Phosphorus trioxide is a chemical compound with the molecular formula P4O6. Although it should properly be named tetraphosphorus hexoxide, the name phosphorus trioxide preceded the knowledge of the compound’s molecular structure, and its usage continues today.
P4Os undergoes combustion according to the following balanced equation:
P 4 O 6(s) +2O 2(g) P 4 O 10(s)
a) Use the following equation to calculate the enthalpy change, AH, for the combustion of P 4 O 6(s) .
P 4(s) +5O 2(g) P 4 O 10(s) P 4(s) +3O 2(s) P 4 O 6(s)
DeltaH = – 2940Kj DeltaH = – 1640KJ
The Correct Answer and Explanation is :
To determine the enthalpy change (ΔH) for the combustion of ( P_4O_6 ), we use Hess’s Law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for individual steps that lead to the overall reaction.
Given reactions and their enthalpy changes:
- Formation of ( P_4O_{10} ) from phosphorus:
[
P_4(s) + 5O_2(g) \rightarrow P_4O_{10}(s) \quad \Delta H = -2940 \text{ kJ}
] - Formation of ( P_4O_6 ) from phosphorus:
[
P_4(s) + 3O_2(g) \rightarrow P_4O_6(s) \quad \Delta H = -1640 \text{ kJ}
]
We need the enthalpy change for:
[
P_4O_6(s) + 2O_2(g) \rightarrow P_4O_{10}(s)
]
Applying Hess’s Law:
- The first reaction forms ( P_4O_{10} ) directly from phosphorus.
- The second reaction forms ( P_4O_6 ) from phosphorus.
- To obtain the desired equation, subtract the second reaction from the first.
[
[P_4(s) + 5O_2(g) \rightarrow P_4O_{10}(s)] – [P_4(s) + 3O_2(g) \rightarrow P_4O_6(s)]
]
Canceling out ( P_4(s) ) from both sides:
[
P_4O_6(s) + 2O_2(g) \rightarrow P_4O_{10}(s)
]
Now, subtract the enthalpy values:
[
\Delta H = (-2940 \text{ kJ}) – (-1640 \text{ kJ})
]
[
\Delta H = -1300 \text{ kJ}
]
Explanation:
This calculation shows that the combustion of ( P_4O_6 ) releases 1300 kJ of energy. Since the reaction involves the oxidation of phosphorus trioxide to phosphorus pentoxide, it is exothermic, meaning heat is released into the surroundings. This is consistent with the general trend that oxidation reactions of phosphorus compounds are highly exothermic due to the formation of strong P=O bonds in ( P_4O_{10} ).
This concept is particularly useful in thermochemistry and industrial applications, where understanding enthalpy changes helps predict energy requirements or outputs in chemical processes.