Consider the following intermediate chemical equations:
[tex]
\begin{array}{ll}
NO(g)+O_3(g) \rightarrow NO_2(g)+O_2(g) & \Delta H_1=-198.9 kJ \\
\frac{3}{2} O_2(g) \rightarrow O_3(g) & \Delta H_2=142.3 kJ \\
O(g) \rightarrow \frac{1}{2} O_2(g) & \Delta H_3=-247.5 kJ
\end{array}
[/tex]
What is the enthalpy of the overall chemical equation [tex]NO ( g )+ O ( g ) \rightarrow NO _2(g)[/tex]?
A. -305 kJ
B. -304.1 kJ
C. -93.7 kJ
D. 588.7 kJ
Answer (2)
Using Hess's Law, the enthalpy change for the reaction NO ( g ) + O ( g ) → N O 2 ( g ) was calculated to be − 588.7 kJ . The steps involved combining the first and third reactions and then subtracting the second to isolate the desired equation. Therefore, the answer is − 588.7 kJ .
;
Add the enthalpy changes of the given reactions to find the enthalpy change of the overall reaction.
Combine the first and third reactions: Δ H 1 + Δ H 3 = − 198.9 kJ + ( − 247.5 kJ ) = − 446.4 kJ .
Subtract the second reaction from the result: Δ H = − 446.4 kJ − 142.3 kJ = − 588.7 kJ .
The enthalpy of the overall reaction is − 588.7 kJ .
Explanation
Understanding the Problem We are given three chemical equations and their corresponding enthalpy changes, and we want to find the enthalpy change for the overall reaction NO ( g ) + O ( g ) i g h t ha r p oo n u pN O 2 ( g ) . We will use Hess's Law, which states that the enthalpy change for an overall reaction is the sum of the enthalpy changes for each step in the reaction.
Listing the Given Reactions The given reactions are:
NO ( g ) + O 3 ( g ) i g h t ha r p oo n u pN O 2 ( g ) + O 2 ( g ) , Δ H 1 = − 198.9 kJ
2 3 O 2 ( g ) i g h t ha r p oo n u p O 3 ( g ) , Δ H 2 = 142.3 kJ
O ( g ) i g h t ha r p oo n u p 2 1 O 2 ( g ) , Δ H 3 = − 247.5 kJ
Combining Reactions We want to manipulate these equations so that when added together, they give us the desired equation: NO ( g ) + O ( g ) i g h t ha r p oo n u pN O 2 ( g ) .
We can add the first and third equations:
NO ( g ) + O 3 ( g ) + O ( g ) i g h t ha r p oo n u pN O 2 ( g ) + O 2 ( g ) + 2 1 O 2 ( g )
The enthalpy change for this combined reaction is Δ H 1 + Δ H 3 = − 198.9 kJ + ( − 247.5 kJ ) = − 446.4 kJ .
Final Calculation Now, we need to get rid of O 3 and O 2 from the equation. We can subtract the second equation from the combined equation:
NO ( g ) + O 3 ( g ) + O ( g ) − 2 3 O 2 ( g ) i g h t ha r p oo n u pN O 2 ( g ) + O 2 ( g ) + 2 1 O 2 ( g ) − O 3 ( g )
This simplifies to:
NO ( g ) + O ( g ) i g h t ha r p oo n u pN O 2 ( g )
The enthalpy change for this overall reaction is Δ H = − 446.4 kJ − 142.3 kJ = − 588.7 kJ .
Conclusion Therefore, the enthalpy of the overall chemical equation NO ( g ) + O ( g ) ⇀ N O 2 ( g ) is − 588.7 kJ .
Examples
Hess's Law is useful in many real-world applications, such as calculating the enthalpy change of a reaction that is difficult or impossible to measure directly. For example, we can use Hess's Law to determine the enthalpy change for the formation of a compound from its elements, even if the reaction does not occur in a single step. This is important in fields like chemical engineering, where enthalpy changes are needed to design and optimize chemical processes. Understanding these concepts helps engineers predict energy requirements and manage chemical reactions safely and efficiently.
