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 find the standard enthalpy change for the reaction:

N2O4(g) + Cl2(g) → 2 NOCl(g) + O2(g)

we'll use the given thermochemical equations to calculate the overall standard enthalpy change.

First, let's reverse the second equation:

NO2(g) → NO(g) + (1/2) O2(g) ΔH = +56.53 kJ

Now, let's multiply the first equation by 2:

2 NO(g) + Cl2(g) → 2 NOCl(g) ΔH = 2 × -37.78 kJ = -75.56 kJ

Let's multiply the third equation by 2:

4 NO2(g) → 2 N2O4(g) ΔH = 2 × -58.03 kJ = -116.06 kJ

Finally, let's sum up all the equations to get our target equation:

2 NOCl(g) + O2(g) + N2O4(g) → 2 NOCl(g) + O2(g)

ΔH(total) = ΔH1 + ΔH2 + ΔH3
ΔH(total) = -75.56 kJ + 56.53 kJ + (-116.06 kJ)
ΔH(total) = -135.63 kJ

Therefore, the standard enthalpy change for the reaction N2O4(g) + Cl2(g) → 2 NOCl(g) + O2(g) is -135.63 kJ (per mol N2O4), 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 the law of conservation of energy.

First, let's write the given equations:
1) NO(g) + (1/2) Cl2(g) -> NOCl(g) ΔH = -37.78 kJ/mol NOCl
2) NO(g) + (1/2) O2(g) -> NO2(g) ΔH = -56.53 kJ/mol NO2
3) 2NO2(g) -> N2O4(g) ΔH = -58.03 kJ/mol N2O4

To get the target equation, we need to manipulate the given equations by multiplying or reversing them as necessary:
Multiplying equation 1 by 2: 2NO(g) + Cl2(g) -> 2NOCl(g) ΔH = -37.78 kJ/mol NOCl
Reversing equation 2: NO2(g) -> NO(g) + (1/2) O2(g) ΔH = +56.53 kJ/mol NO2

Now, let's add up the manipulated equations to obtain the target equation:
2 NO(g) + Cl2(g) + NO2(g) -> 2NOCl(g) + (1/2) O2(g)

Finally, we add up the ΔH values for the manipulated equations to find the ΔH for the target equation:
ΔH = -37.78 kJ/mol NOCl + 56.53 kJ/mol NO2

ΔH = 18.75 kJ/mol

Therefore, the standard enthalpy change for the reaction N2O4(g) + Cl2(g) -> 2NOCl(g) + O2(g) is 18.75 kJ/mol N2O4.