Question

24 Question (1 point) @ See page 236 A 67.0 mL sample of 1.0 M NaOH is mixed with 41.0 mL of 1.0 M H2SO4 in a large Styrofoam coffee cup; the cup is fitted with a lid through which passes a calibrated thermometer. The temperature of each solution before mixing is 27.5°C. After adding the NaOH solution to the coffee cup, the mixed solutions are stirred until reaction is complete. Assume that the density of the mixed solutions is 1.0 g/mL, that the specific heat of the mixed solutions is 4.18 J/g •°C), and that no heat is lost to the surroundings. The AHrxn for the neutralization of NaOH with H2SO4 is -114 kJ/mol H2SO4.

Answer 1

Solution:

Moles of NaOH = Molarity x Volume in L

= 1.0 M x 0.067 L

= 0.067 moles

Moles of H₂SO₄ = Molarity x Volume in L

= 1.0 M x 0.041 L

= 0.041 moles

H₂SO₄(aq) --------> 2 H⁺(aq) + SO₄^2-(aq)

Since, 1 mole H₂SO₄ gives 2 moles H⁺  ions.

Total moles of H⁺ = 2 x moles of H₂SO₄ = 2 x 0.041 moles = 0.082 moles

Stoichiometry: OH^- + H⁺ -----> H₂O

1 mole OH^- neutralizes 1 mole H⁺.

Here, NaOH is the limiting reagent because the number of moles of OH^- is lesser than that of proton.

Thus, total moles of H₂O produced = number of moles of OH^- (limiting reagent)

= 0.067 moles.

Molar enthalpy of neutralization = -57 kJ/mol for the formation of 1 mole H₂O during neutralization.

Here, the given value is -114 kJ/mol for the neutralization of NaOH with H₂SO₄ without mentioning the relative moles. It seems the gives value is for neutralization of n moles of H₂SO₄ (= 2 n H⁺) with excess of NaOH.

Thus, (-114 kJ/mole) / 2 = -57 kJ/mol for formation of 1 mole H₂O.

So,

Heat produced during neutralization = Molar enthalpy of neutralization x Moles of OH^- neutralized

= (- 57 kJ/mol) x 0.067 moles

= - 3.819 kJ

= - 3819 J

Total volume of reaction mixture = 67.0 mL + 41.0 mL = 108.0 mL

Mass of reaction mixture = volume x density = 108.0 mL x (1.0 g/mL) = 108 g

Now,

q = m c ▵T

where,

q = heat gained

m = mass

c = specific heat of solution = 4.18 J/g⁰C

▵T = (final – initial) temperature

3819.0 J = 108 g x (4.18 J/g⁰C) x (T – 27.5 ⁰C)

(T – 27.5 ⁰C) = 3819.0 J / (451.44 J/⁰C)

(T – 27.5 ⁰C) = 8.4596 ⁰C

T = 8.4596 ⁰C + 27.5 ⁰C

T = 35.9596 ⁰C

Thus, the maximum (final) temperature recorded = 35.96 ⁰C