Answer:
The standard molar enthalpy of formation of NO is 180.5 kJ/mol.
Explanation:
[tex]N_2 + 2O_2\rightarrow 2NO_2, \Delta H^{o}_1= 66.4 kJ/mol[/tex]..(1)
[tex]2NO + O_2\rightarrow 2NO_2, \Delta H^{o}_2 = -114.1 kJ/mol[/tex]..(2)
[tex]N_2+O_2\rightarrow 2NO,\Delta H^{o}_3= ?[/tex]..(3)
On Subtraction of equation (1)from equation (2) we get equation (3)
(1) - (2) = (3)
[tex]\Delta H^{o}_3 = \Delta H^{o}_1-\Delta H^{o}_2[/tex]
[tex]\Delta H^{o}_3=66.4 kJ/mol - (-114.1 kJ/mol)=180.5 kJ/mol[/tex]
