In a constant-pressure calorimeter, 60.0 mL of 0.310 M Ba(OH)2 was added to 60.0 mL of 0.620 M HCl. The reaction caused the temperature of the solution to rise from 23.44 °C to 27.66 °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.
q = mass H2O x specific heat H2O x (Tfinal-Tinitial)
You had 0.310 x 0.06 = about 0.0186 mols Ba(OH)2 and that produces 2 mols H2O.
dHrxn = q/0.0186 and divide that by 2 to find per mol H2O.
To determine ΔH for the reaction per mole of H2O produced, we can use the equation:
ΔH = q / n
Where:
- ΔH represents the enthalpy change (in J/mol)
- q represents the heat absorbed or released (in J)
- n represents the number of moles of the substance of interest
To calculate q, we will use the equation:
q = m * c * ΔT
Where:
- q represents the heat absorbed or released (in J)
- m represents the mass of the substance of interest (in g)
- c represents the specific heat capacity (in J/(g·°C))
- ΔT represents the change in temperature (in °C)
Since the volume of the solution is given, and the density and specific heat are the same as water, we can assume that the mass of the solution is equal to its volume (in g).
First, calculate the mass of the solution:
mass of solution = volume of solution * density of water
Next, calculate the heat absorbed or released by the reaction:
q = mass of solution * specific heat capacity * ΔT
Finally, determine the number of moles of H2O produced and calculate ΔH:
moles of H2O = (moles of limiting reactant) * (coefficient of H2O in balanced equation)
ΔH = q / moles of H2O
Let's calculate it step by step:
Step 1: Calculate the mass of the solution
mass of solution = (volume of Ba(OH)2 solution + volume of HCl solution) * density of water
Given:
volume of Ba(OH)2 solution = 60.0 mL
volume of HCl solution = 60.0 mL
density of water = 1.00 g/mL
mass of solution = (60.0 mL + 60.0 mL) * 1.00 g/mL
mass of solution = 120.0 g
Step 2: Calculate the heat absorbed or released by the reaction
q = mass of solution * specific heat capacity * ΔT
Given:
specific heat capacity = specific heat capacity of water = 4.18 J/(g·°C)
ΔT = final temperature - initial temperature = 27.66 °C - 23.44 °C
q = 120.0 g * 4.18 J/(g·°C) * (27.66 °C - 23.44 °C)
q = 120.0 g * 4.18 J/(g·°C) * 4.22 °C
q = 2067.552 J
Step 3: Determine the number of moles of H2O produced
The balanced equation for the reaction between Ba(OH)2 and HCl is:
Ba(OH)2 + 2HCl -> BaCl2 + 2H2O
From the equation, the coefficient of H2O is 2. Therefore, the number of moles of H2O produced is twice the number of moles of Ba(OH)2 used.
moles of H2O = (moles of Ba(OH)2) * 2
To calculate the moles of Ba(OH)2 used, we need to determine its initial moles.
Given:
volume of Ba(OH)2 solution = 60.0 mL
molarity of Ba(OH)2 solution = 0.310 M
moles of Ba(OH)2 = volume of Ba(OH)2 solution * molarity of Ba(OH)2 solution
moles of Ba(OH)2 = 60.0 mL * (0.310 mol/L / 1000 mL/L)
moles of Ba(OH)2 = 0.0186 mol
moles of H2O = 0.0186 mol * 2
moles of H2O = 0.0372 mol
Step 4: Calculate ΔH
ΔH = q / moles of H2O
ΔH = 2067.552 J / 0.0372 mol
ΔH = 55587 J/mol (rounded to 4 significant figures)
Therefore, ΔH for this reaction per mole of H2O produced is 55587 J/mol.
To calculate the enthalpy change (ΔH) for this reaction per mole of H2O produced, we can use the equation:
ΔH = q / n
where:
- ΔH is the enthalpy change,
- q is the heat transferred, and
- n is the number of moles of H2O produced.
To calculate q, the heat transferred, we can use the equation:
q = m × C × ΔT
where:
- q is the heat transferred,
- m is the mass of water,
- C is the specific heat of water, and
- ΔT is the change in temperature.
In this case, we can assume that the total volume of the solution is the sum of the individual volumes. Since the density and specific heat of the solution are the same as water, the mass of the water (m) can be calculated using the equation:
m = V × ρ
where:
- V is the total volume of the solution, and
- ρ is the density of water.
Let's now calculate all the necessary values step by step:
1. Calculate the mass of water (m):
Since the total volume (V) is the sum of the individual volumes, V = 60.0 mL + 60.0 mL = 120.0 mL. However, the volume must be converted to liters for consistency in the equation. So, V = 120.0 mL = 0.120 L.
The density of water is approximately 1 g/mL or 1000 kg/m³. Therefore, ρ = 1000 kg/m³.
m = V × ρ = 0.120 L × 1000 kg/m³ = 0.120 kg.
2. Calculate the heat transferred (q):
Since the solution has the same specific heat as water, we can use the specific heat of water, which is approximately 4.18 J/g°C or 4184 J/kg°C.
ΔT is the change in temperature, which is calculated as the final temperature minus the initial temperature: ΔT = 27.66 °C - 23.44 °C = 4.22 °C.
However, we need to convert ΔT to Kelvin since temperature differences in the equation should be in Kelvin. So, ΔT = 4.22 °C + 273.15 = 277.37 K.
q = m × C × ΔT = 0.120 kg × 4184 J/kg°C × 277.37 K.
3. Calculate the number of moles of H2O produced:
To find the number of moles, we need to determine the limiting reactant in the reaction.
First, determine the moles of Ba(OH)2:
moles of Ba(OH)2 = Volume × Concentration = 60.0 mL × 0.310 M = 0.0186 mol
Next, determine the moles of HCl:
moles of HCl = Volume × Concentration = 60.0 mL × 0.620 M = 0.0372 mol
Since there is a 1:2 molar ratio between Ba(OH)2 and H2O, the limiting reactant is Ba(OH)2 because it is half as many moles as the HCl.
Therefore, the number of moles of H2O produced is equal to twice the mole quantity of Ba(OH)2, which is 2 * 0.0186 mol = 0.0372 mol.
4. Calculate the enthalpy change (ΔH):
ΔH = q / n = (0.120 kg × 4184 J/kg°C × 277.37 K) / 0.0372 mol
Now, plug in the values to calculate ΔH.