Consider the following intermediate reactions:
[tex]
\begin{array}{ll}
CH_4(g)+2 O_2(g) \rightarrow CO_2(g)+2 H_2 O(g) & \Delta H_1=-802 kJ \\
2 H_2 O(g) \rightarrow 2 H_2 O(l) & \Delta H_2=-88 kJ
\end{array}
[/tex]
The overall chemical reaction is as follows:
[tex]CH_4(g)+2 O_2(g) \rightarrow CO_2(g)+2 H_2 O(l) \quad \Delta H_2=-890 kJ[/tex]
What is the correct enthalpy diagram using Hess's law for this system?
[tex]CH_2(a)+2 O_2(g)[/tex]
Answer (1)
The overall reaction is the sum of the two intermediate reactions.
The enthalpy change for the overall reaction is the sum of the enthalpy changes for the intermediate reactions: Δ H = Δ H 1 + Δ H 2 .
Δ H = − 802 k J + ( − 88 k J ) = − 890 k J .
The enthalpy diagram should visually represent these energy changes, showing that the overall reaction's enthalpy change is the sum of the intermediate steps.
Explanation
Understanding the Problem We are asked to determine the correct enthalpy diagram for the given system using Hess's law. We are given two intermediate reactions and the overall reaction, along with their respective enthalpy changes. Hess's law states that the total enthalpy change for a chemical reaction is the same whether the reaction takes place in one step or in several steps.
Analyzing the Reactions The given intermediate reactions are:
C H 4 ( g ) + 2 O 2 ( g ) → C O 2 ( g ) + 2 H 2 O ( g ) \t Δ H 1 = − 802 k J
2 H 2 O ( g ) → 2 H 2 O ( l ) \t Δ H 2 = − 88 k J
The overall reaction is:
C H 4 ( g ) + 2 O 2 ( g ) → C O 2 ( g ) + 2 H 2 O ( l ) \t Δ H = − 890 k J
We can see that the overall reaction is the sum of the two intermediate reactions. Therefore, the enthalpy change for the overall reaction should be the sum of the enthalpy changes for the intermediate reactions.
Verifying Enthalpy Change Let's verify the enthalpy change for the overall reaction:
Δ H = Δ H 1 + Δ H 2 = − 802 k J + ( − 88 k J ) = − 890 k J
This matches the given enthalpy change for the overall reaction.
Describing the Enthalpy Diagram The enthalpy diagram should show the following:
The reactants C H 4 ( g ) + 2 O 2 ( g ) at the initial energy level.
The first intermediate reaction goes to C O 2 ( g ) + 2 H 2 O ( g ) with Δ H 1 = − 802 k J . This represents a decrease in energy.
The second intermediate reaction goes to C O 2 ( g ) + 2 H 2 O ( l ) with Δ H 2 = − 88 k J . This represents another decrease in energy.
The overall reaction goes directly to C O 2 ( g ) + 2 H 2 O ( l ) with Δ H = − 890 k J . This also represents a decrease in energy.
The enthalpy diagram should visually represent these energy changes, showing that the overall reaction's enthalpy change is the sum of the intermediate steps.
Conclusion The correct enthalpy diagram will have the reactants at the top, an intermediate level representing the products of the first reaction, and a final level representing the products of the overall reaction. The energy differences between these levels will correspond to the enthalpy changes for each step.
Examples
Hess's Law is useful in determining the enthalpy change for reactions that are difficult or impossible to measure directly. For example, if you want to find the enthalpy change for the formation of a compound from its elements, but the reaction proceeds through multiple steps, you can use Hess's Law to calculate the overall enthalpy change by summing the enthalpy changes for each step. This is commonly used in industrial processes to optimize reaction conditions and energy efficiency.
