Question

Andrea has been measuring the enthalpy change associated with reaction between HCl and NaOH in a coffee cup calorimeter. Andrea combined 45.24 mL of 1.00 M HCl and 26.25 mL of 1.00 M NaOH in a coffee cup calorimeter (mass of the coffee cups + a stir bar = 15.00 g). If the initial temperature of the acid/base solution was 22.08 oC, and the final observed temperature was 39.67 oC, what is the enthalpy change of the neutralization reaction, in Kilojoules (KJ) per mole of NaOH? Assume that the density (d = 1.00 g/mL) and specific heat capacity of the solution are identical to that of pure water (4.184 J/g-K). Provide your response to two digits after the decimal. You may wish to consult a periodic table. Important note: because this neutralization reaction is an exothermic process, use the negative sign (-) before type the number in your answer with no-space in between (i.e. -125.30). Do not use units (i.e. J) in your answer.

Answer 1

Ans. Step 1: Determine limiting reactant:

Balanced equation:   HCl(aq) + NaOH(aq) ----------> NaCl(aq) + H₂O

According to the stoichiometry of balanced reaction, 1 mol HCL is neutralized by 1 mol NaOH.

# Moles of HCl taken = Molarity of solution x Volume of solution in liters

                                    = 1.0 M x 0.04525 L

                                    =0.04524 mol

Moles of NaOH taken = 1.0 M x 0.02625 L = 0.02625 mol

Since moles of NaOH is less than that of HCl, NaOH is the limiting reactant.

Therefore, total heat released during the process is due to complete neutralization of NaOH.

# Step 2: Total volume of reaction mixture = 45.24 mL + 26.25 mL = 71.49 mL

Taking density of reaction mixture to be 1.00 g/ mL, total mass of reaction mixture = 71.49 g.

Since the calorimetric constant is NOT provided, the heat gained by coffee cup calorimeter is NOT accounted.

# Heat gained by solution during increase in its temperature is given by-

q = m s dT                            - equation 1

Where,

q = heat change

m = mass of solution

s₁ = specific heat of solution

dT = Final temperature – Initial temperature = (T2 – T1)⁰C

Putting the values in equation 1-

            q = 71.49 g x (4.184 J g^-10C^-1) x (39.67 – 22.08)⁰C = 5261.4180744 J

Therefore, total heat released during neutralization of 0.02625 mol NaOH = -5261.4180744 J, the –ve sign indicates that heat is released during neutralization.

# Step 3: Calculate molar enthalpy of neutralization:

Molar enthalpy of neutralization of NaOH = heat released / Moles of NaOH consumed

                                                = -5261.4180744 J / 0.02625 mol

                                                = -200434.97 J/ mol

                                                = -200.43 kJ/ mol

Note: The experimental value is abnormally high. Re-check the values recorded during experiment.