Question

44. For each determination calculate the number of moles of HCl and of Naon that were present when you mixed 50.0 mL of 2.00M HCl with 51.0 mL of 2.00M NaOH, using Equation 5. determination 1 mol HCI determination 2 _mol HCI mol NaOH mol NaOH 5. For each determination, calculate the number of moles of NaOH that reacted when you mixed 51.0 mL of 2.00M NaOH with 50.0 mL of 2.00M HCl. determination 1 determination 2 6. For each determination, calculate AH neut for the reaction of 1 mol of HCl and 1 mol of NaOH, using Equation 6. (Remember that AH is negative for exothermic reactions.) Bulupapun determination 1 determination 2

Answer 1

Answer:

determination 1

1) Temperature change for the reaction $\Delta$ T = T final - T initial

                                                               = 34.25 - 21.27             [ average initial temp of acid and base = 21.27]

                                                               = 12.98 deg C

2) Amount of heat released = Total volume x density x sp.heat x $\Delta$ T

                                       = 101 mL x 1.04 g / mL x 3.89 J / g-deg C x 12.98 deg C

                                       = 5304 J

3)   mol of HCl = Vol of HCl added mL x 1 L / 1000 mL x Molarity of HCl mol / L          [ as per eq.5 ]

                = 50 mL x 1 L / 1000 mL x 2 mol/L

               = 0.1 mole

similarly mol of NaOH = 51 mL x 1 L / 1000 mL x 2 mol / L

                                 = 0.102 mol

4 )   mole of NaOH reacted = 0.1 mol [ since here HCl is the limiting reagent which has 0.1 mole only.]

5) Heat of neutralisation = - amount of energy released / number of moles of NaOH reacted

                                   = - 5304 J / 0.10 mol

                                   = - 53040 J / mol = - 5.3 x 10⁴ J / mol

determination 2

1) Temperature change for the reaction $\Delta$ T = T final - T initial

                                                               = 34.24 - 21.11             [ average initial temp of acid and base = 21.3]

                                                               = 13.13 deg C

2) Amount of heat released = Total volume x density x sp.heat x $\Delta$ T

                                       = 101 mL x 1.04 g / mL x 3.89 J / g-deg C x 13.13 deg C

                                       = 5365 J

3)   mol of HCl = Vol of HCl added mL x 1 L / 1000 mL x Molarity of HCl mol / L          [ as per eq.5 ]

                = 50 mL x 1 L / 1000 mL x 2 mol/L

               = 0.1 mole

similarly mol of NaOH = 51 mL x 1 L / 1000 mL x 2 mol / L

                                 = 0.102 mol

4 )   mole of NaOH reacted = 0.1 mol [ since here HCl is the limiting reagent which has 0.1 mole only.]

5) Heat of neutralisation = - amount of energy released / number of moles of NaOH reacted

                                   = - 5365 J / 0.10 mol

                                   = - 53650 J / mol = - 5.37 x 10⁴ J / mol

Thus

the heat of neutralisation for determination 1 = 5.3 x 10⁴ J / mol

                                    for determination 2 = 5.37 x 10⁴ J / mol