Answer:
-55.8 kJ/mol
Explanation:
There is a part missing from the question.
<em>Assume no heat is lost to the calorimeter or the surroundings, and the density and heat capacity of the resulting solution are the same as water.
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The initial moles of NaOH and HNO₃ are:
0.1000 L × 0.300 mol/L = 3.00 × 10⁻² mol
The neutralization reaction is:
NaOH + HNO₃ → NaNO₃ + H₂O
When 3.00 × 10⁻² moles of NaOH react with 3.00 × 10⁻² moles of HNO₃, they produce 3.00 × 10⁻² moles of NaNO₃ and 3.00 × 10⁻² moles of H₂O.
According to the law of conservation of energy, the sum of the heat released by the reaction and the heat absorbed by the solution is equal to zero.
ΔH°rxn + ΔH°sol = 0
ΔH°rxn = -ΔH°sol [1]
The volume of the solution is 100.0 mL + 100.0 mL = 200.0 mL. Since the density is 1.00 g/mL, the mass of the solution is 200.0 g.
We can calculate the heat absorbed by the solution using the following expression.
ΔH°sol = c × m × ΔT = (4.184 × 10⁻³ kJ/g.°C) × 200.0 g × (37.00°C - 35.00°C) = 1.674 kJ
where,
c: specific heat capacity of the solution
m: mass of the solution
ΔT: change in the temperature
From [1],
ΔH°rxn = -1.674 kJ
We can express the enthalpy of reaction per mole of NaNO₃.
ΔH°rxn = -1.674 kJ / 3.00 × 10⁻² mol = -55.8 kJ/mol
