A student needs to determine the enthalpy change (ΔH) for the reaction between ammonium nitrate (NH4NO3) and water (H2O) using Hess's Law calculations. Given the following information: - ΔHf° (NH4NO3) = -365.4 kJ/mol- ΔHf° (HNO3) = -207.3 kJ/mol- ΔHf° (NH4+) = -132.5 kJ/mol- ΔHf° (NO3-) = -206.1 kJ/mol- ΔHf° (H2O) = -285.83 kJ/molFormulate the thermochemical equation for the reaction by using the given data and Hess's Law calculations, and then calculate the enthalpy change for the reaction.

Question

A student needs to determine the enthalpy change (ΔH) for the reaction between ammonium nitrate (NH4NO3) and water (H2O) using Hess's Law calculations. Given the following information: - ΔHf° (NH4NO3) = -365.4 kJ/mol- ΔHf° (HNO3) = -207.3 kJ/mol- ΔHf° (NH4+) = -132.5 kJ/mol- ΔHf° (NO3-) = -206.1 kJ/mol- ΔHf° (H2O) = -285.83 kJ/molFormulate the thermochemical equation for the reaction by

Answer 1

First, we need to write the balanced chemical equation for the reaction between ammonium nitrate and water:NH4NO3  s  + H2O  l   NH4+  aq  + NO3-  aq  + H2O  l Now, we can use Hess's Law to determine the enthalpy change for the reaction. Hess's Law states that the enthalpy change for a reaction is equal to the sum of the enthalpy changes for the formation of the products minus the sum of the enthalpy changes for the formation of the reactants.H = [Hf  NH4+  + Hf  NO3-  + Hf  H2O ] - [Hf  NH4NO3  + Hf  H2O ]Substitute the given values:H = [ -132.5 kJ/mol  +  -206.1 kJ/mol  +  -285.83 kJ/mol ] - [ -365.4 kJ/mol  +  -285.83 kJ/mol ]H = [ -132.5 - 206.1 - 285.83  kJ/mol] - [ -365.4 - 285.83  kJ/mol]H =  -624.43 kJ/mol  -  -651.23 kJ/mol H = 26.8 kJ/molThe enthalpy change  H  for the reaction between ammonium nitrate and water is 26.8 kJ/mol.