NO2(g) --> 1/2N2(g) + O2(g) [deltaH1= -33.9 kJ] NO2(g) --> NO(g) + 1/2O2(g) [del

Question

NO2(g) --> 1/2N2(g) + O2(g) [deltaH1= -33.9 kJ]
NO2(g) --> NO(g) + 1/2O2(g) [deltaH2= 56.5kJ]

What is deltaH for the reaction 1/2N2(g) + 1/2O2(g) --> NO(g) ?

Explanation / Answer

NO2(g) --> 1/2N2(g) + O2(g) [deltaH1= -33.9 kJ ------ (1) NO2(g) --> NO(g) + 1/2O2(g) [deltaH2= 56.5k ------ (2) now reverse the equation (2) which gives NO(g) + 1/2O2(g) ----- > NO2(g) deltaH3 = -56.5 KJ (3) adding eqaution (1) and (3) gives NO(g) ----> 1/2 N2 + 1/2 O2 delatH = -56.5 + -33.9 = -90.4 KJ --(4) reversing (4) gives you 1/2N2(g) + 1/2O2(g) --> NO(g) and delta H = 90.4 KJ

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