Chemistry: An Atoms-Focused Approach
Chemistry: An Atoms-Focused Approach
14th Edition
ISBN: 9780393912340
Author: Thomas R. Gilbert, Rein V. Kirss, Natalie Foster
Publisher: W. W. Norton & Company
Question
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Chapter 19, Problem 19.57QA
Interpretation Introduction

To estimate:

The Hrxn for hydrogenation of acetylene (C2H2) with one mole of hydrogen gas to make ethylene (C2H4).

Expert Solution & Answer
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Answer to Problem 19.57QA

Solution:

The Hrxn for hydrogenation of acetylene (C2H2) with one mole of hydrogen gas to make ethylene (C2H4) is -174.30 kJ.

Explanation of Solution

1) Concept:

We can calculate the Hrxn from the difference between the standard molar enthalpies of reactants and products. We are given two equations with thermochemical data and a third reaction for which we are asked to find H0. We can manipulate the equations algebraically and then add to give the equation for which H0 is unknown. Then we can calculate the unknown value by applying Hess’s law.

2) Formula:

Hrxn=nproducts×Hf,products0-nreactants×Hf,reactants0

3) Given:

The given reactions are

Equation A: HCCH g+2H2gCH3CH3(g)

Equation B: H2C=CH2 g+H2gCH3CH3(g)

Final reaction: HCCH g+H2gH2C=CH2 g

Hf0(kJ/mol)
HCCH g 226.7
H2g 0.0
CH3CH3(g) -84.67
H2C=CH2 g 52.4

4) Calculation:

Calculation for Hrxn for equation A:

HCCH g+2H2gCH3CH3(g)

Hf0(kJ/mol)
HCCH g 226.7
H2g 0.0
CH3CH3(g) -84.67

Hrxn=nproducts×Hf,products0-nreactants×Hf,reactants0

Hrxn=1 mol×Hf,CH3CH3(g)0-1 mol×Hf,HCCH g0+2 mol×Hf,H2g0

Hrxn=1 mol×-84.67kJmol-1 mol×226.7kJmol+2 mol×0.0kJmol

Hrxn=-84.67-(226.7)

Hrxn=-311.37 kJ

Equation A: HCCH g+2H2gCH3CH3g,     Hrxn=-311.37 kJ

Calculation for Hrxn for equation B:

H2C=CH2 g+H2gCH3CH3(g)

Hf0(kJ/mol)
H2C=CH2 g 52.4
H2g 0.0
CH3CH3(g) -84.67

Hrxn=nproducts×Hf,products0-nreactants×Hf,reactants0

Hrxn=1 mol×Hf,CH3CH3(g)0-1 mol×Hf,H2C=CH2 g0+1 mol×Hf,H2g0

Hrxn=1 mol×-84.67kJmol-1 mol×52.4kJmol+1 mol×0.0kJmol

Hrxn=-84.67-(52.4)

Hrxn=-137.07 kJ

Equation B: H2C=CH2 g+H2gCH3CH3g,     Hrxn=-137.07 kJ

For the final reaction, the ethylene (H2C=CH2) is on the product side. Therefore, we reverse the equation B. Reversing equation B means we need to reverse the sign of Hrxn of equation B. When we add equations A and B, ethane (CH3CH3) gets cancelled out and we get equation C.

Equation A: HCCH g+2H2gCH3CH3g,     Hrxn=-311.37 kJ

Equation B: CH3CH3gH2C=CH2 g+H2g,     Hrxn=137.07 kJ

--------------------------------------------------------------------------------------------------

Final reaction: HCCH g+H2gH2C=CH2 g,     Hrxn=-174.30 kJ

Hrxn for the hydrogenation acetylene with one mole of hydrogen gas to make ethylene is -174.30 kJ

Conclusion:

Applying Hf0 values from Appendix 4 and Hess’s law, the Hrxn for hydrogenation of acetylene (C2H2) with one mole of hydrogen gas to make ethylene (C2H4) is calculated.

