constant pressure calorimeter

Question

In a constant‑pressure calorimeter,  of 70.0 mL of 0.330M Ba(OH) was added to 70mL of 0.660 mm HCL.

The reaction caused the temperature of the solution to rise from 22.90 to 27.4 If the solution has the same density and specific heat as water (1.00g/mL and 4.184J/g respectively), what is Δ𝐻 for this reaction (per mole  produced)? Assume that the total volume is the sum of the individual volumes.

Answer 1

Let us look at the reaction taking place:

Ba(OH)₂ + 2HCl ==> BaCl₂ + 2H₂O ... balanced neutralization reaction [note Ba(OH)₂ and not Ba(OH)]

∆H under constant pressure is the same as q, which is the heat of the reaction. We can write...

q = mC∆T

q = heat = ∆H

m = mass = 70.0 ml + 70.0 ml = 140.0 ml x 1g/ml = 140.0 g (this ignores the mass of reactants)

C = specific heat = 4.184 J/g/deg (not J/g as stated in your question)

∆T = change in temperature = 27.4º - 22.90º = 4.5 degrees (this is positive since temperature went up)

Now, we can solve for q (∆H)

q = (140 g)(4.184 J/g/deg)(4.5 deg)

q = 2636 Joules

This is the ∆H of the reaction as given in the problem, that is, the ∆H for 70 ml 0.330 M Ba(OH)₂ + 70 ml of 0.660 M HCl (again, there is an error in your problem which says "0.660 mm HCl).

So, the question asks for ∆H PER MOLE. Thus, we need to now find moles of reactants and products.

moles Ba(OH)₂ = 70.0 ml x 1 L/1000 ml x 0.330 mol/L = 0.0231 moles

moles HCl = 70.0 l x 1 L/1000 ml x 0.660 mol/L = 0.0462 moles

Note that HCl and Ba(OH)₂ are present in stoichiometric amounts. That is, there are twice as many moles of HCl as Ba(OH)₂, which is exactly what is required by the balanced equation. Thus, there will be 0.0231 moles of BaCl₂ produced. This is seen by... 0.0231 mol Ba(OH)₂ x 1 mol BaCl₂/mol Ba(OH)₂ = 0.0231 mol BaCl₂

∆H = 2636 J / 0.0231 moles = 114,113 Joules/mole = 114 kJ/mole