Question

What is the standard enthalpy change for the reaction N2(g) + 3H2(g) → 2NH3(g) given the following standard enthalpies of formation: ΔHf°(N2)=0 kJ/mol, ΔHf°(H2)=0 kJ/mol, ΔHf°(NH3)=-46.2 kJ/mol? Make sure to show your calculations and units.

Answer 1

To calculate the standard enthalpy change for the reaction, we can use the following equation:H reaction  =  Hf products  -  Hf reactants For the given reaction, N2 g  + 3H2 g   2NH3 g , we have:Hf N2  = 0 kJ/molHf H2  = 0 kJ/molHf NH3  = -46.2 kJ/molNow, we can plug these values into the equation:H reaction  = [2 * Hf NH3 ] - [1 * Hf N2  + 3 * Hf H2 ]H reaction  = [2 *  -46.2 kJ/mol ] - [1 *  0 kJ/mol  + 3 *  0 kJ/mol ]H reaction  =  -92.4 kJ/mol  -  0 kJ/mol H reaction  = -92.4 kJ/molSo, the standard enthalpy change for the reaction is -92.4 kJ/mol.