Chapter 15.4, Problem 32PP

Interpretation Introduction

Interpretation:

The value of $\text{Δ}\text{Η}$ is to be calculated for the given equation.

Concept introduction:

The change in enthalpy of a reaction can be calculated using Hess Law.

According to it, if one equation is the resultant of two different equations, then change in enthalpy of the equation is the sum of enthalpy change of both reactions.

Answer to Problem 32PP

$\text{Δ}\text{Η}$ = $-385.4\text{k}\text{J}$

Explanation of Solution

Given data:

$\begin{array}{l}2\text{C}\text{O}+{\text{O}}_2\to 2\text{C}{\text{O}}_2\text{Δ}\text{Η}=-566\text{k}\text{J}\end{array}$

$\begin{array}{c}{\text{N}}_2+{\text{O}}_2\to 2\text{N}\text{O}\text{Δ}\text{Η}=-180.6\text{k}\text{J}\end{array}$

Calculation:

Adding the first equation to the reverse of the second equation, the resultant equation is obtained.

$2\text{C}\text{O}+2\text{N}\text{O}\to 2\text{C}{\text{O}}_2+{\text{N}}_2$

Therefore, according to Hess Law,

$\begin{array}{c}\text{Δ}\text{Η}=-566-(-180.6)\end{array}$

  $\begin{array}{c}=-385.4\text{k}\text{J}\end{array}$

Conclusion

Thus, $\text{Δ}\text{Η}$ = $-385.4\text{k}\text{J}$ .