In a constant‑pressure calorimeter, 55.0 ml of 0.940 m h 2 so 4 was added to 55.0 ml of 0.370 m naoh . the reaction caused the temperature of the solution to rise from 21.65 ∘ c to 24.17 ∘ c. if the solution has the same density and specific heat as water ( 1.00 g / ml and 4.184 j / ( g ⋅ °c), respectively), what is δ h for this reaction (per mole of h 2 o produced)? assume that the total volume is the sum of the individual volumes.

ΔH = -57 kJ/mol (negative because it's an exothermic reaction)

Explanation:

Step 1: Data given

55.0 mL of 0.940 M H 2 SO 4 was added to 55.0 mL of 0.370 M NaOH

temperature of the solution to rise from 21.65 ∘ C to 24.17 ∘ C.

Density = 1g/mL

Specific heat = 4.184 J/g°C

Step 2: The balanced equation

H2SO4 + 2NaOH → Na2SO4 + 2H2O

Step 3: Calculate mass of the volume

55.0 mL + 55.0 mL = 110.0 mL

Mass = volume * density

Mass = 110 mL * 1g/mL

Mass = 110 grams

Step 4: Calculate moles of H2SO4

Moles H2SO4 = Molarity * volume

Moles H2SO4 = 0.940 M * 0.055 L = 0.0517 moles H2SO4

Step 5: Calculate moles NaOH

Moles NaOH = 0.370 M * 0.055 L

Moles NaOH = 0.02035 moles

Step 6: The limiting reactant

For 1 mol H2SO4, we need 2 moles NaOH to produce 1 mol of Na2SO4 and 2 moles of H2O

NaOH is the limiting reactant. It will completely be consumed (0.02035 moles).

H2SO4 is in excess. There will be reacted 0.02035/2 = 0.010175 moles

There will remain 0.0517 - 0.010175 = 0.041525 moles

Step 6: Calculate the heat

Q = m*c*ΔT

⇒ with m = the mass of solution = 110 grams

⇒ with c= the specific heat = 4.184 J/g°C

⇒ with ΔT = T2- T1 = 24.17 - 21.65 = 2.52 °C

Q = 110g * 4.184 J/g°C * 2.52 °C

Q = 1159.8 J

Step 7: Calculate moles H2O

Moles NaOH consumed = moles H2O produced

Moles H2O produced = 0.02035 mol

Step 8: Calculate ΔH

ΔH = 1159.8 / 0.02035

ΔH = 56992.6 J/ mol = 57 kJ/mol

since this is an exothermic reaction, ΔH is negative: -57 kJ/mol

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explanation: