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 21.91 °C to 26.41 °C. If the solution has the same density and specific heat as water, what is ÄH for this reaction (per mole of H2O produced)? Assume that the total volume is the sum of the individual volumes.
Ba(OH)2 + 2HCl ==> 2H2O + BaCl2
q = mass H2O x specific heat H2O x (Tfinal-Tinitial) = delta H.
mols HCl = 0.065 x 0.660 = 0.0429 mols
delta H in J/mol = q/0.0429 = ?
To calculate the heat of the reaction (∆H), you can use the equation:
∆H = q / n
where q is the heat absorbed or released by the reaction and n is the number of moles of the limiting reactant. In this case, the limiting reactant is Ba(OH)2.
To calculate q, you can use the equation:
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.
First, let's calculate the mass of the solution:
mass = volume × density
The volume of the solution is the sum of the individual volumes:
volume = volume of Ba(OH)2 + volume of HCl
volume = 65.0 mL + 65.0 mL
volume = 130.0 mL
Converting mL to g:
mass = 130.0 mL × 1 g/mL
mass = 130.0 g
Next, let's calculate the heat absorbed or released by the reaction (q):
q = mass × C × ΔT
C is the specific heat capacity of water, which is 4.18 J/g·°C.
ΔT = final temperature - initial temperature
ΔT = 26.41 °C - 21.91 °C
ΔT = 4.5 °C
Plugging in the values:
q = 130.0 g × 4.18 J/g·°C × 4.5 °C
q = 2460 J
Next, let's calculate the number of moles of Ba(OH)2:
moles of Ba(OH)2 = volume of Ba(OH)2 × molarity of Ba(OH)2
moles of Ba(OH)2 = 65.0 mL × (0.330 mol/L)
moles of Ba(OH)2 = 0.065 mol
Finally, calculate the heat of the reaction per mole of H2O produced (∆H):
∆H = q / n
∆H = 2460 J / 0.065 mol
∆H ≈ 37846.15 J/mol or 37.9 kJ/mol (rounded to one decimal place)
Therefore, the heat of the reaction (∆H) is approximately 37.9 kJ/mol of H2O produced.
To determine the enthalpy change (ΔH) for the reaction, we need to use the heat transfer equation:
q = m × c × ΔT
where:
q = heat transfer (in Joules)
m = mass of the solution (in grams)
c = specific heat capacity of water (in J/g·°C)
ΔT = change in temperature (in °C)
Since the total volume is the sum of the individual volumes, we can assume that the density and specific heat of the resulting solution are the same as water. Therefore, we can consider the mass of the solution to be equal to the volume in milliliters (mL).
First, let's calculate the mass (m) of the solution:
m = volume × density
Given that the individual volumes are both 65.0 mL and the density of water is approximately 1 g/mL, we can determine the mass of the solution:
m = 65.0 mL × 1 g/mL = 65.0 g
Next, we calculate the change in temperature (ΔT):
ΔT = final temperature - initial temperature
ΔT = 26.41 °C - 21.91 °C = 4.50 °C
Now we can substitute the values into the heat transfer equation:
q = m × c × ΔT
The value we obtain for q will be in Joules. To calculate ΔH (the enthalpy change) in Joules per mole of H2O produced, we need to multiply q by the molar ratio of H2O to Ba(OH)2, which is 2:1.
Finally, let's put it all together:
1. Calculate q:
q = 65.0 g × (4.18 J/g·°C) × 4.50 °C
2. Calculate ΔH:
ΔH = q × (1 mol H2O / 2 mol Ba(OH)2)
To find the enthalpy change per mole of H2O produced, you need to divide ΔH by the number of moles of H2O produced. This can be done by taking into account the stoichiometry of the reaction, which is not given in the question.