Question

What is the standard enthalpy change for the reaction when 50 ml of 0.2 M hydrochloric acid is mixed with 50 ml of 0.1 M sodium hydroxide solution at 25°C? The specific heat capacity of the mixed solution is 4.18 J/g°C and the density is 1 g/mL.

Answer 1

To calculate the standard enthalpy change for the reaction, we first need to determine the limiting reactant and the amount of heat released during the reaction. The balanced chemical equation for the reaction between hydrochloric acid  HCl  and sodium hydroxide  NaOH  is:HCl aq  + NaOH aq   NaCl aq  + H2O l First, let's find the moles of each reactant:Moles of HCl =  volume in L    molarity  =  50 mL   1 L / 1000 mL    0.2 M = 0.01 molMoles of NaOH =  volume in L    molarity  =  50 mL   1 L / 1000 mL    0.1 M = 0.005 molSince there are fewer moles of NaOH, it is the limiting reactant. The reaction will produce 0.005 mol of water.Next, we need to find the heat released during the reaction. We can use the heat of neutralization, which is the heat released when one mole of water is formed from the reaction between an acid and a base. The heat of neutralization for a strong acid and a strong base is approximately -57.1 kJ/mol.Heat released =  moles of water formed    heat of neutralization  = 0.005 mol  -57.1 kJ/mol = -0.2855 kJNow we can calculate the standard enthalpy change for the reaction. Since the heat released is negative, the reaction is exothermic, and the standard enthalpy change will also be negative.H = -0.2855 kJThe standard enthalpy change for the reaction is -0.2855 kJ.