What is the Molar Enthalpy of Neutralization (ÄHn) in kJ/mol if 5.00 moles of HCl(aq) (acid) neutralized 5.00 moles of NaOH(aq) (base) and released 2033 J of heat?
To find the molar enthalpy of neutralization (ΔHn) in kJ/mol, we can use the equation:
ΔHn = Q / n
Where:
ΔHn is the molar enthalpy of neutralization in kJ/mol.
Q is the heat released or absorbed in Joules (J).
n is the number of moles of the limiting reactant.
In this case, HCl is the limiting reactant because it is completely consumed in the reaction with NaOH. Since both HCl and NaOH have equal moles (5.00 moles), we can use the heat released (2033 J) and the number of moles of the limiting reactant (5.00 moles) to calculate ΔHn.
First, convert the heat released from Joules to kilojoules:
2033 J = 2033 J / 1000 = 2.033 kJ
Now, substitute the values into the equation:
ΔHn = Q / n
ΔHn = 2.033 kJ / 5.00 mol
ΔHn = 0.4066 kJ/mol
Therefore, the molar enthalpy of neutralization (ΔHn) is 0.4066 kJ/mol.