q = mass H2O x specific heat H2O x (Tfinal-Tinitial)
q/mol = delta H for 2 mols H2O. Divide by 2 for J/mol.
q/mol = delta H for 2 mols H2O. Divide by 2 for J/mol.
ΔH = q / n
Where:
ΔH = enthalpy change
q = heat transferred
n = moles of H2O produced
To find q, we can use the formula:
q = m * C * ΔT
Where:
m = mass of the solution (in grams)
C = specific heat capacity of water (4.18 J/g°C)
ΔT = change in temperature (in °C)
First, let's calculate the mass of the solution:
mass = volume * density
Given that the volume is the sum of the individual volumes (50.0 mL + 50.0 mL = 100.0 mL), we need to convert it to grams:
mass = 100.0 mL * 1 g/mL = 100.0 g
Next, we calculate q using the formula:
q = 100.0 g * 4.18 J/g°C * (28.32 °C - 23.69 °C)
Now, let's calculate moles of H2O produced:
moles = volume * concentration
Given that the volume of Ba(OH)2 is 50.0 mL and the concentration is 0.340 M, we can calculate moles of Ba(OH)2:
moles of Ba(OH)2 = 50.0 mL * 0.340 mol/L = 0.017 mol
Similarly, for HCl:
moles of HCl = 50.0 mL * 0.680 mol/L = 0.034 mol
Since the balanced equation is 2HCl(aq) + Ba(OH)2(aq) → BaCl2(aq) + 2H2O(l), we can see that 2 moles of HCl produce 2 moles of H2O.
Therefore, the moles of H2O produced = 0.034 mol * (2 mol H2O / 2 mol HCl) = 0.034 mol
Finally, we can substitute the values into the equation to find ΔH:
ΔH = q / n = (100.0 g * 4.18 J/g°C * (28.32 °C - 23.69 °C)) / 0.034 mol
Calculate the above expression to find the value of ΔH, which will be in J/mol.