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The standard enthalpy of formation of H 2 O( l ) at 298 K is −285.8 kJ/mol. Calculate the change in internal energy for the following process at 298 K and 1 atm: H 2 O ( l ) → H 2 ( g ) + 1 2 O 2 ( g ) Δ E ∘ = ? ( Hint: Using the ideal gas equation, derive an expression for work in terms of n, R, and T. )

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Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

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Section
BuyFindarrow_forward

Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 16, Problem 104CP
Textbook Problem
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The standard enthalpy of formation of H2O(l) at 298 K is −285.8 kJ/mol. Calculate the change in internal energy for the following process at 298 K and 1 atm:

H 2 O ( l ) H 2 ( g ) + 1 2 O 2 ( g ) Δ E = ?

(Hint: Using the ideal gas equation, derive an expression for work in terms of n, R, and T.)

Interpretation Introduction

Interpretation: The dissociation reaction of H2O , standard enthalpy of formation of H2O(l) at 298K is given. The change in internal energy for the given reaction is to be calculated.

Concept introduction: The expression of change in internal energy is given by,

ΔE=ΔHΔngRT

The expression of Δng is,

Δng=(Numberofgaseousmoleculesonproduct)(Numberofgaseousmoleculesonreactant)

Explanation of Solution

The stated reaction is,

H2O(l)H2(g)+12O2(g)

The expression of Δng is,

Δng=(Numberofgaseousmoleculesonproduct)(Numberofgaseousmoleculesonreactant)

The number of gaseous molecules in the product side of the given reaction is 32 .

The number of gaseous molecules in the reactant side of the given reaction is 0 .

Substitute the values of the number of gaseous molecules of reactant and product in the above expression.

Δng=(Numberofgaseousmoleculesonproduct)(Numberofgaseousmoleculesonreactant)Δng=320=32_

Given

Standard enthalpy of formation H2O(l) is 285.8kJ/mol .

The stated reaction is,

H2O(l)H2(g)+12O2(g)

The value of Δng is 32 .

The change in enthalpy of H2(g) and O2(g) is zero.

Formula

Standard enthalpy of stated reaction is calculated as,

ΔH=ΔHproductΔHreactant

Where,

  • ΔH is the change in enthalpy of given reaction.
  • ΔHproduct is the change in enthalpy of product.
  • ΔHreactant is the change in enthalpy of reactant.

Substitute the value of ΔHproduct and ΔHreactant in the above equation.

ΔH=ΔHproductΔHreactant=0(285.8kJ/mol)=285

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Chapter 16 Solutions

Chemistry: An Atoms First Approach
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