Question

Calculate the enthalpy change for the following chemical reaction: 2NH3(g) + 3O2(g) → 2NO(g) + 3H2O(l) Given the following enthalpy values: ΔH°f (NH3) = −46.11 kJ/molΔH°f (O2) = 0 kJ/molΔH°f (NO) = 90.3 kJ/molΔH°f (H2O) = −285.8 kJ/mol

Answer 1

To calculate the enthalpy change  H  for the given chemical reaction, we can use the following formula:H =  Hf products  -  Hf reactants For the products:2 moles of NO: 2  90.3 kJ/mol = 180.6 kJ3 moles of H2O: 3  -285.8 kJ/mol = -857.4 kJFor the reactants:2 moles of NH3: 2  -46.11 kJ/mol = -92.22 kJ3 moles of O2: 3  0 kJ/mol = 0 kJ  since the enthalpy of formation for O2 is 0 Now, substitute these values into the formula:H =  180.6 kJ +  -857.4 kJ   -  -92.22 kJ + 0 kJ H =  -676.8 kJ  -  -92.22 kJ H = -584.58 kJThe enthalpy change for the given chemical reaction is -584.58 kJ.