what is Delta H for this reaction (per mole H2O produced)?

Question

In a constant-pressure calorimeter, 70.0 mL of 0.330 M Ba(OH)2 was added to 70.0 mL of 0.660 M HCl.

The reaction caused the temperature of the solution to rise from 23.84 C to 28.34 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 Delta H for this reaction (per mole H2O produced)? Assume the total volume is the sum of the individual volume.

Answer 1

Let's take a look at the balanced equation for the reaction taking place:

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

Next, we will find the moles of each reactant present:

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

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

Since, from the balanced equation, it takes 2 mols HCl for each mol of Ba(OH)₂, neither is limiting.

We can use either mols of Ba(OH)₂ or mols HCl for the next step. I've chosen to use HCl.

Next, we will calculate moles of H₂O formed in this reaction:

0.0462 mols HCl x 2 mols H₂O / 2 mols HCl = 0.0462 mols H₂O formed

Now, we will determine the heat that is generated from this reaction. For this we will use...

q = mC∆T

q = heat = ?

m = mass = 70.0 mls + 70.0 mls = 140 mls x 1 g / ml = 140 g

C = specific heat = 4.184 J/gº

∆T = change in temperature = 28.34º - 23.84º = 4.50º

Solving for q, we have...

q = (140 g)(4.184 J/gº)(4.50º)

q = 2636 J

Since the question asks for heat per mole of H₂O, we will now divide this value by 0.0462 mols H₂O ...

q = ∆H = 2636 J / 0.0462 mols

∆H = 57,056 J/mol = 57.1 kJ/mol H₂O (negative sign since rxn is exothermic)