   # Calculate Δ r H ° for the reaction 2 C(s) + 3 H 2 (g) + ½ O 2 (g) → C 2 H 5 OH( ℓ ) given the information below. C(s) + O 2 (g) → CO 2 (g) Δ r H ° = −393.5 kJ/mol-rxn 2 H 2 (g) + O 2 (g) → 2 H 2 O( ℓ ) Δ r H ° = −571.6 kJ/mol-rxn C 2 H 5 OH( ℓ ) + 3 O 2 (g) → 2 CO 2 (g) + 3 H 2 O( ℓ ) Δ r H °= −1367.5 kJ/mol-rxn ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 5, Problem 116SCQ
Textbook Problem
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## Calculate ΔrH° for the reaction2 C(s) + 3 H2(g) + ½ O2(g) → C2H5OH(ℓ)given the information below.C(s) + O2(g) → CO2(g)ΔrH° = −393.5 kJ/mol-rxn2 H2(g) + O2(g) → 2 H2O(ℓ)ΔrH° = −571.6 kJ/mol-rxnC2H5OH(ℓ) + 3 O2(g) → 2 CO2(g) + 3 H2O(ℓ)ΔrH°= −1367.5 kJ/mol-rxn

Interpretation Introduction

Interpretation:

The enthalpy change of reaction has to be determined.

Concept Introduction:

The standard enthalpy of formation is the enthalpy change for the formation of 1mol of the compound directly from its component elements in their standard states.

Enthalpy change for the reaction ΔrH0=ΣnΔfH0(products)-ΣnΔfH0(reactants)

### Explanation of Solution

Given equations are solved to get the required equation.

The 1st equation is multiplied by 2, we get,

2C+2O22CO2                                             ΔrH0=-787kJ/mol

Dividing equation 2 by 2, and multiplying by 3 we get

3H2+3/2O23H2O                                      ΔrH0=-857

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