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

To calculate the enthalpy change of the acid-base reaction, the formula q = m × c × ΔT is used. The total heat absorbed minus heat absorbed by the calorimeter gives the net heat, and dividing this by the moles of reactant gives the ΔHrxn as 58.26 kJ/mol.

Step-by-step explanation:

To calculate the enthalpy change (ΔHrxn) in kJ/mol for the reaction of an acid and a base, we first determine the heat absorbed by the solution using the formula q = m × c × ΔT, where m is the mass of the solution, c is the specific heat of the solution, and ΔT is the change in temperature. With the given density of the solution, we can calculate the mass as follows: 100.0 mL × 1.10 g/mL = 110.0 g. Using the given ΔT (6.5 °C), and the specific heat (4.18 J/g°C), we compute the heat absorbed: q = 110.0 g × 4.18 J/g°C × 6.5 °C = 2991 J or 2.991 kJ.

Since the volumes and molarities of acid and base are equal, their moles are also equal, and we can determine the moles of the acid or base used as follows: #moles = Molarity × Volume (L) = 1.0 mol/L × 0.0500 L = 0.0500 mol. The constant volume calorimeter already absorbed part of the heat, which is given by ΔTcalorimeter × Cal constant = 6.5 °C × 12.0 J/°C = 78 J. To get the net heat absorbed by the reaction, we subtract this value from the total heat absorbed: 2.991 kJ - 0.078 kJ = 2.913 kJ.

Finally, the ΔHrxn is the heat change per mole of reactants. Thus, ΔHrxn = 2.913 kJ / 0.0500 mol = 58.26 kJ/mol. This is the enthalpy change per mole of acid (or base) for the neutralization reaction.

Answer 2

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

Step-by-step 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} * \frac{\text{1.0 mol}}{\text{1 L}} = \text{0.0500 mol}\\\\\text{Moles of base} = \text{0.0500 L} * \frac{\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} * \frac{\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.