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Answer :
Final answer:
The molar enthalpy of the neutralization of sodium hydroxide and acetic acid under the given conditions is approximately -55.8 kJ/mol, indicating the reaction is exothermic.
Explanation:
To calculate the molar enthalpy of neutralization of sodium hydroxide with acetic acid, you'll need to first calculate the heat absorbed using the formula q = m*c*ΔT, where m is the mass of the solution, c is the specific heat capacity of the solution, and ΔT is the change in temperature. Using the given data, q = (97.3 g + 0.500 g)*4.18 J g⁻¹ °C⁻¹ * (22.25 °C - 20.55 °C) = 697.1 J.
Next, calculate how many moles of sodium hydroxide were used: 0.500 g NaOH *(1 mol/40.00 g) = 0.0125 mol NaOH.
The heat of neutralization, ΔH, is defined as the heat change that results when one mole of an acid and one mole of a base undergo a neutralization reaction. In this case, it's the heat change for the reaction between acetic acid and sodium hydroxide. Therefore, the molar enthalpy is ΔH = q/n = 697.1 J / 0.0125 mol =-55.8 kJ/mol. The minus sign indicates that the reaction is exothermic.
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Final answer:
The approximate molar enthalpy of neutralization of sodium hydroxide with acetic acid in a constant-pressure calorimeter is -55.4 kJ/mol, indicating that the reaction is exothermic.
Explanation:
To calculate the molar enthalpy of neutralization of sodium hydroxide with acetic acid, we first need to find the amount of heat released during the reaction. From the given temperature change and the assumption of the solution's heat capacity approximated as water (4.18 J g⁻¹ °C⁻¹), we find the total heat (q) released as q = m * c * ΔT = 97.8 g * 4.18 J g⁻¹ °C⁻¹ * (22.25 °C - 20.55 °C) = 692.16 J = 0.69216 kJ.
Sodium hydroxide, with the molar mass of about 40 g/mol, is present in the amount of 0.500 g / 40 g/mol = 0.0125 mol. As it's the limiting reactant in this reaction, the molar enthalpy of neutralization is ΔH = q / n = 0.69216 kJ / 0.0125 mol = -55.4 kJ/mol (the negative sign indicates the reaction is exothermic).
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