Question

What is the standard enthalpy change for the reaction N2(g) + 3H2(g) → 2NH3(g) at 298 K, given that the standard enthalpies of formation of nitrogen dioxide, water and ammonia are 82.05 kJ/mol, -285.83 kJ/mol and -46.19 kJ/mol, respectively?

Answer 1

To calculate the standard enthalpy change for the reaction, we can use the following formula:H =  Hf products  -  Hf reactants where H is the standard enthalpy change, and Hf is the standard enthalpy of formation.For the given reaction, N2 g  + 3H2 g   2NH3 g , we have:H = [2  Hf NH3 ] - [Hf N2  + 3  Hf H2 ]Since N2 and H2 are in their standard states  elemental forms , their enthalpies of formation are zero:Hf N2  = 0 kJ/molHf H2  = 0 kJ/molNow, we can plug in the given values for the enthalpy of formation of ammonia:H = [2   -46.19 kJ/mol ] - [0 + 3  0]H = -92.38 kJ/molSo, the standard enthalpy change for the reaction N2 g  + 3H2 g   2NH3 g  at 298 K is -92.38 kJ/mol.