In a constant-pressure calorimeter, 75.0 mL of 0.810 M H2SO4 was added to 75.0 mL of 0.390 M NaOH. The reaction caused the temperature of the solution to rise from 22.68 °C to 25.34 °C. If the solution has the same density and specific heat as water, what is ΔH for this reaction (per mole of H2O produced)? Assume that the total volume is the sum of the individual volumes.

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

ΔH = q / n

where:
- ΔH is the enthalpy change (in J/mol)
- q is the heat absorbed or released by the reaction (in J)
- n is the number of moles of water produced in the reaction

To calculate q, we can use the equation:

q = m × c × ΔT

where:
- q is the heat absorbed or released (in J)
- m is the mass of the solution (in g)
- c is the specific heat capacity of water (4.18 J/g·°C)
- ΔT is the change in temperature (in °C)

First, let's calculate the mass of the solution. Since the density of the solution is the same as water, the mass and volume are directly proportional.

The total volume of the solution is the sum of the individual volumes:

Volume = 75.0 mL + 75.0 mL = 150.0 mL

Since the density of water is 1 g/mL, the mass of the solution is:

Mass = Volume × Density = 150.0 g

Next, let's calculate the change in temperature:

ΔT = 25.34 °C - 22.68 °C = 2.66 °C

Now we can calculate q:

q = m × c × ΔT = 150.0 g × 4.18 J/g·°C × 2.66 °C = 1,581.78 J

Next, let's calculate the number of moles of water produced. Since the reaction is between H2SO4 and NaOH, one mole of H2SO4 reacts with two moles of NaOH to produce two moles of water.

The initial number of moles of H2SO4 is:

n(H2SO4) = Molarity × Volume = 0.810 M × 0.0750 L = 0.06075 mol

The initial number of moles of NaOH is:

n(NaOH) = Molarity × Volume = 0.390 M × 0.0750 L = 0.02925 mol

Since the reaction is 1:2 between H2SO4 and NaOH, the limiting reagent is NaOH. Therefore, the number of moles of water produced is half the number of moles of NaOH:

n(H2O) = 0.02925 mol / 2 = 0.014625 mol

Finally, we can calculate ΔH:

ΔH = q / n = 1,581.78 J / 0.014625 mol = 108,155.77 J/mol

Therefore, the value of ΔH for this reaction (per mole of H2O produced) is 108,155.77 J/mol.

To calculate the enthalpy change (ΔH) for this reaction per mole of H2O produced, you need to use the equation:

ΔH = q / n

where ΔH is the enthalpy change, q is the heat absorbed or released by the system, and n is the number of moles of H2O produced.

To find q, you can use the equation:

q = m * c * ΔT

where q is the heat absorbed or released by the solution, m is the mass of the solution, c is the specific heat capacity of water, and ΔT is the change in temperature.

First, calculate the mass of the solution:
Since the density of water is approximately 1 g/mL, the volume of the solution (given as 75.0 mL + 75.0 mL) is equal to the mass in grams. Therefore, the mass of the solution is 150.0 g.

Next, calculate the change in temperature, ΔT:
ΔT = final temperature - initial temperature = 25.34 °C - 22.68 °C = 2.66 °C

Now, calculate q using the equation:
q = m * c * ΔT
q = 150.0 g * 4.18 J/g°C * 2.66 °C = 1,110.12 J

To find n, the number of moles of H2O produced, you need to determine which reactant is the limiting reagent. The balanced equation for the reaction is:

H2SO4 + 2NaOH -> Na2SO4 + 2H2O

From the equation, it can be seen that 1 mole of H2SO4 reacts with 2 moles of NaOH to produce 2 moles of H2O.

Since the initial concentrations of H2SO4 and NaOH are given, use the equation:

Molarity = moles / volume

to calculate the number of moles of H2SO4 and NaOH.

For H2SO4:
moles of H2SO4 = molarity * volume
moles of H2SO4 = 0.810 M * 0.0750 L = 0.06075 mol

For NaOH:
moles of NaOH = molarity * volume
moles of NaOH = 0.390 M * 0.0750 L = 0.02925 mol

Comparing the moles of H2SO4 and NaOH, it can be seen that NaOH is the limiting reagent because it has fewer moles.

Since 1 mole of NaOH reacts to produce 2 moles of H2O, the number of moles of H2O produced is twice the number of moles of NaOH.

Therefore, the number of moles of H2O produced is:
moles of H2O = 2 * moles of NaOH = 2 * 0.02925 mol = 0.0585 mol

Now, use the equation ΔH = q / n to find the enthalpy change per mole of H2O produced:
ΔH = 1,110.12 J / 0.0585 mol = 19,014.46 J/mol

Therefore, the enthalpy change (ΔH) for this reaction per mole of H2O produced is 19,014.46 J/mol.