# 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

jerichorayel | College Teacher | (Level 2) Senior Educator

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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

by inspection we will have the folowing observations:

1 is retained
2 is flipped
3 is flipped then multiplied by 6
4 is retained but bultiplied by 10

the resulting equation would be:

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

ΔH = -610.1kJ/mol

Now we have to get the heat in 100 grams of Cl3PO; we will then convert it to moles.

100grams Cl3PO x ( 1mole Cl3PO /153.3328 grams) \

= 0.652176 moles

-610.1 kJ/mol x 0.652176 mol