World of Chemistry, 3rd edition
World of Chemistry, 3rd edition
3rd Edition
ISBN: 9781133109655
Author: Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCoste
Publisher: Brooks / Cole / Cengage Learning
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Chapter 17.1, Problem 4RQ
Interpretation Introduction

Interpretation:

The reason, how grinding the solid in a heterogeneous reaction can speed up the reaction by using the collision theory, needs to be explained.

Concept introduction:

The chemical reaction depends on the collision between the reactants’ molecules.

Expert Solution & Answer
Check Mark

Answer to Problem 4RQ

Whenthe solid reactantis grinding into its smaller particles, the surface area will be increased. So, there is more surface area on which collision can occur,and then the reaction becomes faster. Hence,grinding the solid in a heterogeneous reaction can speed up the reaction on the basis of collision theory.

Explanation of Solution

When a solid reactant is used in a chemical reaction, first of all, it must be ground into smaller particles to increase the surface area. Once the surface area increased, more space is available for its molecules to attack. Hence, the number of collisions is increased and so, the reaction rate is also increased.

Conclusion

By grinding the solid into smaller particles, the surface area will be increased and hence, the reaction rate will also be increased.

Chapter 17 Solutions

World of Chemistry, 3rd edition

Ch. 17.2 - Prob. 4RQCh. 17.2 - Prob. 5RQCh. 17.3 - Prob. 1RQCh. 17.3 - Prob. 2RQCh. 17.3 - Prob. 3RQCh. 17.3 - Prob. 4RQCh. 17.3 - Prob. 5RQCh. 17.3 - Prob. 6RQCh. 17.3 - Prob. 7RQCh. 17 - Prob. 1ACh. 17 - Prob. 2ACh. 17 - Prob. 3ACh. 17 - Prob. 4ACh. 17 - Prob. 5ACh. 17 - Prob. 6ACh. 17 - Prob. 7ACh. 17 - Prob. 8ACh. 17 - Prob. 9ACh. 17 - Prob. 10ACh. 17 - Prob. 11ACh. 17 - Prob. 12ACh. 17 - Prob. 13ACh. 17 - Prob. 14ACh. 17 - Prob. 15ACh. 17 - Prob. 16ACh. 17 - Prob. 17ACh. 17 - Prob. 18ACh. 17 - Prob. 19ACh. 17 - Prob. 20ACh. 17 - Prob. 21ACh. 17 - Prob. 22ACh. 17 - Prob. 23ACh. 17 - Prob. 24ACh. 17 - Prob. 25ACh. 17 - Prob. 26ACh. 17 - Prob. 27ACh. 17 - Prob. 28ACh. 17 - Prob. 29ACh. 17 - Prob. 30ACh. 17 - Prob. 31ACh. 17 - Prob. 32ACh. 17 - Prob. 33ACh. 17 - Prob. 34ACh. 17 - Prob. 35ACh. 17 - Prob. 36ACh. 17 - Prob. 37ACh. 17 - Prob. 38ACh. 17 - Prob. 39ACh. 17 - Prob. 40ACh. 17 - Prob. 41ACh. 17 - Prob. 42ACh. 17 - Prob. 43ACh. 17 - Prob. 44ACh. 17 - Prob. 45ACh. 17 - Prob. 46ACh. 17 - Prob. 47ACh. 17 - Prob. 48ACh. 17 - Prob. 49ACh. 17 - Prob. 50ACh. 17 - Prob. 51ACh. 17 - Prob. 52ACh. 17 - Prob. 53ACh. 17 - Prob. 54ACh. 17 - Prob. 55ACh. 17 - Prob. 56ACh. 17 - Prob. 57ACh. 17 - Prob. 58ACh. 17 - Prob. 59ACh. 17 - Prob. 60ACh. 17 - Prob. 61ACh. 17 - Prob. 62ACh. 17 - Prob. 63ACh. 17 - Prob. 64ACh. 17 - Prob. 65ACh. 17 - Prob. 66ACh. 17 - Prob. 67ACh. 17 - Prob. 68ACh. 17 - Prob. 69ACh. 17 - Prob. 70ACh. 17 - Prob. 71ACh. 17 - Prob. 72ACh. 17 - Prob. 73ACh. 17 - Prob. 74ACh. 17 - Prob. 75ACh. 17 - Prob. 1STPCh. 17 - Prob. 2STPCh. 17 - Prob. 3STPCh. 17 - Prob. 4STPCh. 17 - Prob. 5STPCh. 17 - Prob. 6STPCh. 17 - Prob. 7STP
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