Question

Determine the standard enthalpy of formation of nitrogen dioxide (NO2) given that the balanced equation for the combustion of one mole of NO2 is:4NO2(g) + 6H2O(l) → 4HNO3(aq) + 2NO(g)The standard enthalpies of formation for HNO3(l), H2O(l), and NO(g) are -174.1 kJ/mol, -285.8 kJ/mol, and 90.4 kJ/mol, respectively.

Answer 1

To determine the standard enthalpy of formation of nitrogen dioxide  NO2 , we can use the following equation:Hf  NO2  = Hf  products  - Hf  reactants First, we need to find the enthalpies of formation for the products and reactants in the balanced equation:4HNO3 aq  + 2NO g  - 4NO2 g  - 6H2O l The standard enthalpies of formation for HNO3 l , H2O l , and NO g  are given as -174.1 kJ/mol, -285.8 kJ/mol, and 90.4 kJ/mol, respectively. We can plug these values into the equation:Hf  NO2  = [4 -174.1  + 2 90.4 ] - [- 4 NO2  - 6 -285.8 ]Now, we can solve for Hf  NO2 :Hf  NO2  = [-696.4 + 180.8] - [- 4 NO2  + 1714.8]Hf  NO2  = [-515.6] - [- 4 NO2  + 1714.8]Now, we can isolate Hf  NO2  on one side of the equation:4 NO2  = 1714.8 - 515.64 NO2  = 1199.2Now, we can solve for the standard enthalpy of formation of nitrogen dioxide  NO2 :Hf  NO2  = 1199.2 / 4Hf  NO2  = 299.8 kJ/molTherefore, the standard enthalpy of formation of nitrogen dioxide  NO2  is 299.8 kJ/mol.