Calculate the enthalpy change for the reaction P4O6 + 2O2 -> P4O10 using the information below

Question

P4 + 3O2 -> P4O6 ΔH= -1640.1 kJ
P4 + 5O2 -> P4O10 ΔH= -2940.1 kJ

Answer 1

To calculate the enthalpy change for the reaction P4O6 + 2O2 -> P4O10, we need to use the given information and apply the concept of Hess's Law.

Hess's Law states that if a reaction can be expressed as the sum of two or more reactions, then the enthalpy change for the overall reaction is equal to the sum of the enthalpy changes for the individual reactions.

To find the enthalpy change for the reaction P4O6 + 2O2 -> P4O10, we can break it down into two steps:

1. P4 + 3O2 -> P4O6

Given: ΔH1 = -1640.1 kJ

2. P4 + 5O2 -> P4O10

Given: ΔH2 = -2940.1 kJ

Now, we can think of the desired reaction P4O6 + 2O2 -> P4O10 as the combination of these two reactions:

P4 + 3O2 -> P4O6
2P4O6 + 4O2 -> 2P4O10

Now, we need to multiply the second reaction by a factor so that the number of moles of P4O6 matches in both reactions. In this case, we need to multiply the second reaction by 2.

P4 + 3O2 -> P4O6
2P4O6 + 4O2 -> 2P4O10

Now, we can add the two reactions together:

P4 + 3O2 -> P4O6
2P4O6 + 4O2 -> 2P4O10
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P4 + 3O2 + 2P4O6 + 4O2 -> P4O6 + 2P4O10

By adding up the enthalpy changes for each reaction, we can find the enthalpy change for the overall reaction:

ΔH_total = ΔH1 + ΔH2

ΔH_total = -1640.1 kJ + (-2940.1 kJ)
ΔH_total = -4580.2 kJ

Therefore, the enthalpy change for the reaction P4O6 + 2O2 -> P4O10 is -4580.2 kJ.