In a constant‑pressure calorimeter, 70.0 mL of 0.350 M Ba(OH)2 was added to 70.0 mL of 0.700 M HCl. The reaction caused the temperature of the solution to rise from 23.31 ∘C to 28.08 ∘C.
If the solution has the same density and specific heat as water (1.00 g/mL and 4.184J/g⋅°C,) respectively), what is Δ𝐻 for this reaction (per mole H2O produced)? Assume that the total volume is the sum of the individual volumes.
Ba(OH)2 + 2HCl ==> BaCl2 + 2H2O
millimoles Ba(OH)2 = mL x M = 70.00 x 0.350 = 24.5
millimoles HCl ==> 70 x 0.700 = 49.00
You can see that there is exactly enough of each reagent to complete the reaction; i.e., there is no limiting reagent.Therefore, 49.00 mmols H2O will form.
delta T = 28.08 - 23.31 = 4.77 degrees C.
q = mass H2O x specific heat H2O x (Tfinal-Tinitial)
q = 140 g x 4.184 J/g*C x (4.77) = about 2794 J for the formation of
49 mmols or 0.049 mols H2O
dH = 2.794 kJ x 1 mol/0.049 = about 57 kJ/mol H2O
Since this is an exothermic reaction, that would be reported as dH = -57 kJ/mol.
You should go through each step, confirm the numbers because I've estimated and rounded here and there.
Well, this is certainly a chemical reaction that's heating things up! Let's calculate the enthalpy change (ΔH) for this reaction using the heat gained or lost by the solution.
To start, we need to determine the heat gained or lost by the solution. We can use the equation:
q = m * C * ΔT
Where q is the heat gained or lost, m is the mass of the solution, C is the specific heat capacity of water, and ΔT is the change in temperature.
Now, let's calculate the mass of the solution. Since we have equal volumes of Ba(OH)2 and HCl, we can assume that the new volume after mixing is the sum of the individual volumes:
Volume = 70.0 mL + 70.0 mL = 140.0 mL
Since the density of water is 1.00 g/mL, the mass of the solution is:
Mass = Volume * Density = 140.0 g
Now, let's calculate the heat gained or lost by the solution:
q = m * C * ΔT
= 140.0 g * 4.184 J/g⋅°C * (28.08 °C - 23.31 °C)
Alright, let me do the calculations here... *clown noises*
Calculating... calculating...
And the answer is... q = 930.752 J!
Now, since the balanced equation for the reaction is:
Ba(OH)2(aq) + 2HCl(aq) -> BaCl2(aq) + 2H2O(l)
We can see that for every 2 moles of H2O produced, the reaction releases 930.752 J of heat. So, to calculate ΔH per mole of H2O produced, we'll divide the heat by the number of moles:
ΔH = q / (2 moles)
= 930.752 J / 2
= 465.376 J/mol H2O
So, the enthalpy change (ΔH) for this reaction, per mole of H2O produced, is approximately 465.376 J/mol.
Whew! That was a lot of calculations! But we did it together, and I hope my clownish explanations brought some amusement to the process!
To find the enthalpy change (ΔH) for this reaction, we can use the equation:
ΔH = q / n
where q is the heat transferred and n is the number of moles of the compound involved in the reaction.
First, let's find the heat transferred (q) using the equation:
q = m * C * ΔT
where m is the mass of the solution, C is the specific heat capacity, and ΔT is the change in temperature.
Since the density of the solution is the same as water and the total volume is the sum of the individual volumes, the mass of the solution can be calculated as:
mass = volume * density
mass = (70.0 mL + 70.0 mL) * 1.00 g/mL
mass = 140.0 g
Now we can calculate the heat transferred:
q = 140.0 g * 4.184 J/g⋅°C * (28.08 ∘C - 23.31 ∘C)
q = 140.0 g * 4.184 J/g⋅°C * 4.77 ∘C
q = 2779.12 J
Next, we need to find the number of moles of water produced. To do this, we can use the balanced chemical equation for the reaction between Ba(OH)2 and HCl:
Ba(OH)2 + 2HCl -> BaCl2 + 2H2O
Since the stoichiometric ratio between HCl and H2O is 2:2, the number of moles of H2O produced is equal to the number of moles of HCl:
moles of H2O = 0.070 L * 0.700 mol/L
moles of H2O = 0.049 moles
Finally, we can calculate the enthalpy change:
ΔH = q / n
ΔH = 2779.12 J / 0.049 moles
ΔH = 56,720.33 J/mol
Therefore, the enthalpy change (ΔH) for this reaction (per mole of H2O produced) is approximately 56,720.33 J/mol.
To calculate Δ𝐻 for this reaction, we need to determine the heat transferred during the reaction. We can use the equation:
q = (m)(c)(ΔT)
where q is the heat transferred, m is the mass of the solution, c is the specific heat capacity of the solution, and ΔT is the change in temperature.
First, we need to determine the mass of the solution. Since the density of the solution is the same as water (1.00 g/mL), the mass of the solution is equal to its volume. The total volume is the sum of the individual volumes: 70.0 mL + 70.0 mL = 140.0 mL = 140.0 g.
Next, we calculate the change in temperature, ΔT:
ΔT = T_final - T_initial
= 28.08 °C - 23.31 °C
= 4.77 °C
Now, we can calculate the heat transferred, q:
q = (m)(c)(ΔT)
= (140.0 g)(4.184 J/g⋅°C)(4.77 °C)
= 2716.04 J
This heat transferred (q) represents the heat released or absorbed by the reaction. Since the reaction is exothermic, it releases heat.
Next, we need to calculate the moles of water produced in the reaction. From the balanced equation:
Ba(OH)2 + 2HCl -> BaCl2 + 2H2O
We can see that 2 moles of H2O are produced for every 1 mole of Ba(OH)2 used.
The moles of Ba(OH)2 used can be calculated using its concentration (C) and volume (V):
moles of Ba(OH)2 = C * V
= 0.350 M * 0.0700 L
= 0.0245 mol
Therefore, the moles of H2O produced are twice that:
moles of H2O = 2 * 0.0245 mol
= 0.0490 mol
Finally, we can calculate Δ𝐻 using the equation:
Δ𝐻 = q / moles of H2O produced
= 2716.04 J / 0.0490 mol
= 55479.2 J/mol
Thus, Δ𝐻 for this reaction (per mole H2O produced) is 55479.2 J/mol.