Use Hess' Law to calculate the heat of the following equation. Show all work including the reactions that you use, in the form that you use them.

2 C + 2 NO2 ---> 2 CO 2 + N 2

deltaHrxn = (n*deltaHf products) - (n*deltaHf reactants).

Look up delta Hf values, substitute, and calculate delta H rxn.

To calculate the heat of the given equation using Hess' Law, we need to break down the overall reaction into multiple sub-reactions for which we know the enthalpy changes.

Step 1: Break down the given equation into its constituent reactions:

1) C + O2 ---> CO2 (∆H1)

2) N2 + O2 ---> 2NO2 (∆H2)

3) C + O2 ---> CO2 (reversed) (∆H1 reversed)

Step 2: Rearrange the equations to match the overall reaction:

1) 2C + 2O2 ---> 2CO2

2) N2 + 2O2 ---> 2NO2

3) 2CO2 ---> 2C + 2O2 (reversed)

Step 3: Determine the enthalpy change for each of the reactions:

∆H1: The enthalpy change for the combustion of carbon:

C + O2 ---> CO2 (∆H1)

∆H1 value can be found from standard enthalpy tables or using thermochemical data. Let's assume its value is -393.5 kJ/mol.

∆H2: The enthalpy change for the production of nitrogen dioxide:

N2 + O2 ---> 2NO2 (∆H2)

∆H2 value can be found from standard enthalpy tables or using thermochemical data. Let's assume its value is +66.4 kJ/mol.

∆H1 reversed: The enthalpy change for the reverse of the combustion of carbon:

2CO2 ---> 2C + 2O2 (∆H1 reversed)

Since this reaction is the reverse of reaction ∆H1, its value would be the negative of ∆H1. So, ∆H1 reversed = -(-393.5 kJ/mol) = +393.5 kJ/mol.

Step 4: Apply Hess' Law to calculate the enthalpy change for the overall reaction:

Multiply the given equation by the appropriate coefficients to match the rearranged equations.

2 C + 2 NO2 ---> 2 CO2 + N2

From equation 1: 2C + 2O2 ---> 2CO2

From equation 2: N2 + 2O2 ---> 2NO2

From equation 3: 2CO2 ---> 2C + 2O2 (reversed)

Add the enthalpy changes of the equations to get the enthalpy change of the overall equation:

∆H overall = (2 * ∆H1) + (∆H2) + (2 * ∆H1 reversed)
= (2 * -393.5 kJ/mol) + (+66.4 kJ/mol) + (2 * 393.5 kJ/mol)
= -787.0 kJ/mol + 66.4 kJ/mol + 787.0 kJ/mol
= +66.4 kJ/mol

Therefore, the heat of the given equation is +66.4 kJ/mol.