In a constant-pressure calorimeter, 60.0 mL of 0.310 M Ba(OH)2 was added to 60.0 mL of 0.620 M HCl. The reaction caused the temperature of the solution to rise from 22.19 °C to 26.41 °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.
Ba(OH)2 + 2HCl ==> BaCl2 + 2H2O
mols Ba(OH)2 = M x L = 0.0186
mols HCl = M x L = 0.0372
These neutralize each other exactly and will produce 0.0372 mol H2O
q = mass H2O x specific heat H2O x delta T
q = 120 g x 4.184 J/g*C x (26.41-22.19)
q = approx 2000 J but you need to redo that more accurately.
Then q = dH rxn. Technically the problem only asks for dH for the reaction and that is approx 2 kJ. However, my experience tells me the problem REALLY want kJ/mol so
approx 2 kJ/0.0372 mol = approx ? kJ/mol
To calculate the enthalpy change (ΔH) for the 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).
q = m * c * ΔT
Where:
m = mass of the solution
c = specific heat capacity of water
ΔT = change in temperature
Step 2: Calculate the mass of the solution (m).
Since the density and volume of the solution are the same as water, the mass can be calculated using the equation:
m = volume * density
Step 3: Calculate the change in temperature (ΔT).
ΔT = T final - T initial
Step 4: Calculate the number of moles of water produced (n).
The balanced chemical equation for the reaction between Ba(OH)2 and HCl is:
Ba(OH)2(aq) + 2HCl(aq) -> BaCl2(aq) + 2H2O(l)
Since 1 mole of Ba(OH)2 reacts to produce 2 moles of water, the number of moles of water produced can be calculated using the given concentration and volume:
n = concentration * volume
Now let's calculate step by step.
Given:
Volume of Ba(OH)2 solution (V1) = 60.0 mL
Volume of HCl solution (V2) = 60.0 mL
Initial temperature (T initial) = 22.19 °C
Final temperature (T final) = 26.41 °C
Concentration of Ba(OH)2 (C1) = 0.310 M
Concentration of HCl (C2) = 0.620 M
Step 1: Calculate the heat absorbed or released by the reaction (q).
q = m * c * ΔT
Step 2: Calculate the mass of the solution (m).
m = V1 + V2
Step 3: Calculate the change in temperature (ΔT).
ΔT = T final - T initial
Step 4: Calculate the number of moles of water produced (n).
n = C1 * V1
Finally, calculate ΔH using the equation:
ΔH = q / n
Please provide the values for the specific heat capacity of water and the density of water.
To find the enthalpy change (ΔH) for this reaction per mole of water produced, we need to use the equation:
q = m × c × ΔT
where:
q is the heat exchanged during the reaction,
m is the mass of the solution,
c is the specific heat capacity of water,
ΔT is the change in temperature.
First, let's calculate the mass of the solution. We know that the density and volume are the same as water, so the mass is given by:
mass = volume × density
The total volume of the solution is the sum of the individual volumes:
total volume = volume of Ba(OH)2 + volume of HCl
Now, let's calculate the mass:
mass = total volume × density
Next, we need to calculate the heat exchanged during the reaction. We can assume that the reaction is exothermic because the temperature of the solution increased.
q = m × c × ΔT
In this case, the specific heat capacity, c, refers to the specific heat capacity of water.
Now, let's calculate the heat exchanged during the reaction:
q = mass × c × ΔT
Finally, we can use the equation for enthalpy change to find ΔH:
ΔH = q / moles of water produced
Since the balanced equation for the reaction is:
Ba(OH)2 + 2HCl -> BaCl2 + 2H2O
we can see that 2 moles of water are produced per mole of Ba(OH)2 reacted.
Therefore, ΔH = q / (moles of Ba(OH)2 × 2)