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Account for the relationship between the rate of a reaction and its activation energy.

Short Answer

Expert verified

The rate of reaction depends upon the activation energy; as the activation energy is high reaction rate is low, whereas if the activation energy is low, the reaction rate is high.

Step by step solution

01

Reaction Rate

The reaction involved the effective collision of two reactants to produce the desired products. Reactions can be natural, which occur in the surrounding environment, whereas it can be artificially done in the laboratory to form the desired product.

The reaction rate can be defined as the reaction speed to produce the products. The reaction rate can be slow, fast or moderate. The reaction can take less than a millisecond to produce products, or it can take years to produce the desired product.

02

Activation Energy

'Activation energy' term was used by Svante Arrhenius, a Swedish scientist, in the year 1889. The activation energy can be defined as the threshold energy (which is called the minimum amount of energy) after which the reaction takes place. A chemical reaction can take place because of the effective collision of the two or more reactants, which produces energy more than the threshold energy (minimum energy).

The unit of the Activation energy is Joule or kcal/mole.

03

Relation between Activation energy and Reaction rate

The rate of the reaction depends upon the Activation energy. As the activation energy is high, then the reaction rate is low, whereas if the activation energy is low, then the reaction rate is high. Activation energy can be defined as the 鈥渂arrier鈥 which is crossed by the reactant to produce products.

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Most popular questions from this chapter

Tripling the concentration of a reactant increases the rate of a reaction nine-fold. With this knowledge, answer the following questions:

  1. What is the order of the reaction with respect to that reactant?
  2. Increasing the concentration of a reactant by a factor of four increases the rate of a reaction four-fold. What is the order of the reaction with respect to that reactant?

Use the provided initial rate data to derive the rate law for the reaction whose equation is: \({\bf{OC}}{{\bf{l}}^ - }\)(aq) + \({{\bf{I}}^ - }\)(aq) 鉄禣I(aq) +\({\bf{C}}{{\bf{l}}^ - }\)(aq)

Trial

(\({\bf{OC}}{{\bf{l}}^ - }\)) (mol/L)

(\({{\bf{I}}^ - }\)) (mol/L)

Initial Rate (mol/L/s)

1.

0.0040

0.0020

0.00184

2.

0.0020

0.0040

0.00092

3.

0.0020

0.0020

0.00046

Determine the rate law expression and the value of the rate constant k with appropriate units for this reaction.

Usethe data provided to graphically determine the order and rate constant of the following reaction: \({\bf{S}}{{\bf{O}}_{\bf{2}}}{\bf{C}}{{\bf{l}}_{\bf{2}}} \to {\bf{S}}{{\bf{O}}_{\bf{2}}}{\bf{ + C}}{{\bf{l}}_{\bf{2}}}\)

Time(hr)

0

5.00*\({\bf{1}}{{\bf{0}}^{\bf{3}}}\)

1.00*\({\bf{1}}{{\bf{0}}^{\bf{4}}}\)

1.50*\({\bf{1}}{{\bf{0}}^{\bf{4}}}\)

2.50*\({\bf{1}}{{\bf{0}}^{\bf{4}}}\)

3.00*104

4.00*104

\({\bf{(S}}{{\bf{O}}_{\bf{2}}}{\bf{C}}{{\bf{l}}_{\bf{2}}}{\bf{)}}\)(M)

0.100

0.0896

0.0802

0.0719

0.0577

0.0517

0.0415

Given the following reactions and the corresponding rate laws, in which of the reactions might the elementary reaction and the overall reaction be the same?

\(\begin{array}{c}{\rm{(a) C}}{{\rm{l}}_2}{\rm{ + CO }} \to {\rm{ C}}{{\rm{l}}_2}{\rm{CO}}\\{\rm{rate = }}k{{\rm{(C}}{{\rm{l}}_2}{\rm{)}}^{\frac{3}{2}}}{\rm{(CO)}}\\{\rm{(b) PC}}{{\rm{l}}_3}{\rm{ + C}}{{\rm{l}}_{\rm{2}}}{\rm{ }} \to {\rm{ PC}}{{\rm{l}}_{\rm{5}}}\\{\rm{rate = }}k{\rm{(PC}}{{\rm{l}}_{\rm{3}}}{\rm{) (C}}{{\rm{l}}_{\rm{2}}}{\rm{)}}\\{\rm{(c) 2NO + }}{{\rm{H}}_{\rm{2}}}{\rm{ }} \to {\rm{ }}{{\rm{N}}_{\rm{2}}}{\rm{ + }}{{\rm{H}}_{\rm{2}}}{\rm{O}}\\{\rm{rate = }}k{\rm{(NO)(}}{{\rm{H}}_{\rm{2}}}{\rm{)}}\\{\rm{(d) 2NO + }}{{\rm{O}}_{\rm{2}}}{\rm{ }} \to {\rm{ 2N}}{{\rm{O}}_{\rm{2}}}\\{\rm{rate = }}k{{\rm{(NO)}}^{\rm{2}}}{\rm{(}}{{\rm{O}}_{\rm{2}}}{\rm{)}}\\{\rm{(e) NO + }}{{\rm{O}}_{\rm{3}}}{\rm{ }} \to {\rm{ N}}{{\rm{O}}_{\rm{2}}}{\rm{ + }}{{\rm{O}}_{\rm{2}}}\\{\rm{rate = }}k{\rm{(NO)(}}{{\rm{O}}_{\rm{3}}}{\rm{)}}\end{array}\)

(a) Multiply 2.334 cm and 0.320 cm.(b) Divide 55.8752 m by 56.53 s.

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