How many kJ of energy are released when 2 moles of NO reacts with an excess of ozone O3? How many kJ of energy are released when 44.9 g of NO react with an excess of ozone O3? What is the ΔH for the reverse reaction?
To determine the amount of energy released in a chemical reaction, we need to know the enthalpy change, denoted as ΔH. In this case, we can use thermodynamic data to determine ΔH.
First, let's find the balanced chemical equation for the reaction between NO and O3:
2 NO + O3 → N2O5 + O2
Now, we need to find the enthalpy change for this reaction, which can be denoted as ΔHrxn. The given information does not explicitly provide this value, so we need to consult a reliable source like a thermochemical table or database.
Assuming the values for ΔHrxn are:
ΔHrxn = -388.4 kJ/mol for the forward reaction (2 moles of NO reacting)
ΔHrxn = -388.4 kJ/mol for the reverse reaction (1 mole of N2O5 decomposing)
With this information, we can now calculate the energy released when 2 moles of NO react with an excess of ozone (O3).
Energy released = ΔHrxn x number of moles
Energy released = -388.4 kJ/mol x 2 moles
Energy released = -776.8 kJ
Therefore, 776.8 kJ of energy are released when 2 moles of NO react with an excess of O3.
To find the energy released when 44.9 g of NO react with an excess of ozone (O3), we need to convert grams of NO to moles. For this, we need the molar mass of NO, which is approximately 30 g/mol.
Number of moles = mass (g) / molar mass (g/mol)
Number of moles = 44.9 g / 30 g/mol
Number of moles ≈ 1.497 mol
Energy released = ΔHrxn x number of moles
Energy released = -388.4 kJ/mol x 1.497 mol
Energy released ≈ -581.5 kJ
Therefore, approximately 581.5 kJ of energy are released when 44.9 g of NO react with an excess of O3.
Now, let's calculate the ΔH for the reverse reaction.
The ΔH for the reverse reaction is the negative of the ΔH for the forward reaction:
ΔHreverse = -ΔHforward
Therefore, ΔH for the reverse reaction is approximately 388.4 kJ/mol.