   # The combustion of ethane, C 2 H 6 , has an enthalpy change of −2857.3 kJ for the reaction as written below. Calculate △ H ° for the combustion of 15.0 g of C 2 H 6 . 2 C 2 H 6 (g) + 7 O 2 (g) → 4 CO 2 (g) + 6 H 2 O(g) Δ r H ∘ = − 2857.3 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 5.5, Problem 1CYU
Textbook Problem
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## The combustion of ethane, C2H6, has an enthalpy change of −2857.3 kJ for the reaction as written below. Calculate △H° for the combustion of 15.0 g of C2H6.2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(g) Δ r H ∘ = − 2857.3  kJ/mol-rxn

Interpretation Introduction

Interpretation:

The change in enthalpy of ethane when combustion takes place has to be determined.

Concept Introduction:

Standard enthalpy of the reaction:

ΔrHo is the change in enthalpy that happens when matter is transformed by a given chemical reaction, when all reactants and products are in their standard states.

Enthalpy of the reaction:

ΔrH, is the change in enthalpy that happens when matter is transformed by a given chemical reaction

The change in enthalpy, ΔH in kJ per mole of a given reactant for the reaction can be calculated as:

ΔrH=enthalpy change/1000 number of moles

### Explanation of Solution

Given reaction is

2C2H6(g)+7O2(g)4CO2(g)+6H2O(g)ΔfH0=-2857.3kJ/mol

ΔrHo=-2857

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