A 70.0mL sample of 1.0 M NaOH is mixed with 44.0 mL of 1.0 M H2SO4 in a large St

Question

A 70.0mL sample of 1.0 M NaOH is mixed with 44.0 mL of 1.0 M H2SO4 in a large Styrofoam cup; the cup is fitted with a lid through which passes a calibrated thermometer. The temperature of each solution before mixing is 23.9 Celcius. After adding the NaOH solution to the coffee cup, the mixed solutions are stirred until reaction is complete. Assume that the density of the mixed solutions is 1.00 g/mL, that the specific heat of the mixed solutions is 4.18 J/(g celcius), and that no heat is lost to the surroundings. The deltaHrxn for the neutralization of NaOH with H2SO4 is -114 kJ/mol H2SO4. What is the maximum measured temperature in the Styrofoam cup?

Explanation / Answer

Ans. Moles of NaOH = (Molarity x volume in Liters) of NaOH solution

                                    = 1.0 M x 0.070 L

                                    = 0.070 moles

Moles of H2SO4 = (Molarity x volume in Liters) of NaOH solution

                                    = 1.0 M x 0.044 L

                                    = 0.044 moles

            Since, 1 mol H2SO4 gives 2 moles H+ (or, H3O+) ions (H2SO4(aq) --------> 2 H+(aq) + SO42-(aq)), total moles of H+ from H2SO4 = 2 x moles of H2SO4 = 2 x 0.044 moles = 0.088 moles

Now, we have-

            Moles of OH- (from NaOH) = 0.070 moles

            Moles of H+ (from H2SO4) = 0.088 moles.

Stoichiometry: OH- + H+ -----> H2O

1 mol OH- neutralizes 1 mol H+. here, NaOH is the limiting reagent because the number of moles of OH- is lesser than that of proton.

Thus, total moles of H2O produced = number of moles of OH- (limiting reagent)

                                                = 0.070 moles.

Note: Molar enthalpy of neutralization = -57 kJ/mol for the formation of 1 mol H2O during neutralization. Here the given value is -114 kJ/mol for the neutralization of NaOH with H2SO4 without mentioning the relative moles. It seems the gives value is for neutralization of n moles of H2SO4 (= 2 n H+) with excess of NaOH.

Thus, use the value, (-114 kJ/mol) / 2 = -57 kJ/mol for formation of 1 mol H2O.

So,

Heat produced during neutralization = molar enthalpy of neutralization x moles of OH- neutralized

                                                = (- 57 kJ/mol) x 0.070 moles

                                                = - 3.99 kJ

                                                = - 3990 J

Total volume of reaction mixture = 70.0 mL + 44.0 mL = 114 mL

Mass of reaction mixture = volume x density = 114.0 mL x (1.0 g/mL) = 114 g

Now,

Ans.     q = m c x dT     - equation 1

            where, q = heat gained

                        m = mass

c = specific heat of solution = 4.18 J/g0C

dT = final – initial temperature

                        Let, the final temperature be T

Assuming heat released during neutralization is all absorbed by calorimeter, the amount of heat absorbed, q = 3990 J.

Putting the values in above equation-

            3990.0 J = 114 g x (4.184 J/g0C) x (T – 23.9 0C )

            Or, (T – 23.9 0C ) = 3990.0 J / (476.52 J/0C) = 8.37 0C

            Or, T = 8.37 0C + 23.9 0C = 32.28 0C

Thus, the maximum (final) temperature recorded = 32.28 0C

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.