Calculating Enthalpy of Neutralization in Calorimetry Experiment

[boredooom]

Could I get a solution to this question?

In a laboratory experiment in calorimetry, 100 mL of 0.500 M of acetic acid is mixed with 100 mL of 0.500 M sodium hydroxide in a calorimeter. The temperature rises from 25.00 C to 27.55 C. The heat capacity of the calorimeter is 150.48 J/C, and the density of the resulting solution is 1.034g/mL. Given that the specific heat capacity of 0.250 M of sodium acetate is 4.034 Jg-1C-1, calculate the enthalpy of neutralization of acetic acid.

Answer: -50.2kJ/mol

 

[leright]

Could I get a solution to this question?

In a laboratory experiment in calorimetry, 100 mL of 0.500 M of acetic acid is mixed with 100 mL of 0.500 M sodium hydroxide in a calorimeter. The temperature rises from 25.00 C to 27.55 C. The heat capacity of the calorimeter is 150.48 J/C, and the density of the resulting solution is 1.034g/mL. Given that the specific heat capacity of 0.250 M of sodium acetate is 4.034 Jg-1C-1, calculate the enthalpy of neutralization of acetic acid.

Answer: -50.2kJ/mol

perhaps you should explain what you've done so far. We don't help you with your homework unless you demonstrate you have at least tried to do it yourself.

 

[m2003]

To calculate the enthalpy of neutralization in this calorimetry experiment, we can use the equation Q = m x c x ΔT, where Q is the heat released or absorbed, m is the mass of the solution, c is the specific heat capacity, and ΔT is the change in temperature. In this case, the heat released is equal to the heat gained by the calorimeter and the resulting solution.

First, we need to calculate the mass of the resulting solution. This can be found by adding the masses of acetic acid and sodium hydroxide used, which is 100 mL each, and then multiplying by the density of the solution.

Mass = (100 mL + 100 mL) x (1.034 g/mL) = 206.8 g

Next, we can calculate the heat released by the reaction using the given data:

Q = (206.8 g) x (4.034 J/gC) x (27.55 C - 25.00 C) = 2115.2 J

Since the heat released by the reaction is equal to the heat gained by the calorimeter and the resulting solution, we can use this value to calculate the enthalpy of neutralization:

ΔH = -Q/ n

Where n is the number of moles of acid or base used in the reaction. In this case, n = 0.100 L x 0.500 mol/L = 0.05 mol (since the reaction is 1:1).

ΔH = -2115.2 J / 0.05 mol = -42,304 J/mol = -42.3 kJ/mol

However, this value is for the specific heat capacity of sodium acetate at 0.250 M. To get the enthalpy of neutralization for acetic acid, we need to adjust for the concentration difference.

Since the specific heat capacity is proportional to the concentration, we can use the equation c1/c2 = ΔH1/ΔH2, where c1 and c2 are the concentrations and ΔH1 and ΔH2 are the corresponding enthalpies.

c1/c2 = (0.250 mol/L) / (0.500 mol/L) = 0.5

Thus, we can adjust the enthalpy of neutralization calculated earlier by multiplying by 0.5.

ΔH = -42.3 kJ/mol