In a constant-pressure calorimeter, 55.0 mL of 0.310 M Ba(OH)2 was added to 55.0 mL of 0.620 M HCl. The reaction caused the temperature of the solution to rise from 22.23 °C to 26.45 °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 calculate the enthalpy change (ΔH) for this reaction per mole of H2O produced, we need to use the formula:
ΔH = q / n
Where:
- ΔH is the enthalpy change,
- q is the heat absorbed or released by the reaction, and
- n is the number of moles of H2O produced.
Let's break down the solution into steps:
Step 1: Calculate the heat absorbed or released by the reaction (q).
We can use the equation q = m * C * ΔT, where:
- q is the heat absorbed or released by the reaction,
- m is the mass of the solution (in grams),
- C is the specific heat of the solution (C = 4.18 J/g·°C for water), and
- ΔT is the change in temperature (in °C).
Since the density and specific heat of the solution are assumed to be the same as water, the mass of the solution is equal to its volume.
Density of water = 1 g/mL
The initial volume of the solution = 55.0 mL + 55.0 mL = 110.0 mL
The mass of the solution = 110.0 mL * (1 g/mL) = 110.0 g
Using the equation q = m * C * ΔT, we find:
q = 110.0 g * 4.18 J/g·°C * (26.45 °C - 22.23 °C) = 1839.26 J
Step 2: Calculate the number of moles of H2O produced (n).
To calculate the number of moles of H2O produced, we need to determine the limiting reactant in the reaction. This can be found by determining which reactant will be completely consumed first.
The balanced chemical equation for the reaction is:
Ba(OH)2 + 2HCl → BaCl2 + 2H2O
From the balanced equation, we can see that 1 mole of Ba(OH)2 produces 2 moles of H2O.
Since the initial concentrations of Ba(OH)2 and HCl are given, we can use the formula:
Molarity (M) = moles / volume (L)
For Ba(OH)2:
0.310 M = moles of Ba(OH)2 / (55.0 mL / 1000 mL/L)
moles of Ba(OH)2 = 0.31 M * 0.055 L = 0.01705 moles
For HCl:
0.620 M = moles of HCl / (55.0 mL / 1000 mL/L)
moles of HCl = 0.62 M * 0.055 L = 0.0341 moles
Since the mole ratio between Ba(OH)2 and H2O is 1:2, the number of moles of H2O produced is twice the number of moles of Ba(OH)2 consumed:
2 * (0.01705 moles) = 0.0341 moles
Step 3: Calculate ΔH using the equation ΔH = q / n.
ΔH = 1839.26 J / 0.0341 moles = 53918.12 J/mol
Therefore, the enthalpy change (ΔH) for this reaction per mole of H2O produced is 53918.12 J/mol.
To find ΔH for the reaction, we need to use the equation:
q = m * C * ΔT
where:
q = heat change
m = mass of the solution
C = specific heat of water
ΔT = change in temperature
First, let's calculate the mass of the solution. We can start by finding the volume of the solution:
Volume of the solution = Volume of Ba(OH)2 + Volume of HCl
= 55.0 mL + 55.0 mL
= 110.0 mL
Since the solution has the same density as water, we can convert the volume to mass using the density of water, which is 1.00 g/mL.
Mass of the solution = Volume of the solution * Density of water
= 110.0 mL * 1.00 g/mL
= 110.0 g
Now, let's calculate the heat change (q) using the equation mentioned earlier:
q = m * C * ΔT
C is the specific heat of water, which is 4.18 J/g°C.
ΔT = final temperature - initial temperature
= 26.45 °C - 22.23 °C
= 4.22 °C
q = 110.0 g * 4.18 J/g°C * 4.22 °C
= 1954.418 J
Now, we need to convert the heat change to ΔH per mole of H2O produced. To do this, we need to know the stoichiometry of the reaction:
Ba(OH)2 + 2HCl -> BaCl2 + 2H2O
From the balanced equation, we can see that 2 moles of HCl reacts to produce 2 moles of water.
So, the heat change per mole of water (ΔH) can be calculated as follows:
ΔH = q / (2 * n)
where n is the number of moles of water produced.
n = (55.0 mL * 0.310 M) / 1000 mL/L
= 0.01705 mol
ΔH = 1954.418 J / (2 * 0.01705 mol)
= 57187.2 J/mol
Therefore, ΔH for this reaction (per mole of H2O produced) is 57187.2 J/mol.
volume H2O is 55.0 + 55.0 = 110.0 and that is 110.0g
q = mass H2O x specific heat H2O x (Tfinal-Tinitial)
mols Ba(OH)2 = M x L = approx 0.017 but you need a better number than that and that produces twice that number of mols or approx 0.034.
Ba(OH)2 + 2HCl ==> BaCl2 + 2H2O
Then q from above is the amount of heat produced for 0.034 mols.
q/0.034 gives you q/mol H2O. Convert to kJ/mol