Consider the chemical equations shown here.
[tex]\begin{array}{l}
P_4(s)+3 O_2(g) \rightarrow P_4 O_6(s) \quad \Delta H_1=-1,640.1 kJ \
P_4 O_{10}(s) \rightarrow P_4(s)+5 O_2(g) \quad \Delta H_2=2,940.1 kJ\\
\end{array}[/tex]
What is the overall enthalpy of reaction for the equation shown below?
Round the answer to the nearest whole number.
[tex]P_4 O_6(s)+2 O_2(g) \rightarrow P_4 O_{10}(s)[/tex]
Answer (2)
By applying Hess's Law and manipulating the given reactions, the overall enthalpy change for the reaction P 4 O 6 ( s ) + 2 O 2 ( g ) → P 4 O 10 ( s ) is calculated to be -1300 kJ. This involves reversing the first reaction and summing the enthalpy changes appropriately. The final enthalpy is rounded to the nearest whole number as -1300 kJ.
;
Reverse the first reaction to get P 4 O 6 ( s ) r i g h t a rro w P 4 ( s ) + 3 O 2 ( g ) with Δ H = 1640.1 k J .
Reverse the second reaction to get P 4 ( s ) + 5 O 2 ( g ) r i g h t a rro w P 4 O 10 ( s ) with Δ H = − 2940.1 k J .
Add the two reversed reactions and cancel out common species to obtain the target reaction P 4 O 6 ( s ) + 2 O 2 ( g ) r i g h t a rro w P 4 O 10 ( s ) .
Calculate the overall enthalpy change by summing the enthalpy changes of the reversed reactions: Δ H = 1640.1 − 2940.1 = − 1300 k J . The final answer is − 1300 .
Explanation
Problem Analysis We are given two chemical equations and their corresponding enthalpy changes, and we want to find the enthalpy change for a third reaction. This problem can be solved using Hess's Law, which states that the enthalpy change for a reaction is independent of the pathway taken. In other words, we can manipulate the given equations to arrive at the target equation, and the sum of the enthalpy changes for the manipulated equations will be the enthalpy change for the target equation.
Strategy Our target equation is: P 4 O 6 ( s ) + 2 O 2 ( g ) r i g h t a rro w P 4 O 10 ( s ) We have the following equations:
P 4 ( s ) + 3 O 2 ( g ) r i g h t a rro w P 4 O 6 ( s ) \t Δ H 1 = − 1 , 640.1 k J
P 4 O 10 ( s ) r i g h t a rro w P 4 ( s ) + 5 O 2 ( g ) \t Δ H 2 = 2 , 940.1 k J
We need to manipulate these equations to obtain the target equation.
Equation Manipulation First, we reverse the first equation: P 4 O 6 ( s ) r i g h t a rro w P 4 ( s ) + 3 O 2 ( g ) \t − Δ H 1 = 1 , 640.1 k J
Next, we reverse the second equation: P 4 ( s ) + 5 O 2 ( g ) r i g h t a rro w P 4 O 10 ( s ) \t − Δ H 2 = − 2 , 940.1 k J
Now, we add the reversed equations: P 4 O 6 ( s ) + P 4 ( s ) + 5 O 2 ( g ) r i g h t a rro w P 4 ( s ) + 3 O 2 ( g ) + P 4 O 10 ( s )
We can cancel out P 4 ( s ) from both sides: P 4 O 6 ( s ) + 5 O 2 ( g ) r i g h t a rro w 3 O 2 ( g ) + P 4 O 10 ( s )
Subtracting 3 O 2 ( g ) from both sides gives us the target equation: P 4 O 6 ( s ) + 2 O 2 ( g ) r i g h t a rro w P 4 O 10 ( s )
Enthalpy Calculation The overall enthalpy change is the sum of the enthalpy changes for the manipulated equations: Δ H = − Δ H 2 + ( − Δ H 1 ) = − 2940.1 + 1640.1 = − 1300 k J
So, the overall enthalpy of reaction for the equation P 4 O 6 ( s ) + 2 O 2 ( g ) r i g h t a rro w P 4 O 10 ( s ) is -1300 kJ.
Final Answer Rounding the answer to the nearest whole number, we get -1300 kJ.
Examples
Hess's Law is useful in determining the enthalpy change for reactions that are difficult or impossible to measure directly in a lab. For example, if you want to find the enthalpy change for the formation of a complex molecule from its elements, you can use Hess's Law to calculate it from a series of simpler reactions whose enthalpy changes are known. This is particularly useful in fields like chemical engineering, where knowing the enthalpy changes of reactions is crucial for designing efficient and safe chemical processes. Imagine you're designing a new industrial process to produce a valuable chemical. By applying Hess's Law, you can calculate the overall energy requirements for the process, optimizing conditions to minimize energy consumption and reduce costs. For instance, if a reaction A r i g h t a rro wB has Δ H 1 , and B r i g h t a rro wC has Δ H 2 , then A r i g h t a rro wC has Δ H = Δ H 1 + Δ H 2 .
