In a constant-pressure calorimeter, 60.0 mL of 0.320 M Ba(OH)2 was added to 60.0 mL of 0.640 M HCl. The reaction caused the temperature of the solution to rise from 23.24 °C to 27.60 °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
Well, well, well! Looks like we have a real chemistry enthusiast here! Alright, let's dive into this calorimeter problem, shall we?
To find ΔH for the reaction per mole of H2O produced, we first need to calculate the heat released or absorbed by the reaction. This can be done using the equation:
q = mcΔT
Where:
q = heat (in joules)
m = mass (in grams)
c = specific heat capacity (in J/g°C)
ΔT = change in temperature (in °C)
Since water has the same density and specific heat as the solution, we can assume the mass is the same as the volume. Therefore, the mass of the solution is:
Mass = Volume × Density
Given that the combined volume of Ba(OH)2 and HCl is 60.0 mL + 60.0 mL = 120.0 mL, we can convert it to grams using the density of water (1 g/mL).
Mass = 120.0 mL × (1 g/mL) = 120.0 g
Now, let's calculate the heat released or absorbed by the reaction using the formula:
q = mcΔT
Since the specific heat capacity of water is 4.18 J/g°C, and the change in temperature (ΔT) is (27.60 °C - 23.24 °C) = 4.36 °C, we can substitute these values into the equation:
q = (120.0 g)(4.18 J/g°C)(4.36 °C)
Calculating this gives us:
q ≈ 2051.52 J
Now comes the fun part! We need to calculate the moles of water produced in the reaction. From the balanced chemical equation:
Ba(OH)2 + 2HCl → BaCl2 + 2H2O
We see that for every 1 mole of Ba(OH)2, we get 2 moles of H2O. Since we added 0.0600 L of 0.320 M Ba(OH)2 and 0.0600 L of 0.640 M HCl, we can calculate the moles of Ba(OH)2 used:
Moles of Ba(OH)2 = volume (L) × concentration (M)
Moles of Ba(OH)2 = 0.060 L × 0.320 M = 0.0192 moles
Since the stoichiometry ratio is 1:2 for Ba(OH)2:H2O, this means that 0.0384 moles of H2O are produced.
Finally, we can find ΔH for the reaction per mole of H2O produced:
ΔH = q / moles of H2O
ΔH = 2051.52 J / 0.0384 moles ≈ 53,424.53 J/mol
Voila! The ΔH for this reaction per mole of H2O produced is approximately 53,424.53 J/mol. Keep in mind that this is just an approximation, so don't take it too seriously!
To find the change in enthalpy (ΔH) for the reaction, we need to use the equation:
ΔH = mcΔT
where:
ΔH is the change in enthalpy,
m is the mass of the solution,
c is the specific heat capacity of the solution,
ΔT is the change in temperature.
In this case, since the density and specific heat of the solution are the same as water, we can assume these values to be:
density (ρ) = 1 g/mL
specific heat capacity (c) = 4.18 J/g°C
Now let's calculate the mass of the solution:
mass = volume × density
mass = (60.0 mL + 60.0 mL) × 1 g/mL
mass = 120.0 g
Next, we calculate the change in temperature (ΔT):
ΔT = final temperature - initial temperature
ΔT = 27.60 °C - 23.24 °C
ΔT = 4.36 °C
Now we can plug these values into the equation to calculate the change in enthalpy:
ΔH = (120.0 g) × (4.18 J/g°C) × (4.36 °C)
ΔH = 2160.576 J
Since we want to determine the change in enthalpy per mole of H2O produced, we need to determine the number of moles of H2O produced. From the balanced chemical equation, we know that 2 moles of HCl react with 1 mole of Ba(OH)2 to produce 2 moles of water:
2 HCl + Ba(OH)2 → 2 H2O + BaCl2
The initial moles of HCl can be calculated using the equation:
moles = concentration × volume
moles of HCl = (0.640 M) × (60.0 mL * 10^-3 L/mL)
moles of HCl = 0.0384 mol
The moles of H2O produced will be half of the moles of HCl used, so:
moles of H2O = 0.0384 mol / 2
moles of H2O = 0.0192 mol
Finally, we can calculate the change in enthalpy per mole of H2O produced:
ΔH per mole of H2O = ΔH / moles of H2O
ΔH per mole of H2O = 2160.576 J / 0.0192 mol
ΔH per mole of H2O = 112,806 J/mol
Therefore, the change in enthalpy (per mole of H2O produced) for this reaction is approximately 112,806 J/mol.
To calculate ΔH (the enthalpy change) for this reaction, we need to use the equation:
ΔH = q / n
where ΔH is the enthalpy change per mole of H2O produced, q is the heat exchanged in the reaction, and n is the number of moles of H2O produced.
To determine the heat exchanged (q), we can use the equation:
q = m × c × ΔT
where q is the heat exchanged, m is the mass of the solution, c is the specific heat capacity of water, and ΔT is the change in temperature.
First, let's find the mass of the solution:
The volume of the solution is the sum of the individual volumes, which is 60.0 mL + 60.0 mL = 120.0 mL.
Since the density of water is 1 g/mL, the mass of the solution is 120.0 g.
Next, let's calculate the heat exchanged (q):
q = m × c × ΔT
Here, we assume the specific heat capacity of water (c) is 4.18 J/g°C.
Substituting the values, we get:
q = 120.0 g × 4.18 J/g°C × (27.60 °C - 23.24 °C)
q = 120.0 g × 4.18 J/g°C × 4.36 °C
q = 2067.84 J
Now, let's calculate the number of moles of H2O produced:
The balanced chemical equation for the reaction is:
Ba(OH)2 + 2HCl → BaCl2 + 2H2O
From the equation, we see that 2 moles of H2O are produced per mole of Ba(OH)2 reacted.
Given the concentration and volume of Ba(OH)2, we can calculate the number of moles of Ba(OH)2 reacted.
molarity (M) = moles / volume (L)
0.320 M = moles of Ba(OH)2 / 0.0600 L
moles of Ba(OH)2 = 0.320 M × 0.0600 L
moles of Ba(OH)2 = 0.0192 mol
Since 2 moles of H2O are produced per mole of Ba(OH)2, the number of moles of H2O produced is twice that:
moles of H2O = 2 × 0.0192 mol
moles of H2O = 0.0384 mol
Now that we have the value for q and the number of moles of H2O produced, we can calculate ΔH:
ΔH = q / n
ΔH = 2067.84 J / 0.0384 mol
ΔH ≈ 53760 J/mol
Therefore, the enthalpy change (ΔH) for this reaction, per mole of H2O produced, is approximately 53,760 J/mol.