Consider the chemical equations shown here.
[tex]
\begin{array}{l}
2 H_2(g)+O_2(g) \rightarrow 2 H_2 O(g) \Delta H_1=-483.6 kJ \div 2=-241.8 kJ / mol \\
3 O_2(g) \rightarrow 2 O_3(g) \Delta H_2=284.6 kJ+2=142.3 kJ / mol
\end{array}
[/tex]
What is the overall enthalpy of reaction for the equation shown below?
[tex]$3 H_2(g)+O_3(g) \rightarrow 3 H_2 O(g)$[/tex]
Answer (2)
Using Hess's Law, we modified the given equations to ultimately calculate the overall enthalpy change for the target reaction. The final enthalpy change for the reaction 3 H_2(g) + O_3(g) → 3 H_2O(g) is -583.1 kJ. This demonstrates how enthalpy changes can be combined to find the enthalpy for complex reactions.
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Multiply the first equation by 2 3 and calculate the new enthalpy change: Δ H 1 ′ = 2 3 \t ( − 483.6 k J ) = − 725.4 k J .
Divide the second equation by 2 and calculate the new enthalpy change: Δ H 2 ′ = 2 1 \t ( 284.6 k J ) = 142.3 k J .
Add the two modified equations to obtain the target equation.
Sum the enthalpy changes to find the overall enthalpy change: Δ H = − 725.4 k J + 142.3 k J = − 583.1 k J .
Explanation
Problem Analysis We are given two chemical equations with their enthalpy changes, and we want to find the enthalpy change for a third equation. This is a Hess's Law problem, where we manipulate the given equations to obtain the target equation and then sum the corresponding enthalpy changes.
Given Equations and Target Equation The given equations are:
2 H 2 ( g ) + O 2 ( g ) → 2 H 2 O ( g ) \t Δ H 1 = − 483.6 k J
3 O 2 ( g ) → 2 O 3 ( g ) \t Δ H 2 = 284.6 k J
The target equation is:
3 H 2 ( g ) + O 3 ( g ) → 3 H 2 O ( g )
Modifying the First Equation First, we multiply the first equation by 2 3 to get 3 moles of H 2 on the reactant side:
2 3 \t ( 2 H 2 ( g ) + O 2 ( g ) → 2 H 2 O ( g )) ⇒ 3 H 2 ( g ) + 2 3 O 2 ( g ) → 3 H 2 O ( g )
The enthalpy change for this modified reaction is:
Δ H 1 ′ = 2 3 \t Δ H 1 = 2 3 \t ( − 483.6 k J ) = − 725.4 k J
Modifying the Second Equation Next, we divide the second equation by 2 to get 1 mole of O 3 on the product side:
2 1 \t ( 3 O 2 ( g ) → 2 O 3 ( g )) ⇒ 2 3 O 2 ( g ) → O 3 ( g )
The enthalpy change for this modified reaction is:
Δ H 2 ′ = 2 1 \t Δ H 2 = 2 1 \t ( 284.6 k J ) = 142.3 k J
Combining the Modified Equations Now, we add the two modified equations:
( 3 H 2 ( g ) + 2 3 O 2 ( g ) → 3 H 2 O ( g )) + ( 2 3 O 2 ( g ) → O 3 ( g ))
This gives us:
3 H 2 ( g ) + 2 3 O 2 ( g ) + 2 3 O 2 ( g ) → 3 H 2 O ( g ) + O 3 ( g )
Rearranging to match the target equation:
3 H 2 ( g ) + O 3 ( g ) → 3 H 2 O ( g )
Calculating the Overall Enthalpy Change The overall enthalpy change for the target reaction is the sum of the enthalpy changes for the two modified reactions:
Δ H = Δ H 1 ′ + Δ H 2 ′ = − 725.4 k J + 142.3 k J = − 583.1 k J
Final Answer Therefore, the overall enthalpy of reaction for the equation 3 H 2 ( g ) + O 3 ( g ) → 3 H 2 O ( g ) is − 583.1 k J .
Examples
Hess's Law is useful in determining the enthalpy change of 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 is too slow or produces unwanted side products, you can use Hess's Law and a series of other reactions with known enthalpy changes to calculate the desired value. This is commonly used in industrial chemistry to optimize reaction conditions and improve efficiency.
