C. Use Hess's law and the following equations to calculate ΔH for the reaction N2(g) + 2O2(g) N2O4(g). (Show your work.) (8 points) 2NO2(g) N2O4(g) ΔH = –57.0 kJ/mol

N2(g) + O2(g) 2NO(g) ΔH = 180.6 kJ/mol

2NO2(g) 2NO(g) + O2(g) ΔH = 114.4 kJ/mol

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ΔH=9.2 KJ

Explanation:

2NO₂→N₂O₄  ΔH=-57 KJ/mol

N₂ + O₂→2NO ΔH=180.6 KJ/mol

2NO + O₂→2NO₂ ΔH= -114.4 (notice this was reversed and sign changed

N₂+2O₂→N₂O₄ ΔH= 9.2 KJ

By reversing the third equation, we can get the O₂ on the left side or the (reactant side) so that we can get a total of 2O₂ in the final equation on the bottom. Cross out 2NO₂ on the first equation on the reactant side and 2NO₂ in equation 2 on the product side. You also want to cross out 2NO in the third equation on both the reactant and product side. This leaves the same above the line as below the line.

Since you reverse equation 3, the change in enthalpy will change sign so you input -57, 180.6 and -114.4 to get 9.2 KJ

Hope that helps.

Explanation to question:

2NO₂→N₂O₄  ΔH=-57 KJ/mol

N₂ + O₂→2NO ΔH=180.6 KJ/mol

2NO + O₂→2NO₂ ΔH= -114.4