- Reverse the equation \(P_4(s) + 3O_2(g) \rightarrow P_4O_6(s)\) to get \(\Delta H = 1,640.1 \text{ kJ}\).
- Reverse the equation \(P_4O_{10}(s) \rightarrow P_4(s) + 5O_2(g)\) to get \(\Delta H = -2,940.1 \text{ kJ}\).
- Sum the two equations to obtain the target reaction \(P_4O_6(s) + 2O_2(g) \rightarrow P_4O_{10}(s)\).
- Sum the enthalpies: \(1,640.1 + (-2,940.1) = -1,300\).
- The final enthalpy is \(\boxed{-1300 \text{ kJ}}\).
Explanation
- Problem Analysis and Data Evaluation
To determine the overall enthalpy of reaction for \(P_4O_6(s) + 2O_2(g) \rightarrow P_4O_{10}(s)\), we use Hess's Law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for its individual steps. We are given the following:
- \(P_4(s) + 3O_2(g) \rightarrow P_4O_6(s)\) with \(\Delta H_1 = -1,640.1 \text{ kJ}\)
- \(P_4O_{10}(s) \rightarrow P_4(s) + 5O_2(g)\) with \(\Delta H_2 = 2,940.1 \text{ kJ}\)
- Manipulating Equations to Match the Target Reaction
To obtain the target reaction, we need to manipulate the given equations:
- Since \(P_4O_6(s)\) is a reactant in our target equation, we reverse the first equation. This causes the sign of \(\Delta H_1\) to change:
\(P_4O_6(s) \rightarrow P_4(s) + 3O_2(g)\), \(\Delta H_{1'} = -(-1,640.1 \text{ kJ}) = 1,640.1 \text{ kJ}\).
- Since \(P_4O_{10}(s)\) is a product in our target equation, we reverse the second equation. This causes the sign of \(\Delta H_2\) to change:
\(P_4(s) + 5O_2(g) \rightarrow P_4O_{10}(s)\), \(\Delta H_{2'} = -(2,940.1 \text{ kJ}) = -2,940.1 \text{ kJ}\).
- Calculating the Net Enthalpy Change
Now, we add the two modified equations together:
\((P_4O_6(s) \rightarrow P_4(s) + 3O_2(g)) + (P_4(s) + 5O_2(g) \rightarrow P_4O_{10}(s))\)
Combining the species:
\(P_4O_6(s) + P_4(s) + 5O_2(g) \rightarrow P_4(s) + 3O_2(g) + P_4O_{10}(s)\)
Canceling \(P_4(s)\) from both sides and simplifying the \(O_2(g)\) terms (\(5O_2 - 3O_2 = 2O_2\)) gives the target reaction:
\(P_4O_6(s) + 2O_2(g) \rightarrow P_4O_{10}(s)\).
The enthalpy of this reaction is the sum of the enthalpies of the modified equations:
\(\Delta H_{net} = 1,640.1 \text{ kJ} + (-2,940.1 \text{ kJ}) = -1,300 \text{ kJ}\).
- Conclusion
The calculated enthalpy change for the target reaction is \(-1,300 \text{ kJ}\). As the value is already an integer, no further rounding is needed. The final enthalpy change is \(\boxed{-1300 \text{ kJ}}\).
Examples
Hess's Law is a fundamental concept in chemical thermodynamics used extensively by scientists to calculate reaction enthalpies that cannot be easily measured in a laboratory. For example, it allows chemists to determine the heat released or absorbed when building complex molecules, such as pharmaceutical intermediates or synthetic polymers, by using data from simpler, more easily measured reactions. This ensures safer production processes and helps researchers understand the energy balance of various industrial chemical reactions.