Question

In a calorimeter, you combine 60 mL of a 0.983 M HNO3 solution with 40 mL of a 1.842 M NaOH solution. The reaction releases 3.2439 kJ of heat. What is the standard molar enthalpy of neutralization for this reaction (kJ/mol)?

Answer 1

Answer:

60.094 kJ/mol is the standard molar enthalpy of neutralization for this reaction.

Explanation:

$HNO_3+NaOH\rightarrow NaNO_3+H_2O$

$Moles (n)=Molarity(M)\times Volume (L)$

Moles of nitric acid  = n

Volume of nitric  solution = 60 mL = 0.060 L

Molarity of the hydrogen peroxide = 0.983 M

$n=0.983 M\times 0.060 L=0.05898 mol$

Moles of sodium hydroxide  = n'

Volume of sodium hydroxide solution = 40.0 mL = 0.040 L

Molarity of the sodium hydroxide= 1.842 M

$n'=1.842 M\times 0.040 L=0.07368 mol$

As we can see that moles of sodium hydroxide are in excessive amount and moles of nitric acid are in limiting amount.So, enthalpy of reaction will be calculated with respect to the moles of nitric acid .

Energy released on the reaction for the given amount of acid and base = Q

Q = 3.2439 kJ

0.05898 moles of nitric acid gives 3.2439 kJ of energy.

Energy release when 1 mol of nitric acid reacts:

$\frac{Q}{0.05898 mol}=\frac{3.2439 kJ }{0.05898 }=60.094 kJ$

60.094 kJ/mol is the standard molar enthalpy of neutralization for this reaction.