Chapter 16: Q16.60P (page 731)

Question: At $25^{0} C$ , what is the fraction of collisions with energy equal to or greater than an activation energy of 100. kJ/mol?

Short Answer

Expert verified

The fraction of collisions is equal to $2.96 \times 10^{- 18}$ .

Step by step solution

Step 1: Arrhenius equation

The Arrhenius equation is written like this:

$k = A e^{- E_{a} / R T} k = A f$

Where k is the rate constant, A is the frequency factor, $E_{a}$ is the reaction's activation energy at a certain temperature T, and R is the gas constant .The equation f equals $e^{- E a / R T}$ and represents the proportion of collisions with a given energy.

Determine the fraction of collisions

Determine the fraction of collisions by substituting the values:

$\ begingathered f = e^{- E a / R T} = e^{\frac{100 k J / m o l}{(8.314 J / m o l . k) (273 + 25) K} \times \frac{1000 J}{1 k J}} = e^{- 40.36} = 2.96 \times 10^{- 18}$

The fraction of collisions is equal to $2.96 \times 10^{- 18}$ .

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Question: At $25^{0} C$ , what is the fraction of collisions with energy equal to or greater than an activation energy of 100. kJ/mol?

Short Answer

Step by step solution

Step 1: Arrhenius equation

Determine the fraction of collisions

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Question: At 250C , what is the fraction of collisions with energy equal to or greater than an activation energy of 100. kJ/mol?

Short Answer

Step by step solution

Step 1: Arrhenius equation

Determine the fraction of collisions

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Nuclear Chemistry

Kinetics

Chemistry Branches

Ionic and Molecular Compounds

Chemical Reactions

Study anywhere. Anytime. Across all devices.

Question: At $25^{0} C$ , what is the fraction of collisions with energy equal to or greater than an activation energy of 100. kJ/mol?