In a constant-pressure calorimeter, 65.0 mL of 0.320 M Ba(OH)2 was added to 65.0 mL of 0.640 M HCl. The reaction caused the temperature of the solution to rise from 21.98 °C to 26.34 °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.

Correction - BaCl2 on product side of rxn.

To find ΔH for the reaction, we need to calculate the heat released or absorbed by the reaction. We can use the formula:

q = m * C * ΔT

where:
- q is the heat of the reaction,
- m is the mass of the solution,
- C is the specific heat of the solution,
- ΔT is the change in temperature.

To use this formula, we need to determine the mass of the solution. Since the density of the solution is given as 1.00 g/mL, the mass can be calculated as:

mass = volume * density

mass = (65.0 mL + 65.0 mL) * 1.00 g/mL
mass = 130.0 g

Now, we can calculate the heat, q:

q = (130.0 g) * (4.184 J/g·K) * (26.34 °C - 21.98 °C)
q = 130.0 * 4.184 * 4.36
q = 2343.48 J

Next, we need to determine the number of moles of H2O produced in the reaction. Since we have a 1:1 mole ratio between Ba(OH)2 and H2O, the number of moles of H2O is equal to the number of moles of Ba(OH)2.

moles of Ba(OH)2 = volume * concentration

moles of Ba(OH)2 = (65.0 mL * 0.320 mol/L) / 1000
moles of Ba(OH)2 = 0.0208 mol

Therefore, the number of moles of H2O is also 0.0208 mol.

Finally, we can calculate ΔH using the equation:

ΔH = q / moles of H2O

ΔH = 2343.48 J / 0.0208 mol
ΔH = 112,445.38 J/mol

Therefore, ΔH for this reaction (per mole of H2O produced) is approximately 112,445 J/mol.

To find the enthalpy change (ΔH) for the reaction per mole of water produced, we need to calculate the heat released or absorbed during the reaction. We can use the equation:

q = mcΔT

where:
q = heat absorbed or released
m = mass of the solution (in grams)
c = specific heat capacity of water (4.184 J/g·K)
ΔT = change in temperature (in Kelvin)

First, we need to determine the mass of the solution. Since the density of water is 1.00 g/mL, we can assume that the mass of the solution is equal to its volume. Therefore, the total volume of the solution is the sum of the individual volumes:

Total volume = Volume of Ba(OH)2 + Volume of HCl
= 65.0 mL + 65.0 mL
= 130.0 mL

Now, let's convert the total volume to grams:

Mass of the solution = Total volume * Density of water
= 130.0 mL * 1.00 g/mL
= 130.0 g

Next, we calculate the change in temperature (ΔT):

ΔT = Final temperature - Initial temperature
= 26.34 °C - 21.98 °C
= 4.36 °C

Since we need the temperature in Kelvin, we add 273.15 to ΔT:

ΔT = 4.36 °C + 273.15
= 277.51 K

Now, we can calculate the heat (q) absorbed or released:

q = mcΔT
= 130.0 g * 4.184 J/g·K * 277.51 K

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

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

From the equation, we can see that one mole of Ba(OH)2 produces two moles of water. Therefore, the moles of water produced in the reaction is equal to half the moles of Ba(OH)2.

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

moles = concentration * volume

Moles of Ba(OH)2 = 0.320 M * 0.0650 L
= 0.0208 mol

Moles of water = 0.0208 mol / 2
= 0.0104 mol

Now we can calculate the enthalpy change (ΔH) per mole of water produced:

ΔH = q / Moles of water
= (130.0 g * 4.184 J/g·K * 277.51 K) / 0.0104 mol

Calculate this expression to get the value of ΔH.

(65ml)(0.320M Ba(OH)2)+(65ml)(0.640M HCl)

=>0.065(0.32)mole Ba(OH)2+ 0.065(0.64)mole HCl
=>0.0208mole Ba(OH)2 + 0.0416mole HCl
=>0.0208mole Ba(OH)2 + 0.0416 mole HOH
gram yield HOH = 0.0416mole(18g/mole) = 0.749g HOH
Delta T = (26.34-21.98)C = 4.36C
q = mcT =(0.749g)(4.184J/gC)(4.36C) = 13.7 Joules/0.749g HOH
Molar Delta H = (13.7)/(0.749/18)Joules/mole = 328 Joules/mole HOH yield