Question

Calculate the enthalpy change of mixing 50 mL of 1 M HCl with 50 mL of 1 M NaOH at 25°C. It is known that the standard enthalpies of formation of HCl and NaOH are -92.3 kJ/mol and -469.2 kJ/mol, respectively. Assume that the heat capacity of the mixture is 4.18 J/g°C and the density of the solution is 1 g/mL.

Answer 1

To calculate the enthalpy change of mixing, we need to find the enthalpy change of the reaction between HCl and NaOH. The balanced chemical equation for the reaction is:HCl aq  + NaOH aq   NaCl aq  + H2O l First, we need to find the moles of HCl and NaOH in the solution. Since both solutions have a concentration of 1 M and a volume of 50 mL, we can calculate the moles as follows:moles of HCl =  1 mol/L  *  50 mL  *  1 L/1000 mL  = 0.05 molmoles of NaOH =  1 mol/L  *  50 mL  *  1 L/1000 mL  = 0.05 molSince the moles of HCl and NaOH are equal, the reaction will go to completion, and all of the HCl and NaOH will react to form NaCl and H2O.Next, we need to find the standard enthalpy of formation for NaCl and H2O. The standard enthalpy of formation for NaCl aq  is -407.3 kJ/mol, and for H2O l  is -285.8 kJ/mol.Now, we can calculate the enthalpy change of the reaction using Hess's Law:H_reaction =  H_f products  -  H_f reactants H_reaction = [ -407.3 kJ/mol  +  -285.8 kJ/mol ] - [ -92.3 kJ/mol  +  -469.2 kJ/mol ]H_reaction =  -693.1 kJ/mol  -  -561.5 kJ/mol  = -131.6 kJ/molNow that we have the enthalpy change per mole of reaction, we can calculate the total enthalpy change for the reaction:H_total = H_reaction * moles of reactionH_total = -131.6 kJ/mol * 0.05 mol = -6.58 kJThe enthalpy change of mixing 50 mL of 1 M HCl with 50 mL of 1 M NaOH at 25C is -6.58 kJ.