Question

Based on Table I, what is the AH value for the production of 1.00 mole of NO2(g) from its elements at

101.3 kPa and 298 K?
1) +33.2 k) 2) -33.2 kJ 3) +132.8 kJ 4) -132.8 kJ

Answer 1

The ΔH value for the production of 1.00 mole of NO₂(g) from its elements at 101.3 kPa and 298 K is +33.2 kJ, which is calculated by dividing the given ΔH for 2 moles (66.4 kJ) by 2.

Step-by-step explanation:

Based on the information provided, to find the ΔH value for the production of 1.00 mole of NO₂(g) from its elements at 101.3 kPa and 298 K, we start with the given reaction:

N₂(g) + 2O₂(g) → 2NO₂(g) ΔH = +66.4 kJ

This equation shows the formation of 2 moles of NO₂(g) has an enthalpy change (ΔH) of +66.4 kJ. To determine the ΔH for the formation of 1 mole of NO₂(g), we divide the enthalpy change by 2, because enthalpy is an extensive property and is directly proportional to the number of moles in a reaction:

ΔH for 1 mole of NO₂(g) = +66.4 kJ / 2 = +33.2 kJ

Therefore, the correct ΔH value for the production of 1.00 mole of NO₂(g) from its elements at 101.3 kPa and 298 K is +33.2 kJ, which corresponds to option 1 in the question.

Answer 2

ANSWER: 33.2 kJ

EXPLANATION: This is the heat required to produce two moles of NO2(g). Therefore, it would only require half the amount, or 33.2 kj, to produce just one mole.