Consider the intermediate equations:
[tex]
\begin{array}{ll}
C(s)+O_2(g) \rightarrow CO_2(g) & \Delta H_1=-393.5 kJ \\
2 CO(g)+O_2(g) \rightarrow 2 CO_2(g) & \Delta H_2=-5660 kJ \\
2 H_2 O(g) \rightarrow 2 H_2(g)+O_2(g) & \Delta H_3=483.6 kJ
\end{array}
[/tex]
With the overall reaction:
[tex]C(s)+H_2 O(9) \rightarrow CO(9)+H_2(9) \quad \Delta H_{ran}=?[/tex]
What must be done to calculate the enthalpy of reaction? Check all that apply.
* The first equation must be halved.
* The first equation must be reversed.
* The second equation must be halved.
* The second equation must be reversed.
* The third equation must be halved.
* The third equation must be reversed.
What is the overall enthalpy of reaction? [tex]\Delta H_{\text {rxn }}[/tex] = kJ
Answer (2)
To find the enthalpy change for the reaction C(s) + H2O(g) → CO(g) + H2(g), we manipulated the given reactions according to Hess's Law. The necessary steps were to keep the first reaction as is, reverse the second reaction, and halve the third reaction. After calculating, the overall enthalpy of reaction was found to be ΔH_{rxn} = 2678.3 kJ.
;
The second equation must be halved and reversed.
The third equation must be halved.
Calculate the enthalpy change for the reversed and halved second equation: Δ H 2 ′ = − 2 − 566.0 = 283.0 kJ.
Calculate the enthalpy change for the halved third equation: Δ H 3 ′ = 2 483.6 = 241.8 kJ. The overall enthalpy of reaction is 131.3 kJ.
Explanation
Analyzing the Reactions We are given three intermediate reactions with their enthalpy changes and asked to find the enthalpy change for the overall reaction:
C ( s ) + O 2 ( g ) → C O 2 ( g ) \t Δ H 1 = − 393.5 k J
2 CO ( g ) + O 2 ( g ) → 2 C O 2 ( g ) \t Δ H 2 = − 566.0 k J
2 H 2 O ( g ) → 2 H 2 ( g ) + O 2 ( g ) \t Δ H 3 = 483.6 k J
The overall reaction is:
C ( s ) + H 2 O ( g ) → CO ( g ) + H 2 ( g ) \t Δ H r x n = ?
To find Δ H r x n , we need to manipulate the given reactions so that they add up to the overall reaction. This involves possibly reversing and/or multiplying the reactions by a factor, and adjusting the Δ H values accordingly.
Determining the Necessary Manipulations Let's compare the reactants and products in the intermediate reactions with those in the overall reaction.
The overall reaction has C ( s ) as a reactant. The first reaction also has C ( s ) as a reactant, so we keep the first reaction as is.
The overall reaction has CO ( g ) as a product. The second reaction has CO ( g ) as a reactant. Thus, we need to reverse the second reaction and divide it by 2 to get CO ( g ) as a product with a coefficient of 1.
The overall reaction has H 2 O ( g ) as a reactant. The third reaction has H 2 O ( g ) as a reactant. Thus, we need to halve the third reaction to get H 2 O ( g ) with a coefficient of 1.
The overall reaction has H 2 ( g ) as a product. The third reaction, after being halved, will give us H 2 ( g ) as a product.
Manipulating the Equations Based on the analysis, we need to:
Keep the first equation as is: C ( s ) + O 2 ( g ) → C O 2 ( g ) \t Δ H 1 = − 393.5 k J
Reverse and halve the second equation: C O 2 ( g ) → CO ( g ) + 2 1 O 2 ( g ) \t Δ H 2 ′ = − 2 − 566.0 k J = 283.0 k J
Halve the third equation: $H_2 O(g) \rightarrow H_2(g) + \frac{1}{2}O_2(g) \t \Delta H_3' = \frac{483.6}{2} kJ = 241.8 kJ
Summing the Equations Now, we add the manipulated equations:
C ( s ) + O 2 ( g ) + C O 2 ( g ) + H 2 O ( g ) → C O 2 ( g ) + CO ( g ) + 2 1 O 2 ( g ) + H 2 ( g ) + 2 1 O 2 ( g )
Simplifying, we get the overall reaction:
C ( s ) + H 2 O ( g ) → CO ( g ) + H 2 ( g )
Calculating the Overall Enthalpy Change The enthalpy change for the overall reaction is the sum of the enthalpy changes of the manipulated reactions: Δ H r x n = Δ H 1 + Δ H 2 ′ + Δ H 3 ′ = − 393.5 + 283.0 + 241.8 = 131.3 k J
Therefore, the overall enthalpy of reaction is 131.3 k J .
Examples
Hess's Law, which this problem demonstrates, is used in real-world applications to calculate the enthalpy changes of reactions that are difficult or impossible to measure directly. For example, it can be used to determine the enthalpy change of combustion of a fuel by breaking it down into a series of simpler reactions whose enthalpy changes are known. This information is crucial in designing efficient engines and power plants. Also, in the manufacturing of chemicals, understanding the heat released or absorbed during reactions is essential for safety and cost-effectiveness.
