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Use the PhET Reactions & Rates interactive simulation to simulate a system. On the “Single collision” tab of the simulation applet, enable the “Energy view” by clicking the “+” icon. Select the first A + BC⟶AB + C reaction (A is yellow, B is purple, and C is navy blue). Using the “straight shot” default option, try launching the A atom with varying amounts of energy. What changes when the Total Energy line at launch is below the transition state of the Potential Energy line? Why? What happens when it is above the transition state? Why?

Short Answer

Expert verified

When the total energy line is below the transition state of the potential line, the reaction does not occur. Above the transition potential energy line, the reaction proceeds in the forward direction.

Step by step solution

01

Energy of Activation

For any single step chemical reaction, the peak of the transition state energy represents the Energy of Activation, Eafor the given reaction. For the forward reaction to occur, the reactants must have total energy \({\bf{E}}\left( {\bf{T}} \right) \ge {{\bf{E}}_{\bf{a}}}\).

02

Total energy less than transition state energy

Initially, when the total energy line is below the transition state potential energy the reactants have total energy less than the activation energy, and the reaction \({\bf{A + BC}} \to {\bf{AB + C}}\) does not occur.

03

Total energy greater than transition state energy

Above the transition potential energy line, the reactants have energy greater than the activation energy, so the reaction proceeds in the forward direction.

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Most popular questions from this chapter

The hydrolysis of the sugar sucrose to the sugars glucose and fructose, \({{\bf{C}}_{{\bf{12}}}}{{\bf{H}}_{{\bf{22}}}}{{\bf{O}}_{{\bf{11}}}}{\bf{ + }}{{\bf{H}}_{\bf{2}}}{\bf{O}} \to {{\bf{C}}_{\bf{6}}}{{\bf{H}}_{{\bf{12}}}}{{\bf{O}}_{\bf{6}}}{\bf{ + }}{{\bf{C}}_{\bf{6}}}{{\bf{H}}_{{\bf{12}}}}{{\bf{O}}_{\bf{6}}}\) follows a first-order rate equation for the disappearance of sucrose: \({\bf{Rate = k}}\left( {{{\bf{C}}_{{\bf{12}}}}{{\bf{H}}_{{\bf{22}}}}{{\bf{O}}_{{\bf{11}}}}} \right)\) (The products of the reaction, glucose and fructose, have the same molecular formulas but differ in the arrangement of the atoms in their molecules.)

  1. In neutral solution, \({\bf{k = 2}}{\bf{.1 \times 1}}{{\bf{0}}^{{\bf{ - 11}}}}{{\bf{s}}^{{\bf{ - 1}}}}\) at 27 °C and \({\bf{8}}{\bf{.5 \times 1}}{{\bf{0}}^{{\bf{ - 11}}}}{{\bf{s}}^{{\bf{ - 1}}}}\) at 37 °C. Determine the activation energy, the frequency factor, and the rate constant for this equation at 47 °C (assuming the kinetics remain consistent with the Arrhenius equation at this temperature).
  2. When a solution of sucrose with an initial concentration of 0.150 M reaches equilibrium, the concentration of sucrose is\({\bf{1}}{\bf{.65 \times 1}}{{\bf{0}}^{{\bf{ - 7}}}}{\bf{ M}}\). How long will it take the solution to reach equilibrium at 27 °C in the absence of a catalyst? Because the concentration of sucrose at equilibrium is so low, assume that the reaction is irreversible.
  3. Why does assuming that the reaction is irreversible simplify the calculation in part (b)?

From the given data, use a graphical method to determine the order and rate constant of the following reaction: 2X⟶Y + Z

Time(s)

5.0

10.0

15.0

20.0

25.0

30.0

35.0.

40.0

(X)(M)

0.0990

0.0497

0.0332

0.0249

0.0200

0.0166

0.0143

0.0125

Iodine-131 is a radioactive isotope that is used to diagnose and treat some forms of thyroid cancer. Iodine-131 decays to xenon-131 according to the equation:I-131⟶Xe-131 + electron. The decay is first-order with a rate constant of 0.138 d−1. All radioactive decay is first order. How many days will it take for 90% of the iodine−131 in a 0.500 M solution of this substance to decay to Xe-131?

Why are elementary reactions involving three or more reactants very uncommon?

Experiments were conducted to study the rate of the reaction represented by this equation.(2)\({\rm{2NO(g) + 2}}{{\rm{H}}_{\rm{2}}}{\rm{(g) }} \to {\rm{ }}{{\rm{N}}_{\rm{2}}}{\rm{(g) + 2}}{{\rm{H}}_{\rm{2}}}{\rm{O(g)}}\)Initial concentrations and rates of reaction are given here.

Experiment Initial Concentration

\(\left( {{\bf{NO}}} \right){\rm{ }}\left( {{\bf{mol}}/{\bf{L}}} \right)\)

Initial Concentration, \(\left( {{{\bf{H}}_{\bf{2}}}} \right){\rm{ }}\left( {{\bf{mol}}/{\bf{L}}} \right)\)Initial Rate of Formation of \({{\bf{N}}_{\bf{2}}}{\rm{ }}\left( {{\bf{mol}}/{\bf{L}}{\rm{ }}{\bf{min}}} \right)\)
1\({\bf{0}}.{\bf{0060}}\)\({\bf{0}}.{\bf{00}}1{\bf{0}}\)\({\bf{1}}.{\bf{8}} \times {\bf{1}}{{\bf{0}}^{ - {\bf{4}}}}\)
2\({\bf{0}}.{\bf{0060}}\)\({\bf{0}}.{\bf{00}}2{\bf{0}}\)\({\bf{3}}.{\bf{6}} \times {\bf{1}}{{\bf{0}}^{ - {\bf{4}}}}\)
3\({\bf{0}}.{\bf{00}}1{\bf{0}}\)\({\bf{0}}.{\bf{0060}}\)\({\bf{0}}.{\bf{30}} \times {\bf{1}}{{\bf{0}}^{ - {\bf{4}}}}\)
4\({\bf{0}}.{\bf{00}}2{\bf{0}}\)\({\bf{0}}.{\bf{0060}}\)\({\bf{1}}.{\bf{2}} \times {\bf{1}}{{\bf{0}}^{ - {\bf{4}}}}\)

Consider the following questions:(a) Determine the order for each of the reactants, \({\bf{NO}}\) and \({{\bf{H}}_{\bf{2}}}\), from the data given and show your reasoning.(b) Write the overall rate law for the reaction.(c) Calculate the value of the rate constant, k, for the reaction. Include units.(d) For experiment 2, calculate the concentration of \({\bf{NO}}\)remaining when exactly one-half of the original amount of \({{\bf{H}}_{\bf{2}}}\) had been consumed.(e) The following sequence of elementary steps is a proposed mechanism for the reaction.Step 1:Step 2:Step 3:Based on the data presented, which of these is the rate determining step? Show that the mechanism is consistent with the observed rate law for the reaction and the overall stoichiometry of the reaction.

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