Chemistry: An Atoms-Focused Approach (Second Edition)
Chemistry: An Atoms-Focused Approach (Second Edition)
2nd Edition
ISBN: 9780393614053
Author: Thomas R. Gilbert, Rein V. Kirss, Stacey Lowery Bretz, Natalie Foster
Publisher: W. W. Norton & Company
Question
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Chapter 14, Problem 14.121QA
Interpretation Introduction

To find:

a) Explain if carbon dioxide is a viable source of the fuel  CO.

b) Explain whether the reaction is endothermic.

c) Calculate the value of Kp at each temperature.

d) Explain whether the decomposition of CO2 is an antidote for global warming.

Expert Solution & Answer
Check Mark

Answer to Problem 14.121QA

Solution:

a) Carbon dioxide is not a viable source of the fuel CO.

b) The reaction is endothermic.

c) The value of Kp at each temperature:

Temperature (K) Kp
1500 5.5×10-11
2500 4.0×10-3
3000 0.40

d) The decomposition of CO2 is not an antidote for global warming.

Explanation of Solution

1) Concept

Given the percentage of decomposition of CO2 at different temperatures and partial pressure of the CO2 at 1 atm, we can calculate the Kp and  determine whether reaction is endothermic.

2) Formula:

Kp=PproductscoefficentsPreactantscoefficents

Here,Kp is the equilibrium partial pressure constant,Pproducts is the partial pressure of products and Preactants is the partial pressure of reactants and the coefficient are the coefficient of reactants and product from the balanced reaction.

3) Given:

i) 2CO2g 2CO(g)+O2g

ii) Initial partial pressure of CO2=1 atm

iii) Percent decomposition of CO2 at different temperatures:

Temperature (K) Decomposition (%)
1500 0.048
2500 17.6
3000 54.8

4)  Calculations:

We can say that the amount of CO2 decomposed increases with increasing temperature. This means that K2>K1 in the equation:

lnK2K1= -ΔH0R1T2-1T1

Where T2>T1, so 1T2-1T1<0.

As the temperature increases, value of Kp increases. Therefore,lnK2K1>0.

From 1T2-1T1<0, the value of H must be positive. Therefore, the given reaction is endothermic

RICE table for given reaction:

Reaction 2CO2g                     2CO(g)      +                  O2g
PCO2(atm) PCO(atm) PO2(atm)
Initial 1 0 0
Change -2x +2x +x
Equilibrium (1-2x) 2x x

Change in concentration of CO2  is  2x, the value of x can be calculated as,

At T=1500 K,

2x=0.048100=0.00048

x=0.000482=0.00024

1-2x=(1-0.00048)=0.99952

Therefore, equilibrium constant expression for partial pressures for the given reaction is:

Kp=PCO2PO21PCO22=0.000482(0.00024)0.999522=5.5×10-11

At T=2500 K,

2x=17.6100=0.176

x=0.1762=0.088

1-2x=1-0.0176=0.824

Therefore, equilibrium constant expression for partial pressures for the given reaction is:

Kp=PCO2PO21PCO22=0.1762(0.088)0.8242=4.0×10-3

At, T=3000K

2x=54.8100=0.548

x=0.5482=0.274

1-2x=1-0.548=0.452

Therefore, equilibrium constant expression for partial pressures for the given reaction is:

Kp=PCO2PO21PCO22=0.5482(0.274)0.4522=0.40

The value of the equilibrium constant increases with increasing temperature, but it is still less than 1. Thus this reaction does not favor products even at a very high temperature, and this is not a viable source of  CO. Therefore, it is not an antidote to decrease  CO2 as a contributor to global warming.

Final statement:

Using the given % decomposition, we calculated the Kp values at different temperatures.

