Question

How much energy is released in the production of 37.4g of nitrogen dioxide according to the following equation: NO + 1/2 O2 = NO2

Answer 1

To calculate the energy released in the production of 37.4g of nitrogen dioxide (NO2), we need to use the balanced equation and standard enthalpies of formation.

The balanced equation is: NO + 1/2 O2 → NO2

The standard enthalpies of formation (ΔH) are:
ΔHf(NO) = 90.25 kJ/mol
ΔHf(O2) = 0 kJ/mol (by definition)
ΔHf(NO2) = 33.18 kJ/mol

We can calculate the energy released by subtracting the sum of the standard enthalpies of formation of the reactants from the sum of the products.

Reactants:
1 mol of NO: 90.25 kJ/mol (from ΔHf(NO))
0.5 mol of O2: 0 kJ/mol (by definition)

Products:
1 mol of NO2: 33.18 kJ/mol (from ΔHf(NO2))

Energy released = ΣΔHf(products) - ΣΔHf(reactants)
Energy released = (1 mol)(33.18 kJ/mol) - (1 mol)(90.25 kJ/mol) + (0.5 mol)(0 kJ/mol)
Energy released = 33.18 kJ/mol - 90.25 kJ/mol
Energy released = -57.07 kJ/mol

Now we need to calculate the number of moles of NO2 produced from 37.4g using the molar mass of NO2.

Molar mass of NO2 = (14 + 16 + 16) g/mol = 46 g/mol

Number of moles of NO2 = Mass of NO2 / Molar mass of NO2
Number of moles of NO2 = 37.4g / 46 g/mol = 0.813043478 mol

Finally, we can calculate the total energy released in the production of 37.4g of NO2.

Total energy released = Energy released per mole × Number of moles
Total energy released = (-57.07 kJ/mol) × (0.813043478 mol)
Total energy released = -46.38 kJ

Therefore, approximately 46.38 kJ of energy is released in the production of 37.4g of nitrogen dioxide.