Question

What is the standard enthalpy change for the reaction where 2 moles of hydrochloric acid react with 1 mole of magnesium hydroxide in aqueous solutions to produce 1 mole of magnesium chloride and 2 moles of water at 298 K and 1 atm pressure?

Answer 1

To calculate the standard enthalpy change for the reaction, we need to use Hess's Law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for each step in the reaction. In this case, we need to find the standard enthalpies of formation for each compound involved in the reaction and then use the following equation:H_reaction =  H_products -  H_reactantsThe balanced chemical equation for the reaction is:2 HCl aq  + Mg OH 2 aq   MgCl2 aq  + 2 H2O l Now, we need the standard enthalpies of formation  Hf  for each compound:Hf HCl, aq  = -167.2 kJ/molHf Mg OH 2, aq  = -924.7 kJ/molHf MgCl2, aq  = -801.0 kJ/molHf H2O, l  = -285.8 kJ/molNow, we can plug these values into the equation:H_reaction = [ 1  -801.0  +  2  -285.8 ] - [ 2  -167.2  +  1  -924.7 ]H_reaction =  -801.0 - 571.6  -  -334.4 - 924.7 H_reaction =  -1372.6  -  -1259.1 H_reaction = -113.5 kJTherefore, the standard enthalpy change for the reaction is -113.5 kJ.