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Chapter 14 Solutions

Chemistry: An Atoms-Focused Approach (Second Edition)

Ch. 14 - Prob. 14.11QACh. 14 - Prob. 14.12QACh. 14 - Prob. 14.13QACh. 14 - Prob. 14.14QACh. 14 - Prob. 14.15QACh. 14 - Prob. 14.16QACh. 14 - Prob. 14.17QACh. 14 - Prob. 14.18QACh. 14 - Prob. 14.19QACh. 14 - Prob. 14.20QACh. 14 - Prob. 14.21QACh. 14 - Prob. 14.22QACh. 14 - Prob. 14.23QACh. 14 - Prob. 14.24QACh. 14 - Prob. 14.25QACh. 14 - Prob. 14.26QACh. 14 - Prob. 14.27QACh. 14 - Prob. 14.28QACh. 14 - Prob. 14.29QACh. 14 - Prob. 14.30QACh. 14 - Prob. 14.31QACh. 14 - Prob. 14.32QACh. 14 - Prob. 14.33QACh. 14 - Prob. 14.34QACh. 14 - Prob. 14.35QACh. 14 - Prob. 14.36QACh. 14 - Prob. 14.37QACh. 14 - Prob. 14.38QACh. 14 - Prob. 14.39QACh. 14 - Prob. 14.40QACh. 14 - Prob. 14.41QACh. 14 - Prob. 14.42QACh. 14 - Prob. 14.43QACh. 14 - Prob. 14.44QACh. 14 - Prob. 14.45QACh. 14 - Prob. 14.46QACh. 14 - Prob. 14.47QACh. 14 - Prob. 14.48QACh. 14 - Prob. 14.49QACh. 14 - Prob. 14.50QACh. 14 - Prob. 14.51QACh. 14 - Prob. 14.52QACh. 14 - Prob. 14.53QACh. 14 - Prob. 14.54QACh. 14 - Prob. 14.55QACh. 14 - Prob. 14.56QACh. 14 - Prob. 14.57QACh. 14 - Prob. 14.58QACh. 14 - Prob. 14.59QACh. 14 - Prob. 14.60QACh. 14 - Prob. 14.61QACh. 14 - Prob. 14.62QACh. 14 - Prob. 14.63QACh. 14 - Prob. 14.64QACh. 14 - Prob. 14.65QACh. 14 - Prob. 14.66QACh. 14 - Prob. 14.67QACh. 14 - Prob. 14.68QACh. 14 - Prob. 14.69QACh. 14 - Prob. 14.70QACh. 14 - Prob. 14.71QACh. 14 - Prob. 14.72QACh. 14 - Prob. 14.73QACh. 14 - Prob. 14.74QACh. 14 - Prob. 14.75QACh. 14 - Prob. 14.76QACh. 14 - Prob. 14.77QACh. 14 - Prob. 14.78QACh. 14 - Prob. 14.79QACh. 14 - Prob. 14.80QACh. 14 - Prob. 14.81QACh. 14 - Prob. 14.82QACh. 14 - Prob. 14.83QACh. 14 - Prob. 14.84QACh. 14 - Prob. 14.85QACh. 14 - Prob. 14.86QACh. 14 - Prob. 14.87QACh. 14 - Prob. 14.88QACh. 14 - Prob. 14.89QACh. 14 - Prob. 14.90QACh. 14 - Prob. 14.91QACh. 14 - Prob. 14.92QACh. 14 - Prob. 14.93QACh. 14 - Prob. 14.94QACh. 14 - Prob. 14.95QACh. 14 - Prob. 14.96QACh. 14 - Prob. 14.97QACh. 14 - Prob. 14.98QACh. 14 - Prob. 14.99QACh. 14 - Prob. 14.100QACh. 14 - Prob. 14.101QACh. 14 - Prob. 14.102QACh. 14 - Prob. 14.103QACh. 14 - Prob. 14.104QACh. 14 - Prob. 14.105QACh. 14 - Prob. 14.106QACh. 14 - Prob. 14.107QACh. 14 - Prob. 14.108QACh. 14 - Prob. 14.109QACh. 14 - Prob. 14.110QACh. 14 - Prob. 14.111QACh. 14 - Prob. 14.112QACh. 14 - Prob. 14.113QACh. 14 - Prob. 14.114QACh. 14 - Prob. 14.115QACh. 14 - Prob. 14.116QACh. 14 - Prob. 14.117QACh. 14 - Prob. 14.118QACh. 14 - Prob. 14.119QACh. 14 - Prob. 14.120QACh. 14 - Prob. 14.121QACh. 14 - Prob. 14.122QACh. 14 - Prob. 14.123QACh. 14 - Prob. 14.124QACh. 14 - Prob. 14.125QACh. 14 - Prob. 14.126QACh. 14 - Prob. 14.127QACh. 14 - Prob. 14.128QA
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