First, we need to flip the second equation and multiply it by 2 to get the same number of moles of oxygen on both sides:
2P4O10 (s) → 4P4 (s) + 10O2 (g) ΔH = 2(-2940.1 kJ) = -5880.2 kJ
Now we can cancel out the P4 and O2 on both sides of the equation to get:
2P4O6 (s) + 5O2 (g) → 2P4H10 (s)
And we can use the given ΔH of the first equation to find the ΔH of the overall reaction:
P4 (s) + 3O2 (g) → P4O6 (s) ΔH = -1640.1 kJ
2P4O6 (s) + 5O2 (g) → 2P4 (s) + 6O2 (g) ΔH = 2(1640.1 kJ) = 3280.2 kJ
2P4O6 (s) + 5O2 (g) → 2P4H10 (s) + 6O2 (g) ΔH = 3280.2 kJ - (-5880.2 kJ) = 9160.4 kJ
Therefore, the enthalpy change for the reaction P4O6(s) + 2O2(g) → P4H10(s) is +9160.4 kJ.
Calculate the enthalpy change for the reaction: P4O6(s)+ 2O2(g) → P4H10(s)
Given the following enthalpies of reaction:
P4 (s) + 3 O2 (g)⇾P4O6 (S) incrementH = -1640.1 kJ
P4 (s) + 5 O2 (g)⇾P4O10 (S) incrementH = -2940.1 kJ
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