In a constant-pressure calorimeter, 50.0 mL of 0.340 M Ba(OH)2 was added to 50.0 mL of 0.680 M HCl. The reaction caused the temperature of the solution to rise from 23.69 °C to 28.32 °C. If the solution has the same density and specific heat as water, what is ÄH for this reaction (per mole of H2O produced)? Assume that the total volume is the sum of the individual volumes.
q = mass H2O x specific heat H2O x (Tfinal-Tinitial)
q/mol = delta H for 2 mols H2O. Divide by 2 for J/mol.
no idea
To determine the enthalpy change (ΔH) for the reaction per mole of H2O produced, we can use the equation:
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.