# Calculate the heat energy produced (q) when 100.0 g of Cl3PO (g) is produced in the reaction: P4O10 (s) + 6PCl5 (g) -> 10Cl3PO (g) From the following data: 1. P4 (s) + 6Cl2 (g) -> 4PCl3...

Calculate the heat energy produced (q) when 100.0 g of Cl3PO (g) is produced in the reaction: P4O10 (s) + 6PCl5 (g) -> 10Cl3PO (g)

From the following data:

1. P4 (s) + 6Cl2 (g) -> 4PCl3 (g) ΔH= -1225.6 kJ/mol

2. P4 (s) + 5O2 (g) -> P4O10 (g) ΔH= -2967.3 kJ/mol

3. PCl3 (g) + Cl2 (g) -> PCl5 (g) ΔH= -84.2 kJ/mol

4. PCl3 (g) + 1/2O2 (g) -> Cl3PO (g) ΔH= -285.7 kJ/mol

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Hess's law is the method of obtaining heat values derived from several steps to obtain the standard enthalpy of the reaction. These steps includes several standard enthalpy change of the intermediate reactions. You can simply view it this way: you have to show how the equation P4O10 (s) + 6PCl5 (g) -> 10Cl3PO (g) is formed from the lists of intermediate reaction written (numbers 1-4)

There are several steps to consider when manipulating values via hess's law.

1. when you flip the equation, you change the sign of the delta H

2. when you double the reaction, delta H is doubled as well (same when tripled etc..)

3. add the all the resulting delta H's which are arranged.

With your problem, we have to apply the following rules stated above.

1. P4 (s) + 6Cl2 (g) -> 4PCl3 (g) ΔH= -1225.6 kJ/mol

2. P4 (s) + 5O2 (g) -> P4O10 (g) ΔH= -2967.3 kJ/mol

3. PCl3 (g) + Cl2 (g) -> PCl5 (g) ΔH= -84.2 kJ/mol

4. PCl3 (g) + 1/2O2 (g) -> Cl3PO (g) ΔH= -285.7 kJ/mol

1. P4 (s) + 6Cl2 (g) -> 4PCl3 (g) ΔH= -1225.6 kJ/mol

2. P4O10 -> P4 (s) + 5O2 (g)(g) ΔH= +2967.3 kJ/mol

3. 6PCl5(g) -> 6PCl3 (g) + 6Cl2 (g) ΔH= +505.2 kJ/mol

4. 10PCl3 (g) + 5O2 (g) -> 10Cl3PO (g) ΔH= -2857.0 kJ/mol

**= -397.9kJ ---> final answer**