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 9, Problem 9.117QA
Interpretation Introduction

To find:

The value of the Hrxno for the overall reaction using the given three thermochemical equations.

Expert Solution & Answer
Check Mark

Answer to Problem 9.117QA

Solution:

The value of the overall reaction  Hrxno=-1400 kJ.

Explanation of Solution

1) Concept:

The question is based on Hess’s law. Hess’s law states that the enthalpy of reaction Hrxn  for a process, that is, the sum of two or more reactions is equal to the sum of (Hrxn) values of the constituent reactions. We can manipulate the three equations algebraically for which Hrxno values are known so that they add to give the equation for which Hrxno is unknown. Then by applying Hess’s law, we can calculate the unknown value.

2) Given:

We are given three reactions with their Hrxno values:

Equation (1)                      Pb s+ 12 O2 (g) PbO (s)             Hrxno=-219 kJ     

Equation (2)                     C s+ O2 g CO2(g)                    Hrxno=-394 kJ

Equation (3)                    PbCO3(s) PbO (s)+CO2 (g)          Hrxno=86 kJ

One reaction with unknown Hrxno value:

Equation (4)                   2 Pb s+  2 C s+3 O2g 2 PbCO3(s)

3)  Calculations:

We have to manipulate equations (1), (2) and (3) to get equation (4). In equation (4), there are  2 Pb on the reactants side. In equation (1), Pb is present and is on the left side. But there is only one Pb, so we have to multiply the equation (1) by 2.

[2×eqn (1)]       2 Pb s+  O2 g 2PbO s        2× Hrxno=2 ×-219=438 kJ     

Now, in equation (4), there are  2 C on the reactant side, and in equation (2), there is only one C, so we have to multiply the equation (2) by 2.

[2×equation2]    2C s+ 2O2 g 2CO2g         2×Hrxno=2 ×-394=788 kJ

In equation (4), there are 2 PbCO3 and in equation (3), there is only one PbCO3, so we have to multiply equation (3) by 2. We  also have to reverse the equation (3) to get PbCO3 on the products side.

[2×eqn3, reversed]   2 PbO  (s)+2 CO2 (g)  2 PbCO3(s)     2×(-Hrxno)=-172 kJ

Adding all three equations, we get

2 Pb s+  O2 g 2PbO s                              Hrxno=-438 kJ 

+   

2C s+ 2O2 g 2CO2g                            Hrxno=-788 kJ

+

2 PbO  (s)+2 CO2 (g)  2 PbCO3(s)            Hrxno =-172 kJ

Chemistry: An Atoms-Focused Approach, Chapter 9, Problem 9.117QA  

2 Pb(s)+  2 C (s)+3  O2 (g) 2  PbCO3 (s)             Hrxno=-1400 kJ        

Conclusion:

We have calculated the unknown value of  Hrxno by manipulating the reactions and using Hess’s law.

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Chapter 9 Solutions

Chemistry: An Atoms-Focused Approach

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