Answer:
The standard molar enthalpy of formation of 1 mole of NO gas is 90.25 kJ/mol.
Explanation:
We have :
[tex]N_2(g) + 2O_2 \rightarrow 2NO_2(g), \Delta H^o_{1} = 66.4 kJ[/tex]
[tex]2NO(g) + O_2\rightarrow 2NO_2(g),\Delta H^o_{2} = -114.1 kJ[/tex]
To calculate the standard molar enthalpy of formation
[tex]N_2+O_2\rightarrow 2NO(g),\Delta H^o_{3} = ?[/tex]...[3]
[1] - [2] = [3] (Hess's law)
[tex]N_2+O_2\rightarrow 2NO(g),\Delta H^o_{3} = ?[/tex]
[tex]\Delta H^o_{3} =\Delta H^o_{1} - \Delta H^o_{2} [/tex]
[tex]\Delta H^o_{3}=66.4 kJ - [ -114.1 kJ] = 180.5 kJ[/tex]
According to reaction [3], 1 mole of nitrogen gas and 1 mole of oxygen gas gives 2 mole of nitrogen monoxide.
So, the standard molar enthalpy of formation of 1 mole of NO gas :
[tex]\Delta H^o_{f,NO}=\frac{\Delta H^o_{3}}{2 mol}[/tex]
[tex]\Delta H^o_{f,NO}=\frac{180.5 kJ}{2 mol}=90.25 kJ/mol[/tex]
