3. during an experiment where 50.0 ml of a 1.0 m acid solution was mixed with 50.0 ml of a 1.0 m base solution, the temperature change was measured to be 6.5 oc. if the density of the resulting mixture is 1.10 g/ml, specific heat of the solution is 4.18 j/goc, and the cal constant was 12.0 j/oc, what is the hrxn in kj/mol acid?

Explanation:

Data:

                H₃O⁺ +  OH⁻ ⟶ 2H₂O

    V/mL:  50.0     50.0

c/mol·L⁻¹:   1.0        1.0

    ΔT = 6.5 °C  

      ρ = 1.210 g/mL

      C = 4.18 J·°C⁻¹g⁻¹

C_cal = 12.0 J·°C⁻¹

Calculations:

(a) Moles of acid

(b) Volume of solution

V = 50.0 mL + 50.0 mL = 100.0 mL

(c) Mass of solution

(d) Calorimetry

There are three energy flows in this reaction.

q₁ = heat from reaction

q₂ = heat to warm the water

q₃ = heat to warm the calorimeter

         q₁      +           q₂         +       q₃      = 0

     nΔH      +       mCΔT      + C_calΔT = 0

0.0500ΔH + 1.10×4.18×6.5 + 12.0×6.5 = 0

0.0500ΔH +        2989       + 78.0       = 0

                             0.0500ΔH + 3067 = 0

                                          0.0500ΔH = -3067

                                                      ΔH = -3067/0.0500

                                                            = -61 000 J/mol

                                                            = -61 kJ/mol

Note: The answer can have only two significant figures because that is all you gave for the change in temperature.