In a constant-pressure calorimeter, 75.0 mL of 0.950 M H2SO4 was added to 75.0 mL of 0.460 M NaOH. The reaction caused the temperature of the solution to rise from 21.13 °C to 24.26 °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.
To find the enthalpy change (ΔH) for the reaction per mole of water produced, we need to use the formula:
ΔH = q / n
Where:
- ΔH is the enthalpy change per mole of water produced,
- q is the heat exchanged in the reaction, and
- n is the number of moles of water produced.
To calculate q, we can use the formula:
q = m * c * ΔT
Where:
- q is the heat exchanged in the reaction,
- 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 (final temperature - initial temperature).
First, let's calculate the mass of the solution:
Since the density of water is the same as the solution, the volume of the solution equals the mass of the solution. Therefore, the mass of the solution is:
mass of solution = volume of solution = volume of H2SO4 + volume of NaOH
mass of solution = (75.0 mL) + (75.0 mL) = 150.0 mL = 150.0 g
Next, let's calculate the heat exchanged (q):
q = m * c * ΔT
q = (150.0 g) * (4.18 J/g°C) * (24.26 °C - 21.13 °C)
q = (150.0 g) * (4.18 J/g°C) * (3.13 °C)
q = 1937.91 J
Now, let's calculate the number of moles of water produced:
The balanced chemical equation for the reaction between H2SO4 and NaOH is:
H2SO4 + 2NaOH -> Na2SO4 + 2H2O
From the balanced equation, we can see that for every mole of H2SO4, 2 moles of water are produced. Therefore, the number of moles of water produced is half the number of moles of H2SO4 added.
moles of H2SO4 = volume of H2SO4 * molarity of H2SO4
moles of H2SO4 = (75.0 mL) * (0.950 mol/L) = 71.25 mmol = 0.07125 mol
Since 2 moles of water are produced per mole of H2SO4, the number of moles of water produced is:
moles of water = 0.07125 mol / 2 = 0.035625 mol
Finally, we can calculate the enthalpy change (ΔH) per mole of water produced:
ΔH = q / n
ΔH = 1937.91 J / 0.035625 mol
ΔH = 54394.58 J/mol
Therefore, the enthalpy change (ΔH) for this reaction per mole of water produced is 54394.58 J/mol.