Question

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?

Answer 1

Answer:

$\large \boxed{\text{-61 kJ \cdot mol ^{-1} }}$  

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

$\text{Moles of acid} = \text{0.0500 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{0.0500 mol}\\\\\text{Moles of base} = \text{0.0500 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{0.0500 mol}$

(b) Volume of solution

V = 50.0 mL + 50.0 mL = 100.0 mL

(c) Mass of solution

$\text{Mass of solution} = \text{100.0 mL} \times \dfrac{\text{1.10 g}}{\text{1 mL}} = \text{110.0 g}$

(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

$\text{The enthalpy of reaction is \large \boxed{\textbf{-61 kJ \cdot mol ^{\mathbf{-1}} }} }$

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