Consider the following intermediate chemical equations:
[tex]\begin{array}{ll}
P_4(s)+3 O_2(g) \rightarrow P_4 O_6(s) & \Delta H_1=-1,640 kJ \\
P_4 O_{10}(s) \rightarrow P_4(s)+5 O_2(g) & \Delta H_2=2,940.1 kJ
\end{array}[/tex]
What is the enthalpy of the overall chemical reaction [tex]P _4 O _6(s)+2 O _2(g) \rightarrow P _4 O _{10}(s)[/tex] ?
A. -4,580 kJ
B. -1,300 kJ
C. 1,300 kJ
D. 4,580 kJ
Answer (1)
Reverse the first reaction: P 4 O 6 ( s ) → P 4 ( s ) + 3 O 2 ( g ) , which changes the sign of Δ H 1 to 1640 k J .
Reverse the second reaction: P 4 ( s ) + 5 O 2 ( g ) → P 4 O 10 ( s ) , which changes the sign of Δ H 2 to − 2940.1 k J .
Add the two reversed reactions to obtain the target reaction: P 4 O 6 ( s ) + 2 O 2 ( g ) → P 4 O 10 ( s ) .
Sum the enthalpy changes: Δ H = 1640 k J + ( − 2940.1 k J ) = − 1 , 300 k J .
Explanation
Analyze the Problem We are given two chemical equations and their corresponding enthalpy changes:
Equation 1: P 4 ( s ) + 3 O 2 ( g ) → P 4 O 6 ( s ) Δ H 1 = − 1640 kJ
Equation 2: P 4 O 10 ( s ) → P 4 ( s ) + 5 O 2 ( g ) Δ H 2 = 2940.1 kJ
Our goal is to find the enthalpy change for the following reaction:
Target Equation: P 4 O 6 ( s ) + 2 O 2 ( g ) → P 4 O 10 ( s )
Manipulate the Equations To obtain the target equation, we need to manipulate the given equations using Hess's Law. Hess's Law states that the enthalpy change of an overall reaction is the sum of the enthalpy changes of the individual reactions, regardless of the path taken.
First, we reverse Equation 1:
P 4 O 6 ( s ) → P 4 ( s ) + 3 O 2 ( g ) Δ H = − Δ H 1 = − ( − 1640 kJ ) = 1640 kJ
Next, we reverse Equation 2:
P 4 ( s ) + 5 O 2 ( g ) → P 4 O 10 ( s ) Δ H = − Δ H 2 = − 2940.1 kJ
Combine the Equations Now, we add the reversed equations:
P 4 O 6 ( s ) → P 4 ( s ) + 3 O 2 ( g ) Δ H = 1640 kJ
P 4 ( s ) + 5 O 2 ( g ) → P 4 O 10 ( s ) Δ H = − 2940.1 kJ
Adding these two equations gives:
P 4 O 6 ( s ) + P 4 ( s ) + 5 O 2 ( g ) → P 4 ( s ) + 3 O 2 ( g ) + P 4 O 10 ( s )
Simplifying by canceling out P 4 ( s ) from both sides and subtracting 3 O 2 ( g ) from 5 O 2 ( g ) gives us the target equation:
P 4 O 6 ( s ) + 2 O 2 ( g ) → P 4 O 10 ( s )
Calculate the Enthalpy Change The enthalpy change for the target reaction is the sum of the enthalpy changes of the manipulated equations:
Δ H = 1640 kJ + ( − 2940.1 kJ ) = − 1300.1 kJ
Therefore, the enthalpy of the overall chemical reaction is approximately − 1300 kJ .
Final Answer The enthalpy of the overall chemical reaction P 4 O 6 ( s ) + 2 O 2 ( g ) → P 4 O 10 ( s ) is − 1 , 300 k J .
Examples
Hess's Law is useful in determining the enthalpy change of a reaction that cannot be measured 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 has side reactions, you can use Hess's Law to calculate the enthalpy change from a series of reactions that can be measured. This is commonly used in industrial processes to optimize reaction conditions and improve efficiency.
