Question

What is the standard enthalpy change for the reaction that occurs when 50 ml of 0.1 M HCl is mixed with 50 ml of 0.1 M NaOH at 25°C? The equation for the reaction is: HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l).

Answer 1

To calculate the standard enthalpy change for the reaction, we need to use the heat of neutralization value for the reaction between a strong acid  HCl  and a strong base  NaOH . The heat of neutralization for this reaction is approximately -57.32 kJ/mol.First, we need to determine the number of moles of HCl and NaOH in the solution:Moles of HCl =  Volume of HCl  x  Concentration of HCl  =  0.050 L  x  0.1 mol/L  = 0.005 molMoles of NaOH =  Volume of NaOH  x  Concentration of NaOH  =  0.050 L  x  0.1 mol/L  = 0.005 molSince the moles of HCl and NaOH are equal, the reaction goes to completion, and all the HCl and NaOH will react to form NaCl and H2O.Now, we can calculate the standard enthalpy change for the reaction:H =  moles of HCl  x  heat of neutralization  =  0.005 mol  x  -57.32 kJ/mol  = -0.2866 kJThe standard enthalpy change for the reaction is approximately -0.29 kJ.