Question

Use Hess's Law to calculate ΔH for the following equation:

P4O10(g) + 6PCl5(g) --> 10Cl3PO(g)

1/4P4(s) + 3/2Cl2(g) --> PCl3(g) ΔH=-306.4kJ/mol
P4(s) + 5O2(g) --> P4O10(g) ΔH=-2967.3kJ/mol
PCl3(g) + Cl2(g) --> PCl5(g) ΔH=-84.2kJ/mol
PCl3(g) + 1/2O2(g) -->Cl3PO(g) ΔH=-285.7kJ/mol


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Answer 1

The enthalpy change (ΔH) for the given reaction is calculated as -4319.1 kJ/mol using Hess's Law, demonstrating that the reaction is exothermic.

Step-by-step explanation:

To calculate the enthalpy change (ΔH) for the target reaction using Hess's Law, we need to manipulate the given reactions so that, when combined, they yield the target reaction:

1. 
   `\rm P_4(s) + 5O_2(g) \rightarrow P_4O_(10)(g) \Delta H = -2967.3 \ kJ/mol` (reverse this reaction)
2. 
   `\rm 10(PCl_3(g) + 1/2O_2(g) \rightarrow Cl_3PO(g)) \DeltaH = -285.7 \ kJ/mol` each (multiply by 10)
3. 
   `\rm 6(PCl_3(g) + Cl_2(g) \rightarrow PCl_5(g)) \Delta H = -84.2 \ kJ/mol` each (multiply by 6 and reverse)

Reversing a reaction changes the sign of ΔH. Multiply the ΔH of each individual reaction by the number of times the reaction is utilized. Then, add all ΔH values together for the final enthalpy change of the target reaction:

ΔH = [1(-2967.3 kJ/mol)] + [10(-285.7 kJ/mol)] + [6(84.2 kJ/mol)]

ΔH = -2967.3 kJ/mol - 2857 kJ/mol + 505.2 kJ/mol

ΔH = -4319.1 kJ/mol

This calculation reveals that the reaction is exothermic, as indicated by the negative enthalpy change.

Answer 2

To calculate ΔH for the given equation using Hess's Law, we can use a series of known reactions and their enthalpy changes. Here's the step-by-step process:

1) Reverse and adjust the first equation:

- Reverse the first equation: PCl3(g) + Cl2(g) --> 1/4P4(s) + 3/2Cl2(g)

- Multiply the first equation by -1/2 to get the desired coefficients:

-1/2PCl3(g) - 1/2Cl2(g) --> 1/8P4(s) + 3/4Cl2(g)

- The enthalpy change for this reaction is ΔH = -(-306.4 kJ/mol) = 153.2 kJ/mol

2) Adjust the second equation:

- Multiply the second equation by 2 to get the desired coefficients:

2P4(s) + 10O2(g) --> 2P4O10(g)

- The enthalpy change for this reaction is ΔH = -2967.3 kJ/mol

3) Reverse and adjust the third equation:

- Reverse the third equation: PCl5(g) --> PCl3(g) + Cl2(g)

- The enthalpy change for this reaction is ΔH = -(-84.2 kJ/mol) = 84.2 kJ/mol

4) Reverse and adjust the fourth equation:

- Reverse the fourth equation: Cl3PO(g) --> PCl3(g) + 1/2O2(g)

- Multiply the fourth equation by -10 to get the desired coefficients:

-10Cl3PO(g) --> 10PCl3(g) + 5O2(g)

- The enthalpy change for this reaction is ΔH = -(-285.7 kJ/mol) = 285.7 kJ/mol

5) Sum up the adjusted equations:

- Add the adjusted equations from steps 1, 2, 3, and 4 to get the desired equation:

10Cl3PO(g) + 60PCl5(g) --> 80PCl3(g) + 5O2(g) + 8P4O10(g)

- Sum up the enthalpy changes of the adjusted equations:

ΔH = 10(285.7 kJ/mol) + 60(84.2 kJ/mol) - 80(153.2 kJ/mol) - 5(-2967.3 kJ/mol)

ΔH = 2857 kJ/mol + 5052 kJ/mol - 12256 kJ/mol + 14836.5 kJ/mol

ΔH = -2350.5 kJ/mol

Therefore, ΔH for the given equation is -2350.5 kJ/mol.