Chemical Principles
Chemical Principles
8th Edition
ISBN: 9781305581982
Author: Steven S. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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Chapter 15, Problem 25E
Interpretation Introduction

Interpretation:

The rate law and the value of the rate constant for the rection H2SeO3(aq)+6I(aq)+4H+(aq)Se(s)+2I3(aq)+3H2O(l) should be calculated.

Concept Introduction:

Rate of reaction represents the change of concentration of a reactant or a product with respect to time. It can be expressed either by reduce amount of reactant in per unit time or increase amount of product in per unit time.

Expert Solution & Answer
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Answer to Problem 25E

The rate law and value of rate constant for the given rection is k[H2SeO3][I]3[H+]2 and 5.2×105mol5L5s1 respectively.

Explanation of Solution

The given reaction is H2SeO3(aq)+6I(aq)+4H+(aq)Se(s)+2I3(aq)+3H2O(l).

The initial concentration and rate constant is given as follows:

  Chemical Principles, Chapter 15, Problem 25E

The rate law for the given reaction is represented as follows:

  Rate=k[H2SeO3]a[I]b[H+]c ...(1)

Where,

  • k is the rate constant.
  • a, b and c is the order of the reaction with respect to H2SeO3 , I and H+.

Substitute the values of 1st row and 2nd in equation (1) and divide 1st row by 2nd row values as shown below:

   ( Rate )1 ( Rate )2= ( k [ H 2 SeO 3 ] a [ I ] b [ H + ] c )1 ( k [ H 2 SeO 3 ] a [ I ] b [ H + ] c )21.66× 10 73.33× 10 7= ( k ( 1.0× 10 4 ) a ( 2.0× 10 2 ) b ( 2.0× 10 2 ) c )1 ( k ( 2.0× 10 4 ) a ( 2.0× 10 2 ) b ( 2.0× 10 2 ) c )20.5=(0.5)a(0.5)1=(0.5)aa=1

Therefore, the order of reaction with respect to H2SeO3 is 1.

Substitute the values of 1st row and 6th in equation (1) and divide 1st row by 6th row values as shown below:

   ( Rate )1 ( Rate )6= ( k [ H 2 SeO 3 ] a [ I ] b [ H + ] c )1 ( k [ H 2 SeO 3 ] a [ I ] b [ H + ] c )61.66× 10 713.2× 10 7= ( k ( 1.0× 10 4 ) a ( 2.0× 10 2 ) b ( 2.0× 10 2 ) c )1 ( k ( 1.0× 10 4 ) a ( 4.0× 10 2 ) b ( 2.0× 10 2 ) c )60.125=(0.5)b(0.5)3=(0.5)bb=3

Therefore, the order of reaction with respect to I is 3.

Substitute the values of 6th row and 7th in equation (1) and divide 6th row by 7th row values as shown below:

   ( Rate )6 ( Rate )7= ( k [ H 2 SeO 3 ] a [ I ] b [ H + ] c )6 ( k [ H 2 SeO 3 ] a [ I ] b [ H + ] c )713.2× 10 73.36× 10 7= ( k ( 1.0× 10 4 ) a ( 4.0× 10 2 ) b ( 2.0× 10 2 ) c )6 ( k ( 1.0× 10 4 ) a ( 4.0× 10 2 ) b ( 1.0× 10 2 ) c )74.0=(2.0)c(2.0)2=(2.0)cc=2

Therefore, the order of reaction with respect to H+ is 2.

Substitute a = 1, b = 3 and c = 2 in equation (1).

  Rate=k[H2SeO3][I]3[H+]2

Therefore, the rate law for the reaction is k[H2SeO3][I]3[H+]2.

Substitute 1st row values of given table in equation (1).

  1.66×107molL1s1=k(1.0× 10 4molL 1)(2.0× 10 2mol L 1)3(2.0× 10 2mol L 1)21.66×107molL1s1=k(1.0× 10 4molL 1)(2.0× 10 2mol L 1)5k=1.66× 10 7molL 1s 1( 1.0× 10 4 mol L 1 ) ( 2.0× 10 2 mol L 1 )5=5.2×105mol5L5s1

Therefore, the value of rate constant for the given reaction is 5.2×105mol5L5s1.

Conclusion

The rate law and value of rate constant for the given rection is k[H2SeO3][I]3[H+]2 and 5.2×105mol5L5s1 respectively.

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Chapter 15 Solutions

Chemical Principles

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