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In a constant‑pressure calorimeter, 55.0 mL of 0.310 M Ba(OH)2 was added to 55.0 mL of 0.620 M HCl.
The reaction caused the temperature of the solution to rise from 24.25 ∘C to 28.47 ∘C. If the solution has the same density and specific heat as water ( 1.00 g/mL and 4.184J/g⋅°C,) respectively), what is Δ for this reaction (per mole H2O produced)? Assume that the total volume is the sum of the individual volumes.

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Eduard22sly

The enthalpy change (ΔH) for the reaction (per mole H₂O produced) is 56.96 KJ/mol

How to determine the mass of the solution

  • Volume = 55 + 55 = 110 mL
  • Density = 1 g/mL
  • Mass = ?

Mass = density × volume

Mass = 1 × 110

Mass = 110 g

How to determine the heat

  • Mass (M) = 110 g
  • Initial temperature (T₁) = 24.25 °C
  • Final temperature (T₂) = 28.47 °C
  • Change in temperature (ΔT) = 28.47 – 24.25 = 4.22 °C
  • Specific heat capacity (C) = 4.184 J/gºC
  • Heat (Q) =?

Q = MCΔT

Q = 110 × 4.184 × 4.22

Q = 1942.2128 J

How to determine the mole of H₂O produced

Ba(OH)₂ + 2HCl --> BaCl₂ + 2H₂O

Since the reaction occurs in a constant‑pressure calorimeter, it means both reactants are sufficient enough for the reaction.

Mole of Ba(OH)₂ = molarity × volume

Mole of Ba(OH)₂ = 0.310 × (55 / 1000)

Mole of Ba(OH)₂ = 0.01705 mole

From the balanced equation above,

1 mole of Ba(OH)₂ reacted to produce 2 moles of H₂O.

Therefore,

0.01705 mole of Ba(OH)₂ will react to produce = 0.01705 × 2 = 0.0341 mole of H₂O.

How to determine the enthalpy change ΔH per mole of H₂O produced

  • Heat (Q) = 1942.2128 J
  • Mole of H₂O (n) = 0.0341 mole
  • Enthalpy change (ΔH) =?

ΔH = Q / n

ΔH = 1942.2128 / 0.0341

ΔH = 56956.29 J/mol

Divide by 1000 to express in KJ/mol

ΔH = 56956.29 / 1000

ΔH = 56.96 KJ/mol

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