Specific heat- Chemistry

Question

In a constant‑pressure calorimeter, 60.0 mL60.0 mL of 0.300 M Ba(OH)20.300 M Ba(OH)2 was added to 60.0 mL60.0 mL of 0.600 M HCl.0.600 M HCl.

The reaction caused the temperature of the solution to rise from 23.28 ∘C23.28 ∘C to 27.37 ∘C.27.37 ∘C. If the solution has the same density and specific heat as water (1.00 g/mL1.00 g/mL and 4.184J/g⋅°C,)4.184J/g⋅°C,) respectively), what is Δ𝐻ΔH for this reaction (per mole H2OH2O produced)? Assume that the total volume is the sum of the individual volumes.

Answer 1

∆Hrxn / mole H2O produced

q = mC∆T

q = heat = ?

m = mass = 60 ml + 60 ml = 120 ml x 1 g / ml = 120 g

C = specific heat = 4.184 J/gº

∆T = change in temperature = 27.37º - 23.28º = 4.09º

q = (120 g)(4.184 J/gº)(4.09º)

q = 2054 J This is the heat generated from 60 ml Ba(OH)2 reacting with 60 ml HCl

So we now need to find the moles of H2O produced in the reaction:

Ba(OH)₂ + 2HCl ==> BaCl₂ + 2H2O ... balanced equation for the reaction taking place

moles Ba(OH)₂ present = 60 ml x 1 L / 1000 ml x 0.300 mol / L = 0.018 mols Ba(OH)₂

moles HCl present = 60 ml x 1 L / 1000 ml x 0.600 mol / L = 0.036 mols HCl

moles H2O formed = 0.036 mols HCl x 2 mol H2O / 2 mols HCl = 0.036 mols H2O formed

∆Hrxn / mole H2O = 2054 J / 0.036 mols H2O = 57,055 J = 57.1 kJ / mol H2O