Given the following thermochemical equations,

NO(g) + (1/2) Cl2(g) --> NOCl(g) -37.78 kJ (per mol NOCl)
NO(g) + (1/2) O2(g) --> NO2(g) -56.53 kJ (per mol NO2)
2 NO2(g) --> N2O4(g) -58.03 kJ (per mol N2O4)
what is the standard enthalpy change for the reaction below?
N2O4(g) + Cl2(g) --> 2 NOCl(g) + O2(g)
Give your answer in kJ (per mol N2O4), accurate to two decimal places.

To determine the standard enthalpy change for the reaction N2O4(g) + Cl2(g) --> 2 NOCl(g) + O2(g), we can use Hess's Law.

We need to rearrange the given thermochemical equations and their corresponding enthalpy changes to obtain the desired reaction:

1. Reverse the third equation:
N2O4(g) --> 2 NO2(g) ΔH = +58.03 kJ/mol

2. Multiply the first equation by 2 and reverse it:
2 NOCl(g) --> 2 NO(g) + Cl2(g) ΔH = (+37.78 kJ/mol) x 2 = +75.56 kJ/mol

3. Reverse the second equation:
NO2(g) --> NO(g) + (1/2) O2(g) ΔH = +56.53 kJ/mol

Now, we add the modified equations together to obtain the desired reaction:

N2O4(g) + Cl2(g) --> 2 NOCl(g) + O2(g)

Adding the equations:

N2O4(g) + Cl2(g) + 2 NOCl(g) + O2(g) ΔH = (+58.03 kJ/mol) + (+75.56 kJ/mol) + (+56.53 kJ/mol)

Simplifying:

N2O4(g) + Cl2(g) + 2 NOCl(g) + O2(g) ΔH = +189.12 kJ/mol

Therefore, the standard enthalpy change for the reaction N2O4(g) + Cl2(g) --> 2 NOCl(g) + O2(g) is +189.12 kJ/mol, accurate to two decimal places.

To find the standard enthalpy change for the reaction N2O4(g) + Cl2(g) → 2NOCl(g) + O2(g), we need to use the given thermochemical equations and apply Hess's Law. Hess's Law states that if a reaction can be expressed as the sum of a series of other reactions, then the enthalpy change for the overall reaction is equal to the sum of the enthalpy changes of the individual reactions.

1. Reverse the first equation: NOCl(g) → NO(g) + (1/2) Cl2(g). Since the enthalpy change of this reaction is -37.78 kJ (per mol NOCl), the reversed reaction will have an enthalpy change of +37.78 kJ (per mol NOCl).

2. Multiply the second equation by 2 to make the coefficient of NO2 equal to that in the overall reaction: 2 NO(g) + O2(g) → 2NO2(g). The enthalpy change of this reaction is -56.53 kJ (per mol NO2).

3. Reverse and multiply the third equation by 2 to make the coefficient of N2O4 equal to that in the overall reaction: N2O4(g) → 2NO2(g). The enthalpy change of this reaction is +58.03 kJ (per mol N2O4).

4. Add the equations together and cancel the common species:
N2O4(g) → 2NO2(g) ΔH = +58.03 kJ (multiply this equation by 1)
2NO(g) + O2(g) → 2NO2(g) ΔH = -56.53 kJ (multiply this equation by 1)
2NOCl(g) → 2NO(g) + Cl2(g) ΔH = +37.78 kJ (multiply this equation by 1/2)

5. Add the equations together to get the overall equation:
N2O4(g) + Cl2(g) → 2 NOCl(g) + O2(g) ΔH = +58.03 kJ - 56.53 kJ + 0.5(37.78 kJ)

Simplifying the equation, we get:
ΔH = +0.50 kJ (per mol N2O4)

Therefore, the standard enthalpy change for the reaction N2O4(g) + Cl2(g) → 2 NOCl(g) + O2(g) is +0.50 kJ (per mol N2O4), accurate to two decimal places.

I did this in my head and I BELIEVE (but you need to try it to see)

eqn 3 reversed. Add
to 2x eqn 1
and add to the reverse of 2 x equn 2.

Add the dH values. When you multiply a rxn by a coefficient do the same for dH. When you reverse and equation change the sign on dH.