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In a constant‑pressure calorimeter, 70.0 mL70.0 mL of 0.320 M Ba(OH)20.320 M Ba(OH)2 was added to 70.0 mL70.0 mL of 0.640 M HCl.0.640 M HCl. The reaction caused the temperature of the solution to rise from 23.00 ∘C23.00 ∘C to 27.36 ∘C.27.36 ∘C. If the solution has the same density and specific heat as water ( 1.00 g/mL1.00 g/mL and 4.184J/g⋅K,)4.184J/g⋅K,) respectively), what is ΔHΔH for this reaction (per mole H2OH2O produced)

Answer :

jufsanabriasa

Answer:

57.0kJ/mol is ΔH of the reaction

Explanation:

The reaction is:

1/2 Ba(OH)₂ + HCl → 1/2 BaCl₂ + H₂O + ΔH.

Where  ΔH is the heat of reaction per mole of water.

Moles of water produced are equal to moles of HCl that are:

70.0mL = 0.070L * (0.640mol / L) = 0.0448moles HCl = Moles of water produced.

Now, heat produced is determined using coffee-cup calorimeter equation:

Q = m×ΔT×C

Where Q is heat released

m is mass of solution (70mL + 70mL = 140mL = 140g -Density of 1g/mL-)

ΔT is change in temperature (27.36°C - 23.00°C = 4.36°C)

And C is specific heat of the solution (4.184J/gK)

Replacing:

Q = 140g×4.36°C×4.184J/gK

Q = 2553.9J

This is the heat released when 0.0448 moles of water are produced, that means ΔH is:

2553.9J / 0.0448moles

ΔH = 57000J/mol =

57.0kJ/mol is ΔH of the reaction

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