In a constant-pressure calorimeter, 65.0 mL of 0.330 M Ba(OH)2 was added to 65.0 mL of 0.660 M HCl. The reaction caused the temperature of the solution to rise from 24.28 °C to 28.78 °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.
Why did the solution go to a therapist? Because it needed a ΔH! Ba-dum-tsss!
In order to find ΔH for this reaction, we can use the heat transfer equation:
q = m × c × ΔT
Where:
q is the heat transferred (in joules)
m is the mass of the solution (in grams)
c is the specific heat of the solution (in J/g°C)
ΔT is the change in temperature (in °C)
Since the solution has the same density and specific heat as water, we can assume its mass is equal to its volume. The individual volumes of Ba(OH)2 and HCl are both 65.0 mL, so the total volume is 2 × 65.0 mL = 130.0 mL = 130.0 g.
Now let's calculate the heat transferred:
q = (130.0 g) × (4.18 J/g°C) × (28.78 °C - 24.28 °C)
Calculating this gives us:
q = 130.0 g × 4.18 J/g°C × 4.50 °C
q = 2421.54 J
Since the reaction is exothermic, the heat released by the reaction is equal to the heat transferred. So, ΔH for this reaction can be calculated by dividing q by the number of moles of H2O produced.
To find the number of moles of H2O produced, we need to determine the limiting reactant. From the balanced chemical equation for the reaction, we know that 1 mole of Ba(OH)2 reacts with 2 moles of HCl to produce 2 moles of H2O.
Using the given concentrations and volumes, we can calculate the moles of each reactant:
moles of Ba(OH)2 = (0.330 M) × (0.0650 L) = 0.02145 moles
moles of HCl = (0.660 M) × (0.0650 L) = 0.0429 moles
Since we have a 1:2 mole ratio between Ba(OH)2 and HCl, Ba(OH)2 is the limiting reactant because it is present in a smaller amount. Therefore, we will use the moles of Ba(OH)2 to calculate ΔH.
ΔH = q / moles of H2O
ΔH = 2421.54 J / 0.02145 moles = 112988 J/mol
Note: We multiplied ΔH by 1000 to convert it from J to kJ/mol. So, ΔH for this reaction is approximately 112.99 kJ/mol.
I hope this answer didn't drag you down!
To calculate the enthalpy change (ΔH) for this reaction per mole of water produced, we can use the equation:
ΔH = q / n
where q is the heat absorbed or released by the reaction and n is the number of moles of water produced.
Step 1: Calculate the heat absorbed or released by the reaction
q = m * C * ΔT
where m is the mass of the solution, C is the specific heat capacity of water, and ΔT is the change in temperature.
Since we are told that the solution has the same density and specific heat as water, and the volumes of the two solutions are equal, we can assume that the mass of the solution is equal to the total volume of the solutions multiplied by the density of water (1 g/mL).
Step 2: Calculate the number of moles of H2O produced
To find the number of moles of H2O produced, we need to use the balanced chemical equation for the reaction. From the equation, we can see that 1 mole of Ba(OH)2 reacts with 2 moles of HCl to produce 2 moles of water, so the ratio of moles of water to moles of Ba(OH)2 is 2:1.
Now we can begin the calculations:
Step 1: Calculate the heat absorbed or released by the reaction
m = (V1 + V2) * ρ = (65.0 mL + 65.0 mL) * 1 g/mL = 130 g
C = 4.184 J/g°C (specific heat of water)
ΔT = (28.78 °C - 24.28 °C) = 4.5 °C
q = m * C * ΔT = 130 g * 4.184 J/g°C * 4.5 °C = 2,852.76 J
Step 2: Calculate the number of moles of H2O produced
n(H2O) = n(Ba(OH)2) * (2 moles H2O / 1 mole Ba(OH)2)
n(Ba(OH)2) = M * V = 0.330 M * 65.0 mL * (1 L / 1000 mL) = 0.02145 moles
n(H2O) = 0.02145 moles * (2 moles / 1 mole) = 0.0429 moles
Step 3: Calculate ΔH per mole of H2O produced
ΔH = q / n(H2O) = 2,852.76 J / 0.0429 moles = 66,486 J/mol
Therefore, the enthalpy change (ΔH) for this reaction per mole of H2O produced is 66,486 J/mol.
To determine the enthalpy change (ΔH) for the reaction per mole of water 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 water produced.
To calculate q, we need to determine the heat transferred during the reaction. This can be determined using 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, and ΔT is the temperature change.
First, let's determine the mass of the solution:
mass of solution = volume of solution * density of water
Given that the volume of the solution is the sum of the individual volumes (65.0 mL + 65.0 mL = 130.0 mL), and assuming the density of the solution is the same as water (which is 1.00 g/mL), the mass of the solution can be calculated as:
mass of solution = 130.0 g
Next, we can calculate the heat transferred (q) using the equation:
q = m * C * ΔT
Since we assume the solution has the same specific heat capacity as water, which is 4.18 J/g·°C, and the temperature change (ΔT) is the final temperature minus the initial temperature (28.78 °C - 24.28 °C), we can calculate q as:
q = 130.0 g * 4.18 J/g·°C * (28.78 °C - 24.28 °C)
q = 130.0 g * 4.18 J/g·°C * 4.50 °C
q = 2477 J
Now, we need to calculate the number of moles of water produced, which can be determined from the balanced chemical equation for the reaction:
Ba(OH)2 + 2HCl -> BaCl2 + 2H2O
From the equation, we can see that for every mole of Ba(OH)2, 2 moles of H2O are produced. Therefore, the number of moles of H2O produced can be calculated as:
moles of H2O = moles of Ba(OH)2 * 2
To determine the number of moles of Ba(OH)2, we can use the equation:
moles of Ba(OH)2 = volume of Ba(OH)2 * concentration of Ba(OH)2
Given that the volume of Ba(OH)2 is 65.0 mL and the concentration of Ba(OH)2 is 0.330 M, the moles of Ba(OH)2 can be calculated as:
moles of Ba(OH)2 = 65.0 mL * (0.330 mol/L / 1000 mL)
moles of Ba(OH)2 = 0.02145 mol
Therefore, the number of moles of H2O produced is:
moles of H2O = 0.02145 mol * 2
moles of H2O = 0.0429 mol
Now, we can calculate the enthalpy change (ΔH) using the equation:
ΔH = q / n
where q is the heat transferred (2477 J) and n is the number of moles of water produced (0.0429 mol):
ΔH = 2477 J / 0.0429 mol
ΔH = 57700 J/mol
Thus, the enthalpy change (ΔH) for this reaction per mole of water produced is 57700 J/mol.
Ba(OH)2 + 2HCl ==> 2H2O + BaCl2
mols Ba(OH)2 = M x L = ?
mols HCl = M x L = ? which is twice that of Ba(OH)2 so the two exactly neutralize each other.
q = heat generated = mass H2O x specific heat H2O x (Tfinal-Tinitial)
mass H2O = 65 + 65 = 130 mL = 130 g
specific heat H2O = 4.184 J/g*C
Tfinal given
Tinitial given
solve for q = delta H for the reaction.
delta H per mols H2O is delta H rxn/mols H2O produced