What is the heat produced in the above combustion? It may be necessary to find t

Question

What is the heat produced in the above combustion? It may be necessary to find the appropriate reference material in the text. If 50, mL of 0.20 M HCl(aq) are mixed with 50, mL of 0.20 M NaOH(aq), each at 22.2 degree C, what is the temperature of the final solution after thermal equilibrium is established? The heat of neutralization -56 kJ/mol. Assume there is no heat lost to the surroundings. Also assume the specific heat of the solution is 1.0 cal/g degree C and the density of all solutions is 1.0 g/mL

Explanation / Answer

Ans. Moles of HCl = Molarity x Volume of solution in liters

                                    = 0.20 M x 0.050 L                                       ; [1 L = 1000 mL]

                                    = (0.20 mol/L) x 0.050 L

                                    = 0.01 mol

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

                                    = 0.20 M x 0.050 L                                       ; [1 L = 1000 mL]

                                    = (0.20 mol/L) x 0.050 L

                                    = 0.01 mol

Total volume of reaction mixture = 50.0 mL (HCl) + 50.0 mL (NaOH) = 100.0 mL

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

Stoichiometry: 1 mol HCl reacts with 1 mol NaOH to produce 1 mol each of NaCl and H2O.

Since moles of HCl and NaOH are equal, they completely react to form 0.01 mol H2O.

Given,

Initial temperature of reaction mixture = 22.20C

Specific heat of reaction mixture = 1.0 cal g-1 0C-1 = 4.184 J g-1 0C-1   

Heat of neutralization = -56 kJ/mol. It means that formation of 1 mol H2O during neutralization reaction releases 56 kJ energy. Note that the -ve sign simply indicates the releases of heat (exothermic reaction) during the process.

So,

Total heat liberated = Molar heat of neutralization x moles of H2O formed

                                    = (-56 kJ/mol) x 0.01 mol

                                    = - 0.56 kJ

                                    = - 560 J                                                        ; [1 kJ = 1000 J]

Amount of heat gained by solution = Amount of heat produced during neutralization

                                                            = 560 J

Now, Total heat gained by solution, q = m x s x dT                  - equation 1   

Where,

m = mass in gram,

c = specific heat of solution = 4.184 J g-10C-1

dT = change in temperature = (final – initial) temperature

Let the final temperature be T2

            or, 560 J = 100.0 g x (4.18 J g-10C-1) x (T2 – 22.20C)

            or, 560 J / (418.4 J 0C-1) = T2 – 22.20C

            or, 1.30C + 22.20C = T2

            hence, T2 = 23.50C

Therefore, temperature at thermal equilibrium = 23.50C

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