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.

Step-by-step 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 :

=
`(\Delta H^o_(3))/(2 mol)`

`=(180.5 kJ)/(2 mol)=90.25 kJ/mol`