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Answer :
The enthalpy of neutralization for acetic acid is approximately 5.3792 kJ/mol.
To calculate the enthalpy of neutralization for acetic acid, we need to use the equation:
ΔH = q / n
where ΔH is the enthalpy change, q is the heat absorbed or released, and n is the number of moles of the limiting reactant.
First, we need to find the moles of acetic acid and NaOH used in the reaction.
Using the formula: moles = concentration x volume, we can calculate the moles of acetic acid:
moles of acetic acid = 2.00 mol/L x 0.100 L = 0.200 mol
Since acetic acid and NaOH react in a 1:1 ratio, the moles of NaOH used will also be 0.200 mol.
Next, we can calculate the heat absorbed or released using the equation:
q = mcΔT
where q is the heat absorbed or released, m is the mass of the solution, c is the specific heat capacity of the solution, and ΔT is the change in temperature.
The specific heat capacity of the solution can be assumed to be the same as water, which is 4.18 J/g°C.
The mass of the solution can be calculated by multiplying the density of water (1.00 g/mL) by the volume (100 mL + 100 mL = 200 mL = 200 g).
So, the mass of the solution is 200 g.
The change in temperature (ΔT) is 36.3°C - 23.5°C = 12.8°C.
Now, we can substitute the values into the equation:
q = (200 g) x (4.18 J/g°C) x (12.8°C) = 1075.84 J
Finally, we can calculate the enthalpy of neutralization:
ΔH = q / n = 1075.84 J / 0.200 mol = 5379.2 J/mol
Converting this to kilojoules per mole:
ΔH = 5379.2 J/mol = 5.3792 kJ/mol
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