Calculate the Molar Enthalpy of Neutralization (ΔHn) in kJ/mol of the reaction between a monoprotic acid and a monoprotic base, given the following information:
The temperature change equals 6.92°C,
50.0 mL of 1.00 M concentration of Acid
50.0 mL of 1.00 M concentration of Base
Heat capacity of the calorimeter is 6.50 J/°C.
The specific heat of water is 4.180 J/g°C
50 mL x 1.00M = 0.05 mols acid and 0.05 mols base.
q = delta H = dH = (mass H2O x specific Heat H2O x delta T) + (Ccal*delta T) where Ccal = heat capacity of the calorimeter.
Then q/n = dH/mol
57.8512kjml-
57851.2 jml-
To calculate the molar enthalpy of neutralization (ΔHn) in kJ/mol, we need to use the following equation:
ΔHn = q / (moles of acid or base)
where q is the heat exchanged in the reaction and moles of acid or base is the number of moles of acid or base involved in the reaction.
To calculate q, we first need to determine the heat absorbed or released by the reaction, which can be calculated using the following equation:
q = C * ΔT
where q is the heat absorbed or released, C is the total heat capacity, and ΔT is the change in temperature.
In this case, the heat capacity (C) of the calorimeter is given as 6.50 J/°C.
To calculate the moles of acid or base, we need to use the following equation:
moles = (volume * concentration) / 1000
where volume is the volume of acid or base used in liters and concentration is the molarity of the acid or base.
Given the information provided:
1. Convert the temperature change from Celsius to Kelvin.
ΔT(K) = ΔT(°C) + 273.15
ΔT(K) = 6.92 + 273.15
ΔT(K) = 280.07 K
2. Calculate the moles of acid and base used in the reaction.
moles of acid = (volume of acid * concentration of acid) / 1000
moles of base = (volume of base * concentration of base) / 1000
moles of acid = (50.0 mL * 1.00 M) / 1000
moles of base = (50.0 mL * 1.00 M) / 1000
moles of acid = 0.050 mol
moles of base = 0.050 mol
3. Calculate the heat absorbed or released (q).
q = C * ΔT
q = 6.50 J/°C * 280.07 K
q ≈ 1820.46 J
4. Calculate the molar enthalpy of neutralization (ΔHn).
ΔHn = q / (moles of acid or base)
ΔHn = 1820.46 J / 0.050 mol
ΔHn ≈ 36409.2 J/mol
Finally, convert ΔHn from Joules to kilojoules (kJ).
ΔHn(kJ/mol) = ΔHn(J/mol) / 1000
ΔHn(kJ/mol) ≈ 36.41 kJ/mol
Therefore, the molar enthalpy of neutralization (ΔHn) is approximately 36.41 kJ/mol.