Consider the following intermediate chemical 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=-566.0 kJ \\
2 H_2 O(g) \rightarrow 2 H_2(g)+O_2(g) & \Delta H_3=483.6 kJ
\end{array}
[/tex]
The overall chemical equation is [tex]$C ( s )+ H _2 O ( g ) \rightarrow CO ( g )+ H _2(g)$[/tex]. To calculate the final enthalpy of the overall chemical equation, which step must occur?
A. Reverse the first equation, and change the sign of the enthalpy. Then, add.
B. Reverse the second equation, and change the sign of the enthalpy. Then, add.
C. Multiply the first equation by three, and triple the enthalpy. Then, add.
D. Divide the third equation by two, and double the enthalpy. Then, add.
Answer (1)
Reverse the second equation and change the sign of the enthalpy.
Reverse the third equation and change the sign of the enthalpy.
Divide the second and third equations by 2.
Add the equations to obtain the overall equation and calculate the final enthalpy: − 352.3 k J .
Explanation
Analyzing the Problem Let's analyze the given chemical equations and determine the necessary steps to obtain the overall chemical equation and calculate its enthalpy change.
Target Equation The overall chemical equation we want to achieve is: C ( s ) + H 2 O ( g ) → CO ( g ) + H 2 ( g ) .
Analyzing Equation 1 The first intermediate equation is: C ( s ) + O 2 ( g ) → C O 2 ( g ) Δ H 1 = − 393.5 k J . We need C ( s ) on the left side, which is already the case.
Analyzing Equation 2 The second intermediate equation is: 2 CO ( g ) + O 2 ( g ) → 2 C O 2 ( g ) Δ H 2 = − 566.0 k J . We need CO ( g ) on the right side. To achieve this, we reverse the equation and divide by 2: C O 2 ( g ) → CO ( g ) + 2 1 O 2 ( g ) . The enthalpy change becomes 2 − Δ H 2 = 2 − ( − 566.0 k J ) = 283.0 k J .
Analyzing Equation 3 The third intermediate equation is: 2 H 2 O ( g ) → 2 H 2 ( g ) + O 2 ( g ) Δ H 3 = 483.6 k J . We need H 2 O ( g ) on the left side and H 2 ( g ) on the right side. To achieve this, we reverse the equation and divide by 2: H 2 O ( g ) → H 2 ( g ) + 2 1 O 2 ( g ) . The enthalpy change becomes 2 − Δ H 3 = 2 − 483.6 k J = − 241.8 k J .
Combining the Equations Now, let's add the manipulated equations:
C ( s ) + O 2 ( g ) → C O 2 ( g ) C O 2 ( g ) → CO ( g ) + 2 1 O 2 ( g ) H 2 O ( g ) → H 2 ( g ) + 2 1 O 2 ( g )
Adding these gives:
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 equation: C ( s ) + H 2 O ( g ) → CO ( g ) + H 2 ( g ) .
Calculating Enthalpy Change The overall enthalpy change is the sum of the enthalpy changes of the manipulated equations:
Δ H = Δ H 1 + 2 − Δ H 2 + 2 − Δ H 3 = − 393.5 k J + 283.0 k J − 241.8 k J = − 352.3 k J .
Conclusion Based on the analysis, we need to reverse the second equation, change the sign of the enthalpy, and divide by 2. We also need to reverse the third equation, change the sign of the enthalpy, and divide by 2.
Examples
Hess's Law, which is applied here, is useful in many real-world applications. For example, it can be used to calculate the enthalpy change of a reaction that is difficult or impossible to measure directly. This is particularly useful in industrial chemistry, where knowing the enthalpy change of a reaction can help optimize reaction conditions and improve efficiency. It also helps in understanding energy changes in complex systems like combustion engines or biological processes, where direct measurement is not feasible.
