In a constant-pressure calorimeter, 70.0 mL of 0.340 M Ba(OH)2 was added to 70.0 mL of 0.680 M HCl. The reaction caused the temperature of the solution to rise from 22.00 °C to 26.63 °C. If the solution has the same density and specific heat as water (1.00 g/mL and 4.184 J/g·K, respectively), what is ΔH for this reaction (per mole of H2O produced)? Assume that the total volume is the sum of the individual volumes.

q = mass H2O x 4.184 x 4.63 x 1/1000 = ? kJ heat produced.

That is for 0.680 x 0.07 mols H2O = ?
So dHrxn is q/mols = ?. I think the answer is about 56 kJ/mol or so.

To calculate the enthalpy change (ΔH) for this reaction, you need to determine the amount of heat released or absorbed by the reaction. This can be done using the equation:

q = m × c × ΔT

where:
q = heat
m = mass of the solution
c = specific heat capacity of water
ΔT = change in temperature of the solution

First, let's calculate the mass of the solution. Since you have 70.0 mL of solution and the density is 1.00 g/mL, the mass can be calculated by multiplying the volume by the density:

mass = volume × density
mass = 70.0 mL × 1.00 g/mL
mass = 70.0 g

Next, let's calculate the heat (q) released or absorbed by the reaction:

q = (mass of solution) × c × ΔT
q = (70.0 g) × (4.184 J/g·K) × (26.63 °C - 22.00 °C)
q = 11689.84 J

The heat released (q) is equal to the enthalpy change (ΔH) of the reaction. However, we need ΔH per mole of H2O produced. To find that, we need to determine the number of moles of H2O produced in the reaction.

The balanced chemical equation for the reaction is:

Ba(OH)2 + 2HCl -> BaCl2 + 2H2O

From the balanced equation, we can see that for every one mole of Ba(OH)2, two moles of H2O are produced. So, the number of moles of H2O can be calculated as half the number of moles of Ba(OH)2 used.

To find the number of moles of Ba(OH)2 used, we can use the formula:

moles = concentration × volume

moles of Ba(OH)2 = (0.340 M) × (0.0700 L)
moles of Ba(OH)2 = 0.0238 mol

Therefore, the number of moles of H2O produced is:

moles of H2O = 0.0238 mol / 2
moles of H2O = 0.0119 mol

Now, we can calculate the ΔH per mole of H2O produced:

ΔH = q / moles of H2O
ΔH = 11689.84 J / 0.0119 mol
ΔH = 981,507 J/mol

So, the value of ΔH for this reaction (per mole of H2O produced) is approximately 981,507 J/mol.