In a constant-pressure calorimeter, 50.0 mL of 0.320 M Ba(OH)2 was added to 50.0 mL of 0.640 M HCl. The reaction caused the temperature of the solution to rise from 23.47 °C to 27.83 °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.
mols Ba(OH)2 = M x L = ? about 0.016
mols HCl = M x L = about 0.032
The reaction is stoihciometric.
Ba(OH)2 + 2HCl ==> BaCl2 + 2H2O
qrxn = mass H2O x specific heat H2O x (Tfinal-Tinitial) = about 1800 J
qrxn = delta H for 0.032 mols H2O formed
therefore, delta H/mol = about 1800 J/0.032 = ? joule/mol. Generally these are quoted as kJ/mol.
To find the enthalpy change (ΔH) per mole of H2O produced in this reaction, we need to use the equation:
ΔH = q / n
where q is the heat transferred and n is the number of moles.
First, we need to find the heat transferred (q) during the reaction. We can do this using the equation:
q = msΔT
where m is the mass, s is the specific heat capacity, and ΔT is the change in temperature.
In this case, the reaction takes place in a constant-pressure calorimeter, so the heat transferred is equal to the heat absorbed (or released) by the solution.
Assuming that the total volume is the sum of the individual volumes, we have a total volume of 100.0 mL (50.0 mL Ba(OH)2 + 50.0 mL HCl). Since the solution has the same density and specific heat as water, we can assume that its mass is equal to its volume. Therefore, the mass of the solution is 100.0 grams.
The specific heat capacity of water (s) is 4.18 J/g°C.
Now we can calculate the heat transferred:
q = msΔT
= (100.0 g) x (4.18 J/g°C) x (27.83 °C - 23.47 °C)
≈ 1799 J
Next, we need to find the number of moles of H2O produced. The balanced equation for the reaction between Ba(OH)2 and HCl is:
Ba(OH)2(aq) + 2HCl(aq) -> BaCl2(aq) + 2H2O(l)
From the balanced equation, we can see that for every mole of Ba(OH)2 reacting, 2 moles of H2O are produced.
The initial concentration and volume of Ba(OH)2 are given as 0.320 M and 50.0 mL, respectively. To find the number of moles of Ba(OH)2, we can use the equation:
moles = concentration x volume
moles of Ba(OH)2 = (0.320 M) x (0.050 L)
= 0.016 mol
Since the reaction is 1:2 between Ba(OH)2 and H2O, the number of moles of H2O produced is twice the number of moles of Ba(OH)2:
moles of H2O = 2 x (0.016 mol)
= 0.032 mol
Now we have all the necessary values to calculate the enthalpy change (ΔH):
ΔH = q / n
= 1799 J / 0.032 mol
≈ 56247.5 J/mol
Therefore, the ΔH per mole of H2O produced in this reaction is approximately 56247.5 J/mol.