In a constant-pressure calorimeter, 70.0 mL of 0.340 M Ba(OH)2 was added to 70.0 mL of 0.680 M HCl. The reaction caused the temperature of the solution to rise from 22.00 °C to 26.63 °C. If the solution has the same density and specific heat as water (1.00 g/mL and 4.184 J/g·K, respectively), what is ΔH for this reaction (per mole of H2O produced)? Assume that the total volume is the sum of the individual volumes.
To calculate ΔH for the reaction per mole of H2O produced, we need to find the change in enthalpy (ΔH) for the reaction that occurred in the constant-pressure calorimeter.
The formula to calculate the change in enthalpy is:
ΔH = q / n
where:
ΔH is the change in enthalpy,
q is the heat absorbed or released by the reaction, and
n is the number of moles of H2O produced in the reaction.
To find the heat absorbed or released (q), we will use the equation:
q = m * C * ΔT
where:
m is the mass of the solution (in grams),
C is the specific heat capacity of water (4.184 J/g·K), and
ΔT is the change in temperature (final temperature - initial temperature).
Now let's calculate the values step-by-step:
Step 1: Calculate the mass of the solution:
Given that the density of the solution is 1.00 g/mL and the volumes of both solutions are equal, the mass of 70.0 mL of the solution is:
mass = 70.0 mL * 1.00 g/mL = 70.0 g
Step 2: Calculate the change in temperature:
ΔT = final temperature - initial temperature = 26.63 °C - 22.00 °C = 4.63 °C
Step 3: Calculate the heat absorbed or released (q):
q = m * C * ΔT = 70.0 g * 4.184 J/g·K * 4.63 °C = 1366.45 J
Step 4: Calculate the number of moles of H2O produced:
To determine the number of moles of H2O produced, we need to identify the limiting reactant. From the balanced equation:
Ba(OH)2 + 2HCl → BaCl2 + 2H2O
The stoichiometric ratio between Ba(OH)2 and H2O is 1:2. Therefore, if we assume that all of the Ba(OH)2 reacts, we can say that 1 mole of Ba(OH)2 will produce 2 moles of H2O.
Given that the initial concentration of Ba(OH)2 is 0.340 M and the initial volume is 70.0 mL (which is equal to 0.0700 L), we can calculate the number of moles of Ba(OH)2:
moles of Ba(OH)2 = concentration * volume = 0.340 M * 0.0700 L = 0.0238 moles
Since the stoichiometric ratio is 1:2, we can say that the number of moles of H2O produced is twice the number of moles of Ba(OH)2:
moles of H2O = 2 * 0.0238 moles = 0.0476 moles
Step 5: Calculate the change in enthalpy (ΔH):
ΔH = q / n = 1366.45 J / 0.0476 moles = 28700 J/mol
The ΔH for this reaction per mole of H2O produced is 28700 J/mol.
To calculate the enthalpy change (ΔH) for this reaction per mole of H2O 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.
To find q, we can use the equation:
q = mcΔT
where m is the mass of the solution, c is the specific heat capacity of water, and ΔT is the change in temperature of the solution.
First, let's determine the mass of the solution. Since both the volumes of Ba(OH)2 and HCl are given in milliliters, we need to convert them to grams using their respective densities.
Mass of Ba(OH)2 solution (m1) = volume of Ba(OH)2 × density of water
= 70.0 mL × 1.00 g/mL
= 70.0 g
Mass of HCl solution (m2) = volume of HCl × density of water
= 70.0 mL × 1.00 g/mL
= 70.0 g
Total mass of the solution (m) = m1 + m2
= 70.0 g + 70.0 g
= 140.0 g
Next, let's calculate the heat absorbed or released by the reaction:
q = mcΔT
Here, c is the specific heat capacity of water, which is given as 4.184 J/g·K.
q = (140.0 g) × (4.184 J/g·K) × (26.63 °C - 22.00 °C)
= 23624.0128 J
Now, we need to determine the number of moles of water produced.
The balanced equation for the reaction is:
Ba(OH)2 + 2HCl → BaCl2 + 2H2O
From the balanced equation, we can see that for every mole of Ba(OH)2 reacted, 2 moles of water are produced.
The number of moles of Ba(OH)2 reacted is given by:
moles of Ba(OH)2 = concentration × volume
moles of Ba(OH)2 = (0.340 M) × (0.070 L)
= 0.0238 mol
Therefore, the number of moles of water produced (n) is twice the number of moles of Ba(OH)2:
n = 2 × (0.0238 mol)
= 0.0476 mol
Finally, we can calculate the enthalpy change (ΔH) for the reaction per mole of H2O produced:
ΔH = q / n
ΔH = 23624.0128 J / 0.0476 mol
= 496000 J/mol
So, the ΔH for this reaction per mole of H2O produced is 496000 J/mol.