Chemistry Problem

Question

In a constant‑pressure calorimeter, 55.0 mL of 0.350 M Ba(OH)2 was added to 55.0 mL of 0.700 M HCl.

The reaction caused the temperature of the solution to rise from 23.16 ∘C to 27.93 ∘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 Δ𝐻 for this reaction (per mole H2O produced)? Assume that the total volume is the sum of the individual volumes.

Answer 1

Ba(OH)₂ + 2HCl ==> BaCl₂ + 2H₂O ... balanced equation

q = mC∆T

q = heat = ?

m = mass = 55.0 ml + 55.0 ml = 110 ml x 1.00 g / ml = 110 g

C = specific heat = 4.184 J/gº

∆T = change in temperature = 27.93º - 23.16º = 4.77º

q = (110 g)(4.184 J/gº)(4.77º)

q = 2195 J

Moles Ba(OH)₂ present = 55.0 ml x 1 L / 1000 ml x 0.350 mol/L = 0.01925 mols

Moles HCl present = 55.0 ml x 1 L / 1000 ml x 0.700 mol/L = 0.0385 mols

Since mol ratio in balanced equation is 2HCl : 1 Ba(OH)₂, these are present in stoichiometric amounts.

Moles H₂O produced = 0.0385 mols [1:1 with HCl and 2:1 with Ba(OH)₂]

∆H_rxn/mol H₂O = 2195 J / 0.0385 mols H₂O = 57,013 J / mol H₂O = 57.0 kJ / mol H₂O