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Consider the following equations: 3 A + 6 B → 3 D Δ H = − 403 KJ/mol E + 2 F → A Δ H = − 105.2 KJ/mol C → E+3D Δ H = 64.8 K J / m o l Suppose the first equation is reversed and multiplied by 1 6 , the second and third equations are divided by 2, and the three adjusted equations are added. What is the net reaction and what is the overall heat of this reaction?

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Chemistry

10th Edition
Steven S. Zumdahl + 2 others
Publisher: Cengage Learning
ISBN: 9781305957404

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Section
BuyFindarrow_forward

Chemistry

10th Edition
Steven S. Zumdahl + 2 others
Publisher: Cengage Learning
ISBN: 9781305957404
Chapter 6, Problem 115AE
Textbook Problem
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Consider the following equations:

3 A + 6 B 3 D Δ H = 403 KJ/mol E + 2 F A Δ H = 105.2 KJ/mol C E+3D Δ H = 64.8 K J / m o l

Suppose the first equation is reversed and multiplied by 1 6 , the second and third equations are divided by 2, and the three adjusted equations are added. What is the net reaction and what is the overall heat of this reaction?

Interpretation Introduction

Interpretation: The enthalpy change ΔH of given reaction has to be calculated.

Concept Introduction:

Hess's Law:

  • The enthalpy change accompanying a chemical reaction is self-determining of the route by which the chemical reaction occurs.
  • Hess's Law is saying that if you convert reactants to productsthe overall enthalpy change will be exactly the same whether you do it in one or multiple steps.

    Formula:

    ΔH(total)ΔH(number of steps)......(1)

Rules:

  • The sign of ΔH reversedwhenthe reaction is reversed.
  • The quantity of reactants and products is directly proportional to the magnitude of ΔH in a reaction. If the coefficients in a balanced reaction are multiplied by an integer, the value of ΔHis multiplied by the same integer.

Explanation of Solution

Explanation

3A+6B3DΔH1=-403KJ/mol......(1)E+2FAΔH2=-105.2KJ/mol......(2)CE+3DΔH3=64.8KJ/mol......(3)

  • The first equation is reversed and multiplied by 16 .
  • Second and third equations are divided by 2.

Formula:

ΔH=ΔH1+ΔH2+ΔH3+.....(2)

ΔH=-16(-403)+12(-105.5)+12(64

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