Question

Calculate the standard molar enthalpy of formation, in kJ/mol, of NO(g) from the following data:

Answer 1

The above question is incomplete, here is the complete question:

Calculate the standard molar enthalpy of formation of NO(g) from the following data at 298 K:

$N_2(g) + 2O_2 \rightarrow 2NO_2(g), \Delta H^o = 66.4 kJ$

$2NO(g) + O_2\rightarrow 2NO_2(g),\Delta H^o= -114.1 kJ$

Answer:

The standard molar enthalpy of formation of NO is 90.25 kJ/mol.

Explanation:

$N_2(g) + 2O_2 \rightarrow 2NO_2(g), \Delta H^o_{1} = 66.4 kJ$

$2NO(g) + O_2\rightarrow 2NO_2(g),\Delta H^o_{2} = -114.1 kJ$

To calculate the standard molar enthalpy of formation

$N_2+O_2\rightarrow 2NO(g),\Delta H^o_{3} = ?$...[3]

Using Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

[1] - [2] = [3]

$N_2+O_2\rightarrow 2NO(g),\Delta H^o_{3} = ?$

$\Delta H^o_{3} =\Delta H^o_{1} - \Delta H^o_{2}$

$\Delta H^o_{3}=66.4 kJ - [ -114.1 kJ] = 180.5 kJ$

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 :

=$\frac{\Delta H^o_{3}}{2 mol}$

$=\frac{180.5 kJ}{2 mol}=90.25 kJ/mol$