When 15.0 mL of 3.00 M NaOH was mixed in a calorimeter with 13.0 mL of 3.00 M HCl, both initially at room temperature (22.0 °C), the temperature increased to 29.1 °C. The resultant salt solution had a mass of 28.00 g and a specific heat capacity of 3.74 J K-1 g-1. What is the heat capacity of the calorimeter (in J/°C)? and the molar enthalpy of neutralization per mole of HCl is -55.84 kJ mol-1
oops. Hit the wrong button:
HCl + NaOH ==> NaCl + H2O
millimoles HCl = mL x M = 15.00 x 3.00 = 45.00 = 0.04500 moles HCl
millimiles NaOH = 13.00 x 3.00 = 39.00 = 0.03900 moles NaOH
Total volume of the solution = 15 + 13 = 28 mL and 28 g and NaOH is the limiting regent with 0.039 moles NaOH used in the neutralization.
Delta T = 29.1 - 22.0 = 7.1
There are three heats involved; i.e., q1 + q2 + q3 = 0 where
q1 = heat of neutralization = n*delta H
q2 = heat to warm the acid/base mixture = m*c*delta T
q3 = heat to warm the calorimeter = Ccal x delta T
n*dH + mc*dT + Ccal*dT
0.039*55,840 + 28*3.74*7.1 + Ccal*7.1 = 0
Solve for Ccal.
HCl + NaOH ==> NaCl + H2O
millimoles HCl = mL x M = 15.00 x 3.00 = 45.00 = 0.04500 moles HCl
millimiles NaOH = 13.00 x 3.00 = 39.00 = 0.03900 moles NaOH
Total volume of the solution = 15 + 13 = 28 mL amd 28 g and NaOH is the limiting regent.
Delta T = 29.1 - 22.0 = 7.1
There are three heats involved; i.e., q
Well, it seems like we've got some chemistry equations here. Let's see if I can lighten things up a bit!
To find the heat capacity of the calorimeter, we can use the equation:
q = C × ΔT
Where q is the heat absorbed or released, C is the heat capacity, and ΔT is the change in temperature.
We already know that the temperature change (ΔT) is 29.1 °C - 22.0 °C = 7.1 °C. But we still need to find the heat (q) absorbed by the solution.
To find q, we can use the formula:
q = m × c × ΔT
Where m is the mass of the solution and c is its specific heat capacity.
We know that the mass of the solution is 28.00 g. And the specific heat capacity (c) is 3.74 J K-1 g-1. So let's plug in those numbers:
q = (28.00 g) × (3.74 J K-1 g-1) × (7.1 °C)
Now we have the heat (q) absorbed by the solution. But remember, energy is conserved, so the heat absorbed by the solution is equal to the heat released by the calorimeter, which is equal to the heat absorbed by the system (the 15.0 mL of NaOH and 13.0 mL of HCl).
So, we can set up the equation:
q = C_calorimeter × ΔT
Where C_calorimeter is the heat capacity of the calorimeter.
Therefore, we can rearrange the equation to solve for C_calorimeter:
C_calorimeter = q / ΔT
Now, let's plug in the values we found for q and ΔT:
C_calorimeter = (28.00 g × 3.74 J K-1 g-1 × 7.1 °C) / 7.1 °C
Simplifying, we find:
C_calorimeter = 28.00 g × 3.74 J K-1 g-1 = 104.72 J/°C
So, the heat capacity of the calorimeter is 104.72 J/°C.
Now, as for the molar enthalpy of neutralization per mole of HCl being -55.84 kJ mol-1, well, that's a bit of a serious number. But hey, at least it's negative, right? Let's just say it's one way chemistry can be a bit of a downer!
To find the heat capacity of the calorimeter, we can use the equation:
q(calorimeter) = -q(solution)
The heat gained by the solution (q(solution)) can be calculated using the formula:
q(solution) = (mass of solution) × (specific heat capacity of solution) × (change in temperature)
Given information:
- The mass of the resultant salt solution = 28.00 g
- The specific heat capacity of the resultant salt solution = 3.74 J K-1 g-1
- The change in temperature = (final temperature - initial temperature) = (29.1 °C - 22.0 °C)
1. Calculate q(solution):
q(solution) = (28.00 g) × (3.74 J K-1 g-1) × (29.1 °C - 22.0 °C)
2. Calculate q(calorimeter):
q(calorimeter) = -q(solution)
3. Calculate the heat capacity of the calorimeter:
q(calorimeter) = (heat capacity of the calorimeter) × (change in temperature)
The change in temperature is the same as before, 29.1 °C - 22.0 °C.
4. Rearrange the equation to find the heat capacity of the calorimeter:
(heat capacity of the calorimeter) = q(calorimeter) / (change in temperature)
Hope this helps! Let me know if you have any other questions.
To find the heat capacity of the calorimeter, we can use the principle of conservation of energy and the formula for calculating heat:
q = m * c * Δt
where:
q = heat absorbed or released
m = mass of the substance
c = specific heat capacity of the substance
Δt = change in temperature
In this case, the heat capacity of the calorimeter can be calculated by rearranging the formula:
q_calorimeter = C_calorimeter * Δt
where:
q_calorimeter = heat absorbed or released by the calorimeter
C_calorimeter = heat capacity of the calorimeter
Δt = change in temperature
The heat released by the reaction can be calculated using the following formula:
q_reaction = m * c * Δt
where:
q_reaction = heat released by the reaction
m = mass of the resultant salt solution
c = specific heat capacity of the resultant salt solution
Δt = change in temperature
Now, let's calculate the heat absorbed by the reaction:
q_reaction = (28.00 g) * (3.74 J K-1 g-1) * (29.1 °C - 22.0 °C)
q_reaction = 756.168 J
Since the reaction is exothermic, the heat released by the reaction is equal to the heat absorbed by the calorimeter:
q_reaction = q_calorimeter
756.168 J = C_calorimeter * (29.1 - 22.0)
Simplifying the equation:
756.168 J = C_calorimeter * 7.1
Rearranging the equation to solve for C_calorimeter:
C_calorimeter = 756.168 J / 7.1 °C
C_calorimeter ≈ 106.504 J/°C
Therefore, the heat capacity of the calorimeter is approximately 106.504 J/°C.
Now, let's calculate the molar enthalpy of neutralization per mole of HCl:
The heat released by the reaction can be calculated using the formula:
q_reaction = n * ΔH
where:
q_reaction = heat released by the reaction
n = number of moles of HCl
ΔH = molar enthalpy of neutralization per mole of HCl
We can rearrange the equation to solve for ΔH:
ΔH = q_reaction / n
To calculate the number of moles of HCl, we can use the formula:
n = c * V
where:
c = molar concentration of HCl
V = volume of HCl (in liters)
In this case, the volume of HCl is given as 13.0 mL (or 0.013 L) and the molar concentration of HCl is given as 3.00 M.
n = (3.00 M) * (0.013 L)
n = 0.039 mol
Now, let's calculate ΔH:
ΔH = 756.168 J / 0.039 mol
ΔH ≈ -19,359.28 J/mol
Converting the unit to kJ/mol:
ΔH ≈ -19.36 kJ/mol
Therefore, the molar enthalpy of neutralization per mole of HCl is approximately -19.36 kJ/mol.