈥淪how Bonds鈥� under Options.

  1. Leave the Initial Temperature at the default setting. Observe the reaction. Is the rate of reaction fast or slow?
  2. Click 鈥淧ause鈥� and then 鈥淩eset All,鈥� and then enter 15 molecules of A and 10 molecules of BC once again. Select 鈥淪how Bonds鈥� 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

Short Answer

Expert verified
  1. At default temperature, the rate of reaction is slow.
  2. On increasing temperature, the rate of reaction becomes fast.

Step by step solution

01

 Default temperature 

At the default temperature, the rate of reaction is slow. Molecules of A collide with BC molecules quite frequently; less molecule of them has sufficient energy to make a bond.

At default temperature, reactant molecules have low kinetic energy and low velocity, so the collision of reactant molecules are not much effective in making more bonds. Hence, the formation of product molecules occurs at a slow rate. Therefore, the rate of reaction is slow.

02

High temperature 

With the increase in temperature, the particle average velocity increases. The average kinetic energy of particles is also increased. The result is that the particle will collide more frequently and move around faster. This results in the encounter of more reactant particles.

Thus, on increasing temperature, the reaction rate increases because the reactant molecules collide at a faster rate at the high temperature, and their colloids result in the formation of products.

The molecule of reactants decreases when the molecule of products increases. After a while, there is ideally an equal amount of BC, AB, and C in the mixture with a slightly large amount of A.

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Most popular questions from this chapter

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 鈥淩eload Launcher鈥� and change to 鈥淎ngled shot鈥� 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.

What is the rate equation for the elementary termolecular reaction A + 2B鉄秔roducts? For 3A鉄秔roducts?

What is the difference between average rate, initial rate, and instantaneous rate?

Regular flights of supersonic aircraft in the stratosphere are of concern because such aircraft produce nitric oxide, NO, as a by-product in the exhaust of their engines. Nitric oxide reacts with ozone, and it has been suggested that this could contribute to depletion of the ozone layer. The reaction \({\bf{NO + }}{{\bf{O}}_{\bf{3}}} \to {\bf{N}}{{\bf{O}}_{\bf{2}}}{\bf{ + }}{{\bf{O}}_{\bf{2}}}\) is first order with respect to both NO and \({{\bf{O}}_{\bf{3}}}\) with a rate constant of \({\bf{2}}{\bf{.20 \times 1}}{{\bf{0}}^{\bf{7}}}{\bf{mol}}{{\bf{L}}^{{\bf{ - 1}}}}{{\bf{s}}^{{\bf{ - 1}}}}\). What is the instantaneous rate of disappearance of NO when \(\left( {{\bf{NO}}} \right){\bf{ = 3}}{\bf{.3 \times 1}}{{\bf{0}}^{{\bf{ - 6}}}}{\bf{ M}}\) and \({\bf{(}}{{\bf{O}}_{\bf{3}}}{\bf{) = 5}}{\bf{.9 \times 1}}{{\bf{0}}^{{\bf{ - 7}}}}{\bf{ M}}\)?

The rate constant for the rate of decomposition of \({{\bf{N}}_{\bf{2}}}{{\bf{O}}_{\bf{5}}}\)to\({\bf{NO}}\) and \({{\bf{O}}_{\bf{2}}}\)in the gas phase is 1.66 L/mol/s at 650 K and 7.39 L/mol/s at 700 K:

\({\bf{2}}{{\bf{N}}_{\bf{2}}}{{\bf{O}}_{\bf{5}}}{\bf{(g) - - - 4NO(g) + 3}}{{\bf{O}}_{\bf{2}}}{\bf{(g)}}\)

Assuming the kinetics of this reaction are consistent with the Arrhenius equation, calculate the activation energy for this decomposition.
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