Asked by mara
Calculate Hrxn for the reaction below.
2 NOCl(g) N2(g) + O2(g) + Cl2(g)
You are given the following set of reactions.
1/2 N2(g) + 1/2 O2(g) NO(g) H = 90.30 kJ
NO(g) + 1/2 Cl2(g) NOCl(g) H = -38.6 kJ
2 NOCl(g) N2(g) + O2(g) + Cl2(g)
You are given the following set of reactions.
1/2 N2(g) + 1/2 O2(g) NO(g) H = 90.30 kJ
NO(g) + 1/2 Cl2(g) NOCl(g) H = -38.6 kJ
Answers
Answered by
Anonymous
103.4
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