Question

Michael has been measuring the enthalpy change associated with reaction between HCl and NaOH in a coffee cup calorimeter. Michael combined 50.77 mL of 1.00 M HCl and 37.83 mL of 1.00 M NaOH in a coffee cup calorimeter; mass of the coffee cups and a stir bar is 15.00 g. If the initial temperature of the acid/base solution was 18.51 oC, and the final observed temperature was 32.06 oC, what is the enthalpy change of the neutralization reaction, in KJ per mole of NaOH? Assume that the density is 1.00 g/mL and specific heat capacity of the solution are identical to that of pure water, 4.184 J/g-K.

Answer 1

ns. # Balanced reaction:      NaOH(aq) + HCl(aq) ----> NaCl(aq) + H₂O(l).

One mol NaOH is neutralized by 1 mol HCl.

# Moles of NaOH = Molarity x Volume of solution in liters

                                    = 1.00 M x 0.03783 L

                                    = 0.03783 mol

Moles of HCl = 1.0 M x 0.05077 L = 0.05077 mol

Since moles of NaOH (0.03783 mol) is less than that of HCl (0.05077 mol), NaOH is the limiting reactant in this reaction.

The formation of product follows the stoichiometry of limiting reactant.

So,

Moles of H2O formed – Moles of NaOH neutralized = 0.03783 mol

# Total volume of reaction mixture = 50.77 mL (HCl) + 37.83 mL (NaOH)

                                                            = 88.60 mL

Mass of reaction mixture = Volume x Density of reaction mixture

                                                = 88.60 mL x (1.00 g/ mL

= 88.60 g

# Ans. Amount of heat released during neutralization reaction is given by-

            q = m s dT                            - equation 1

Where,

q = heat change

m = mass

s = specific heat

dT = Final temperature – Initial temperature = (32.06 – 18.51)⁰C = 13.55⁰C

Putting the values in equation 1-

            q = 88.60 g x (4.184 J g^-10C^-1) x 13.55⁰C = 5023.01752 J

Hence, amount of heat released during neutralization = 5023.01752 J

# Enthalpy change per mol NaOH = Amount of heat released / Moles of NaOH neutralized

                                                            = 5023.01752 J / 0.03783 mol

                                                            = 132778.68 J/ mol

                                                            = 132.778 kJ/ mol

Hence, molar enthalpy of neutralization of NaOH = - 132.778 kJ/ mol (the –ve sign indicates heat is being released)

Note: The experimental value is abnormally high. Check if the calorimeter has its own heat capacity value. Or, all other measurements are recorded properly.