In a constant-pressure calorimeter, 50.0 mL of 0.320 M Ba(OH)2 was added to 50.0 mL of 0.640 M HCl. The reaction caused the temperature of the solution to rise from 21.80 °C to 26.16 °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 = 0.320 x 0.050 = 0.016
heat released = mass H2O x specific heat H2O x (Tfinal-Tinitial)
q = 100 x 4.184 x (26.16-21.80) = ?
q/0.016 = q/2 mol H2O so divide by 2 for q/1 mol H2O produced and add a negative sign since heat is released. Divide by 1000 to convert to kJ/mol H2O produced. The answer should be about 57 kJ/mol
To find the enthalpy change (ΔH) for this reaction per mole of H2O produced using a constant-pressure calorimeter, we need to use the equation:
ΔH = q / n
where:
q is the heat exchanged in the reaction
n is the number of moles of H2O produced
To determine q, we can 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 (4.18 J/g°C)
ΔT is the change in temperature of the solution (in °C)
First, we need to find the mass of the solution by considering the density of water. Since the volumes of the Ba(OH)2 and HCl solutions are additive, the total volume of the solution is 100.0 mL. The density of water is approximately 1 g/mL, so the mass of the solution is 100.0 g.
Next, we calculate the change in temperature:
ΔT = final temperature - initial temperature
= 26.16 °C - 21.80 °C
= 4.36 °C
Now, we can calculate the heat exchanged (q):
q = m * c * ΔT
= 100.0 g * 4.18 J/g°C * 4.36 °C
= 1823.68 J
To find the number of moles of H2O produced, we need to consider the balanced chemical equation for the reaction between Ba(OH)2 and HCl:
Ba(OH)2 + 2HCl -> BaCl2 + 2H2O
From the balanced equation, we can see that for every mole of Ba(OH)2, 2 moles of H2O are produced. Since the Ba(OH)2 concentration is 0.320 M and the initial volume is 50.0 mL, the number of moles of Ba(OH)2 is:
moles of Ba(OH)2 = concentration * volume
= 0.320 M * 0.0500 L
= 0.0160 mol
Therefore, the number of moles of H2O produced is twice that:
moles of H2O = 2 * 0.0160 mol
= 0.0320 mol
Finally, we can calculate the enthalpy change (ΔH) per mole of H2O produced:
ΔH = q / n
= 1823.68 J / 0.0320 mol
= 56990 J/mol
So, the enthalpy change (ΔH) for this reaction per mole of H2O produced is 56990 J/mol.