Assuming that the specific heat of the solution is 4.18 J/(g⋅∘C) and that the calorimeter itself absorbs a negligible amount of heat, calculate ΔH in kilojoules for the reaction
CaO(s)+H2O(l)→Ca(OH)2(aq)
When 1.045 g of CaO is added to 50.0 mL of water at 25.0 ∘C in a calorimeter, the temperature of the water increases to 32.3 ∘C.
Why did the CaO go to the therapist? Because it had some serious H2O issues! Let's calculate the enthalpy change and bring peace to this reaction.
Given:
Mass of CaO = 1.045 g
Specific heat of solution = 4.18 J/(g⋅∘C)
Initial temperature (water) = 25.0 ∘C
Final temperature (solution) = 32.3 ∘C
Volume of water = 50.0 mL
First, let's calculate the heat absorbed by the water:
q_water = mass_water * specific heat_water * ΔT_water
The mass of water can be calculated using its density, ρ_water:
mass_water = volume_water * ρ_water
The density of water is approximately 1 g/mL:
mass_water = 50.0 mL * 1 g/mL = 50.0 g
Substituting the values:
q_water = 50.0 g * 4.18 J/(g⋅∘C) * (32.3 ∘C - 25.0 ∘C)
Now, let's calculate the moles of CaO used:
moles_CaO = mass_CaO / molar_mass_CaO
The molar mass of CaO is approximately 56.08 g/mol:
moles_CaO = 1.045 g / 56.08 g/mol
Next, we can calculate the heat released by the reaction using the moles of CaO:
q_reaction = moles_CaO * ΔH
Now, let's relate the heat absorbed by the water and the heat released by the reaction:
q_water = -q_reaction (since the heat released by the reaction is absorbed by the water)
Finally, let's express ΔH in kilojoules:
ΔH = q_reaction / 1000
Now let's put all the pieces together, solve the equations, and find the answer!
To calculate the enthalpy change (ΔH) for this reaction, you need to use the equation:
ΔH = (q / n)
Where:
q = heat absorbed or released by the reaction
n = number of moles of the limiting reactant
First, let's calculate the heat (q) absorbed or released by the reaction using the equation:
q = m × c × ΔT
Where:
m = mass of the water
c = specific heat of the solution (4.18 J/(g⋅∘C))
ΔT = change in temperature (Tf - Ti)
Given data:
Mass of CaO (m) = 1.045 g
Mass of water (m) = 50.0 mL (assume the density of water is 1 g/mL)
Initial temperature (Ti) = 25.0 ∘C
Final temperature (Tf) = 32.3 ∘C
1. Convert the volume of water to mass:
Mass of water (m) = Volume × Density
= 50.0 mL × 1 g/mL
= 50.0 g
2. Calculate the change in temperature (ΔT):
ΔT = Tf - Ti
= 32.3 ∘C - 25.0 ∘C
= 7.3 ∘C
3. Calculate the heat (q) absorbed or released by the reaction:
q = m × c × ΔT
= 50.0 g × 4.18 J/(g⋅∘C) × 7.3 ∘C
= 1518 J
Next, we need to calculate the number of moles of CaO (n) used in the reaction.
4. Calculate the molar mass of CaO:
Molar mass of CaO = atomic mass of Ca + atomic mass of O
= 40.08 g/mol + 16.00 g/mol
= 56.08 g/mol
5. Calculate the number of moles (n) of CaO:
n = Mass / Molar mass
= 1.045 g / 56.08 g/mol
= 0.0186 mol
Now we can calculate the enthalpy change (ΔH) for the reaction.
ΔH = q / n
= 1518 J / 0.0186 mol
= 81613 J/mol
Finally, convert the unit from joules to kilojoules:
ΔH = 81613 J/mol × (1 kJ/1000 J)
= 81.6 kJ/mol
Therefore, the enthalpy change (ΔH) for the reaction CaO(s) + H2O(l) → Ca(OH)2(aq) is approximately 81.6 kJ/mol.
To calculate the enthalpy change (ΔH) for the reaction, you can use the equation:
ΔH = q / n
Where:
- ΔH is the enthalpy change
- q is the heat transferred in the reaction
- n is the amount of substance in moles involved in the reaction
First, let's calculate the heat transferred (q) in the reaction:
q = m * C * ΔT
Where:
- q is the heat transferred
- m is the mass of the water
- C is the specific heat of the solution
- ΔT is the change in temperature
Since the water is absorbing the heat in the reaction, we can assume that the heat transferred is equal to the heat absorbed by the water.
The mass of the water can be calculated using its density:
density_of_water = mass_of_water / volume_of_water
Since we are given the volume of the water (50.0 mL), we need to convert it to grams. The density of water is approximately 1 g/mL.
mass_of_water = volume_of_water * density_of_water
mass_of_water = 50.0 mL * 1 g/mL
mass_of_water = 50.0 g
Now, let's calculate the change in temperature (ΔT):
ΔT = final_temperature - initial_temperature
ΔT = 32.3 ∘C - 25.0 ∘C
ΔT = 7.3 ∘C
Substituting the values into the equation for q:
q = (50.0 g) * (4.18 J/(g⋅∘C)) * (7.3 ∘C)
Calculate q to obtain the value of the heat transferred.
Once you have the heat transferred value (q) and the amount of substance (n) in moles, you can calculate the enthalpy change (ΔH):
Convert the mass of CaO to moles by dividing by the molar mass of CaO.
n = mass_of_CaO / molar_mass_of_CaO
Substitute the calculated value of n and the calculated value of q into the equation for ΔH.
ΔH = q / n
Calculate ΔH to find the answer in kilojoules.
dH for THIS reaction (just the 1.045g) is
q = 50g x 4.184 x (Tfinal-Tinitial).
MOST of the time these problems want the answer expressed in kJ/mol. If that is the case here that would
dH in kJ/mol = q from above x (molar mass CaO/1.05g) x (1 kJ/1000 J) = ? aprox 82 kJ/mol