A 100.0 mL sample of 0.300 M NaOH is mixed with a 100.0 mL sample of 0.300 M HNO3 in a coffee cup calorimeter. If both solutions

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A 100.0 mL sample of 0.300 M NaOH is mixed with a 100.0 mL sample of 0.300 M HNO3 in a coffee cup calorimeter. If both solutions

were initially at 35.00°C and the temperature of the resulting solution was recorded as 37.00°C, determine the ΔH°rxn (in units of kJ/mol NaNO3) for the neutralization reaction between aqueous NaOH and HNO3

Chemistry

2 answers:

PIT_PIT [208] · Answer 1 · 2021-11-07T12:44:34+03:00

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. </em>

<em />

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.