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Chapter 19 Solutions

Chemistry: An Atoms-Focused Approach

Ch. 19 - Prob. 19.11VPCh. 19 - Prob. 19.12VPCh. 19 - Prob. 19.13QACh. 19 - Prob. 19.14QACh. 19 - Prob. 19.15QACh. 19 - Prob. 19.16QACh. 19 - Prob. 19.17QACh. 19 - Prob. 19.18QACh. 19 - Prob. 19.19QACh. 19 - Prob. 19.20QACh. 19 - Prob. 19.21QACh. 19 - Prob. 19.22QACh. 19 - Prob. 19.23QACh. 19 - Prob. 19.24QACh. 19 - Prob. 19.25QACh. 19 - Prob. 19.26QACh. 19 - Prob. 19.27QACh. 19 - Prob. 19.28QACh. 19 - Prob. 19.29QACh. 19 - Prob. 19.30QACh. 19 - Prob. 19.31QACh. 19 - Prob. 19.32QACh. 19 - Prob. 19.33QACh. 19 - Prob. 19.34QACh. 19 - Prob. 19.35QACh. 19 - Prob. 19.36QACh. 19 - Prob. 19.37QACh. 19 - Prob. 19.38QACh. 19 - Prob. 19.39QACh. 19 - Prob. 19.40QACh. 19 - Prob. 19.41QACh. 19 - Prob. 19.42QACh. 19 - Prob. 19.43QACh. 19 - Prob. 19.44QACh. 19 - Prob. 19.45QACh. 19 - Prob. 19.46QACh. 19 - Prob. 19.47QACh. 19 - Prob. 19.48QACh. 19 - Prob. 19.49QACh. 19 - Prob. 19.50QACh. 19 - Prob. 19.51QACh. 19 - Prob. 19.52QACh. 19 - Prob. 19.53QACh. 19 - Prob. 19.54QACh. 19 - Prob. 19.55QACh. 19 - Prob. 19.56QACh. 19 - Prob. 19.57QACh. 19 - Prob. 19.58QACh. 19 - Prob. 19.59QACh. 19 - Prob. 19.60QACh. 19 - Prob. 19.61QACh. 19 - Prob. 19.62QACh. 19 - Prob. 19.63QACh. 19 - Prob. 19.64QACh. 19 - Prob. 19.65QACh. 19 - Prob. 19.66QACh. 19 - Prob. 19.67QACh. 19 - Prob. 19.68QACh. 19 - Prob. 19.69QACh. 19 - Prob. 19.70QACh. 19 - Prob. 19.71QACh. 19 - Prob. 19.72QACh. 19 - Prob. 19.73QACh. 19 - Prob. 19.74QACh. 19 - Prob. 19.75QACh. 19 - Prob. 19.76QACh. 19 - Prob. 19.77QACh. 19 - Prob. 19.78QACh. 19 - Prob. 19.79QACh. 19 - Prob. 19.80QACh. 19 - Prob. 19.81QACh. 19 - Prob. 19.82QACh. 19 - Prob. 19.83QACh. 19 - Prob. 19.84QACh. 19 - Prob. 19.85QACh. 19 - Prob. 19.86QACh. 19 - Prob. 19.87QACh. 19 - Prob. 19.88QACh. 19 - Prob. 19.89QACh. 19 - Prob. 19.90QACh. 19 - Prob. 19.91QACh. 19 - Prob. 19.92QACh. 19 - Prob. 19.93QACh. 19 - Prob. 19.94QACh. 19 - Prob. 19.95QACh. 19 - Prob. 19.96QACh. 19 - Prob. 19.97QACh. 19 - Prob. 19.98QACh. 19 - Prob. 19.99QACh. 19 - Prob. 19.100QACh. 19 - Prob. 19.101QACh. 19 - Prob. 19.102QACh. 19 - Prob. 19.103QACh. 19 - Prob. 19.104QACh. 19 - Prob. 19.105QACh. 19 - Prob. 19.106QACh. 19 - Prob. 19.107QACh. 19 - Prob. 19.108QACh. 19 - Prob. 19.109QACh. 19 - Prob. 19.110QACh. 19 - Prob. 19.111QACh. 19 - Prob. 19.112QACh. 19 - Prob. 19.113QACh. 19 - Prob. 19.114QACh. 19 - Prob. 19.115QACh. 19 - Prob. 19.116QACh. 19 - Prob. 19.117QACh. 19 - Prob. 19.118QACh. 19 - Prob. 19.119QACh. 19 - Prob. 19.120QACh. 19 - Prob. 19.121QACh. 19 - Prob. 19.122QACh. 19 - Prob. 19.123QACh. 19 - Prob. 19.124QACh. 19 - Prob. 19.125QACh. 19 - Prob. 19.126QACh. 19 - Prob. 19.127QACh. 19 - Prob. 19.128QACh. 19 - Prob. 19.129QACh. 19 - Prob. 19.130QACh. 19 - Prob. 19.131QACh. 19 - Prob. 19.132QACh. 19 - Prob. 19.133QACh. 19 - Prob. 19.134QACh. 19 - Prob. 19.135QACh. 19 - Prob. 19.136QACh. 19 - Prob. 19.137QACh. 19 - Prob. 19.138QACh. 19 - Prob. 19.139QACh. 19 - Prob. 19.140QACh. 19 - Prob. 19.141QACh. 19 - Prob. 19.142QA
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