In a constant-pressure calorimeter, 65.0 mL of 0.330 M Ba(OH)2 was added to 65.0 mL of 0.660 M HCl. The reaction caused the temperature of the solution to rise from 23.85 °C to 28.35 °C. If the solution has the same density and specific heat as water, what is ΔH for this reaction (per mole of H2O produced)? answer in kj/mol H20

Ba(OH)2 + 2HCl ==> BaCl2 + 2H2O or

2OH^- + 2H^+ ==> 2H2O

You added 0.02145 mol OH^- to 0.0429 mols H^+ to produce 0.02145 mol H2O.
q = mass H2O x specific heat H2O x (Tfinal-Tinitial)
Substitute and solve for q. That gives you q for 0.0215 mols reacted or
q/0.02145 = delta H/mol rxn (but there two mols for this problem) so divide that by 2, then convert to kJ/mol by dividing by 1000. You should get about 57 kJ/mol.

@drBob222,

How did you find the moles of OH and H?

To find ΔH for this reaction, you can use the equation:

ΔH = q/moles of water

where ΔH is the enthalpy change, q is the heat absorbed or released during the reaction, and moles of water is the number of moles of water produced or consumed in the reaction.

To calculate q, you need to use the equation:

q = mcΔT

where q is the heat absorbed or released, m is the mass of the solution (which we can assume is the sum of the volumes of Ba(OH)2 and HCl solutions due to the fact that their densities and specific heats are the same as water), c is the specific heat capacity of the solution (which is also the same as water), and ΔT is the change in temperature.

First, let's find the mass of the solution. Since the densities of Ba(OH)2 and HCl are not given, we can assume that their densities are the same as water, which is 1 g/mL.

The volume of Ba(OH)2 solution added is 65.0 mL, so the mass of Ba(OH)2 solution is 65.0 g.
The volume of HCl solution added is also 65.0 mL, so the mass of HCl solution is 65.0 g.

Therefore, the total mass of the solution is:

Mass = mass of Ba(OH)2 solution + mass of HCl solution
= 65.0 g + 65.0 g
= 130.0 g

Next, let's calculate q using the equation q = mcΔT.

The specific heat capacity of water (c) is 4.18 J/g°C.
The change in temperature (ΔT) is the final temperature (28.35 °C) minus the initial temperature (23.85 °C):

ΔT = 28.35 °C - 23.85 °C
= 4.50 °C

Now we can calculate q:

q = (130.0 g)(4.18 J/g°C)(4.50 °C)
= 2,602.40 J

Now, we need to find the moles of water produced in the reaction. The balanced chemical equation for the reaction between Ba(OH)2 and HCl is:

Ba(OH)2 + 2HCl → BaCl2 + 2H2O

From the equation, we can see that for every 2 moles of HCl, we get 2 moles of water produced.

Since the volume of HCl is 65.0 mL and the concentration of HCl is 0.660 M (moles per liter), we can calculate the moles of HCl:

Moles of HCl = volume (in liters) x concentration
= 0.065 L x 0.660 mol/L
= 0.0429 mol

Therefore, moles of water produced = 0.0429 mol.

Finally, we can calculate ΔH:

ΔH = q/moles of water
= 2,602.40 J / 0.0429 mol
≈ 60,573 J/mol

To convert J to kJ, divide by 1000:

ΔH ≈ 60,573 J/mol / 1000
= 60.57 kJ/mol

Therefore, ΔH for this reaction (per mole of H2O produced) is approximately 60.57 kJ/mol.