Question

Given the following reactions

N2 (g) + O2 (g) - 2NO (9) AH = +180.7 kJ
2NO(g) + O2 (g) - 2NO2 (g) AH = -113.1 kJ
the enthalpy for the decomposition of nitrogen dioxide into molecular nitrogen and oxygen
2NO2 (g) - N2 (g) + 202 (g)
is ____
kJ.

Answer 1

The enthalpy for the decomposition of nitrogen dioxide (NO2) into molecular nitrogen (N2) and oxygen (O2) is +146.9 kJ, derived from the sum of the reversed reactions' enthalpy changes.

Step-by-step explanation:

To determine the enthalpy for the decomposition of nitrogen dioxide (NO2) into molecular nitrogen (N2) and oxygen (O2), we need to reverse the given reaction and change the sign of the enthalpy change. We are given two reactions:

• N2 (g) + O2 (g) → 2NO (g), ΔH = +180.7 kJ
• 2NO(g) + O2 (g) → 2NO2 (g), ΔH = -113.1 kJ

The equation for the decomposition of 2NO2 is the reverse of the second reaction:

• 2NO2 (g) → 2NO (g) + O2 (g), ΔH = +113.1 kJ

However, we need to consider the formation reaction of NO and adjust our enthalpy changes accordingly. The overall enthalpy change (ΔH) for the decomposition process can be determined by adding the enthalpies of the reversed NO2 formation reaction (+113.1 kJ) and the NO formation reaction (+180.7 kJ), yielding:

ΔH for decomposition of 2NO2 = +113.1 kJ + 180.7 kJ = +293.8 kJ

Since we want the enthalpy for the decomposition of 2NO2 into N2 and 2O2, not just for the conversion to NO and O2, the correct answer would involve halving this value, as the formation of NO is an intermediate step:

ΔH for decomposition of 2NO2 into N2 and 2O2 = +293.8 kJ / 2 = +146.9 kJ