Question

Calculate the standard molar enthalpy of formation, in kJ/mol, of NO(g) from the following data: N2 (g) + 2 O2 (g) → 2 NO2 (g) △H = 66.4 kJ 2 NO (g) + O2 (g) → 2 NO2 (g) △H = −114.1 kJ

Answer 1

The standard molar enthalpy of formation for NO(g) can be calculated using Hess's Law. It is the sum of the enthalpy changes for the given reactions, divided by two to account for the 2 moles of NO(g), which gives +90.25 kJ/mol.

Step-by-step explanation:

To calculate the standard molar enthalpy of formation for NO(g), we need to consider the provided reactions and apply Hess's Law.

Given:

• N2(g) + 2O2(g) → 2NO2(g), ΔH = +66.4 kJ
• 2NO(g) + O2(g) → 2NO2(g), ΔH = −114.1 kJ

We add the first reaction as it is and reverse the second reaction:

• 2NO2(g) ← 2NO(g) + O2(g), ΔH = +114.1 kJ

Now, we can cancel out the 2NO2 on both sides and combine the enthalpies:

N2(g) + 2O2(g) - 2NO2(g) + 2NO2(g) → 2NO(g) + O2(g)

This simplifies to:

N2(g) + O2(g) → 2NO(g), ΔH = 66.4 kJ + 114.1 kJ

Thus, the standard molar enthalpy of formation for NO(g) is half of the combined enthalpy change:

(66.4 kJ + 114.1 kJ)/2 = 90.25 kJ/mol

So, the enthalpy of formation ΔHf for NO(g) is +90.25 kJ/mol.