Question

1.00 M aqueous solution of NaOH, a 0.50 M aqueous solution of H2SO4 and a coffee cup calorimeter were allowed to stand at a room temperature of 25.4 C until the tempurature of all three reached 25.4 C. A 50.0-mL sample of the 1.00 M NaOH was then placed in the calorimeter. 50.0 mL if the 0.50 M H2SO4 was added as rapidly as possible and the two solutions were mixed thrououghly, the temperature rose to 31.9 C. What is the heat of neutralizaion in kJ for one mole of sulfuric acid? Assume that the densities of the NaOH and the H2SO4 solutions are both 1.00 g/mL, and that the specific heat of the solution after reaction is 4.18 J/g-oC.

Answer 1

Answer:

-108.7 kJ / mol H₂SO₄

Explanation:

The reaction of NaOH with H₂SO₄ is:

2 NaOH + H₂SO₄ → 2H₂O + 2NaSO₄

2 Moles of NaOH reacts per mole of H₂SO₄

This reaction, as all acid-base reactions, produce heat. Moles of NaOH and H₂SO₄ reacting are:

0.0500L × (1.00mol NaOH / L) = 0.0500 moles NaOH

0.0500L × (0.50mol H₂O₄ / L) = 0.0250 moles H₂SO₄

Coffer cup calorimeter formula is:

Q = -C×m×ΔT

Where Q is heat, C is specific heat of solution (4.18J/g°C), m is mass of solution (50g NaOH + 50g H₂SO₄ = 100g of solution -Assuming densities of 1.00g/mL), and ΔT is temperature change (31.9°C-25.4°C = 6.5°C).

Replacing:

Q = -4.18J/g°C × 100g × 6.5°C

Q = -2717J

This is the heat produced in neutralization reaction when 0.0250 moles H₂SO₄ react. Now, heat per mole of H₂SO₄ is:

-2717J / 0.0250 moles H₂SO₄

= -108680 J / mol H₂SO₄ ≡ -108.7 kJ / mol H₂SO₄

Answer 2

Answer:

ΔH = -108.7 kJ/mol

Explanation:

Step 1: Data given

Molarity of a NaOH solution = 1.00 M

Volume of NaOH = 50.0 mL = 0.050 L

Molarity of a H2SO4 solution = 0.50 M

Volume of H2SO4 = 50.0 mL = 0.050 L

Initial temperature of 25.4 °C

Final temperature = 31.9 °C

Density of NaOH and H2SO4 = 1.00 g/mL

Specific heat of the solution = 4.18 J/g°C

Step 2: The balanced equation

2NaOH + H2SO4 → Na2SO4 + 2H2O

Step 3: Calculate moles

Moles = molarity * volume

Moles NaOH = 1.00 M * 0.050 L

Moles NaOH = 0.050 moles

Moles H2SO4 = 0.50 M * 0.050 L

Moles H2SO4 = 0.0250 moles

Step 4: Calculate the mass

Mass of NaOH = 50 mL * 1g/mL = 50 grams

Mass of H2SO4 = 50 mL * 1g/mL = 50 grams

Step 5: Calculate heat trnasfer

Q = m*C*ΔT

⇒with Q = the heat transfer = TO BE DETERMINED

⇒with m = total mass of the solution = 50 g + 50 g = 100 grams

⇒with c= the specific heat of the solution = 4.18 J/g°C

⇒with ΔT = The change of temperature = T2 - T1 = 31.9 °C - 25.4 °C = 6.5 °C

Q = 100 g * 4.18 J/g°C * 6.5 °C

Q = 2717 J

Since this is an exothermic reaction ΔH is negative (-2717 J)

Step 6: calculate the heat of neutralizaion in kJ for one mole of sulfuric acid

ΔH = -2717 J / 0.0250 moles

ΔH = -108680 J/mol = -108.7kJ/mol