In a constant-pressure calorimeter, 50.0 mL of 0.300 M Ba(OH)2 was added to 50.0 mL of 0.600 M HCl. The reaction caused the temperature of the solution to rise from 23.77 °C to 27.86 °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.

To calculate ΔH for the reaction, we need to first determine the heat absorbed or released by the solution.

1. Calculate the mass of the solution:
Density of the solution = Density of water = 1.00 g/mL
Volume of the solution = Volume of Ba(OH)2 + Volume of HCl
Volume of Ba(OH)2 = 50.0 mL
Volume of HCl = 50.0 mL
Total volume of the solution = 50.0 mL + 50.0 mL = 100.0 mL = 100.0 g

2. Calculate the heat absorbed or released by the solution using the formula:
q = m * c * ΔT
where:
q = heat (in joules)
m = mass of the solution (in grams)
c = specific heat capacity of water (4.184 J/g·K)
ΔT = change in temperature (in Kelvin)

ΔT = (27.86 °C - 23.77 °C) = 4.09 °C = 4.09 K

q = 100.0 g * 4.184 J/g·K * 4.09 K
= 1713.888 J

3. Convert the heat value to moles of water produced.
The balanced equation for the reaction is:
Ba(OH)2 + 2HCl → BaCl2 + 2H2O

From the balanced equation, we can see that 1 mole of Ba(OH)2 produces 2 moles of water.

ΔH per mole of H2O = ΔH per mole of Ba(OH)2 * (2 moles of H2O / 1 mole of Ba(OH)2)

ΔH per mole of H2O = q / (number of moles of Ba(OH)2 * 2)

To find the number of moles of Ba(OH)2:
moles of Ba(OH)2 = volume of Ba(OH)2 (in liters) * Molarity of Ba(OH)2
= 0.0500 L * 0.300 mol/L
= 0.015 mol

ΔH per mole of H2O = 1713.888 J / (0.015 mol * 2)
= 57092 J/mol

4. Convert the ΔH value to kilojoules:
ΔH (kJ/mol) = 57092 J/mol / 1000
= 57.092 kJ/mol

Therefore, the ΔH for this reaction (per mole of H2O produced) is 57.092 kJ/mol.

To calculate the enthalpy change (ΔH) for this reaction, we need to use the equation:

ΔH = q / n

where ΔH is the enthalpy change, q is the heat transferred, and n is the number of moles.

To solve this problem, we need to find the heat transferred (q) first. We can calculate q using the equation:

q = mcΔT

where q is the heat transferred, m is the mass, c is the specific heat, and ΔT is the change in temperature.

Since the total volume is the sum of the individual volumes, the total volume of the solution is 50.0 mL + 50.0 mL = 100.0 mL.

To find the mass of the solution, we'll use its density:

mass = volume × density
mass = 100.0 mL × 1.00 g/mL = 100.0 g

Now that we have the mass, we can calculate the heat transferred (q) using the equation mentioned earlier. The change in temperature (ΔT) is the final temperature minus the initial temperature:

ΔT = 27.86 °C - 23.77 °C = 4.09 °C

Now we can calculate q:

q = (100.0 g) × (4.184 J/g·K) × (4.09 °C)
q = 1721.056 J

The next step is to determine the number of moles of water (H₂O) produced in the reaction. Since the reaction between Ba(OH)₂ and HCl produces water, the number of moles of water is equal to the number of moles of the limiting reactant.

From the balanced chemical equation:

Ba(OH)₂ + 2HCl → BaCl₂ + 2H₂O

We can see that 1 mole of Ba(OH)₂ reacts with 2 moles of HCl to produce 2 moles of water.

Initially, we have 50.0 mL of a 0.300 M Ba(OH)₂ solution, which is equal to:

50.0 mL × 0.300 mol/L = 0.015 mol of Ba(OH)₂

Initially, we have 50.0 mL of a 0.600 M HCl solution, which is equal to:

50.0 mL × 0.600 mol/L = 0.030 mol of HCl

Since we have an excess of HCl, all 0.015 mol of Ba(OH)₂ will react with 0.015 mol of HCl, producing 0.015 mol of water.

Finally, we can calculate the enthalpy change (ΔH):

ΔH = q / n
ΔH = 1721.056 J / 0.015 mol
ΔH = 114737.067 J/mol or 114.74 kJ/mol (rounded to 3 significant figures)

Therefore, the enthalpy change (ΔH) for this reaction, per mole of water produced, is 114.74 kJ/mol.