Chapter 9, Problem 9.117QA

Interpretation Introduction

To find:

The value of the $∆H_{rxn}^o$ for the overall reaction using the given three thermochemical equations.

Answer to Problem 9.117QA

Solution:

The value of the overall reaction $∆H_{rxn}^o=-1400 kJ$.

Explanation of Solution

1) Concept:

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

2) Given:

We are given three reactions with their $∆H_{rxn}^o$ values:

Equation (1) $\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }{ Pb}_{ }\left(s\right)+ {\frac{1}{2} O}_{2 }\left(g\right)\to {PbO}_{ }\left(s\right) ∆H_{rxn}^o=-219 kJ\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }$

Equation (2) $\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }{C}_{ }\left(s\right)+{ O}_{2 }\left(g\right)\to C{O}_2\left(g\right) ∆H_{rxn}^o=-394 kJ$

Equation (3) $\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }{PbCO}_3\left(s\right)\to {PbO}_{ }\left(s\right)+{CO}_{2 }\left(g\right) ∆H_{rxn}^o=86 kJ$

One reaction with unknown $∆H_{rxn}^o$ value:

Equation (4) $\mathrm{ }\mathrm{ }\mathrm{ } 2 Pb \left(s\right)+ 2{ C}_{ }\left(s\right)+3 O_2\left(g\right)\to 2 {PbCO}_3\left(s\right)$

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\times eqn (1\left)\right]$$\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }{2 Pb}_{ }\left(s\right)+ { O}_{2 }\left(g\right)\to {2PbO}_{ }\left(s\right) 2\times ∆H_{rxn}^o=\left(2 \times -219\right)=438 kJ\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }$

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.

$[2\times equation\left(2\right)]$  $\mathrm{ }\mathrm{ }{2C}_{ }\left(s\right)+{ 2O}_{2 }\left(g\right)\to 2C{O}_2\left(g\right) 2\times ∆H_{rxn}^o=\left(2 \times -394\right)=788 kJ$

In equation (4), there are $2 {PbCO}_3$ and in equation (3), there is only one ${PbCO}_3$, so we have to multiply equation (3) by $2.$ We  also have to reverse the equation (3) to get ${PbCO}_3$ on the products side.

$[2\times eqn\left(3\right), reversed]$${2 PbO}_{ }\left(s\right)+{2 CO}_{2 }\left(g\right)\to 2 {PbCO}_3\left(s\right) 2\times (-∆H_{rxn}^o)=-172 kJ$

Adding all three equations, we get

${2 Pb}_{ }\left(s\right)+ { O}_{2 }\left(g\right)\to {2PbO}_{ }\left(s\right) ∆H_{rxn}^o=-438 kJ$

$+$

${2C}_{ }\left(s\right)+{ 2O}_{2 }\left(g\right)\to 2C{O}_2\left(g\right) ∆H_{rxn}^o=-788 kJ$

$+$

${2 PbO}_{ }\left(s\right)+{2 CO}_{2 }\left(g\right)\to 2 {PbCO}_3\left(s\right) ∆H_{rxn}^o =-172 kJ$

 

$2 {Pb}_{\left(s\right)}+ 2{ C}_{ \left(s\right)}+3 { O}_{2 \left(g\right)}\to 2 { PbCO}_{3 \left(s\right)} ∆H_{rxn}^o=-1400 kJ$

Conclusion:

We have calculated the unknown value of $\mathrm{ }∆H_{rxn}^o$ by manipulating the reactions and using Hess’s law.