In a constant-pressure calorimeter, 60.0 mL of 0.930 M H2SO4 was added to 60.0 mL of 0.320 M NaOH. The reaction caused the temperature of the solution to rise from 22.68 °C to 24.86 °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. answer in KJ/mol of H2O
q = mass x specific heat x (Tfinal-Tinitial) = approx 1.1 kJ but you should get a better answer than this estimate. That is for approx 0.02 mols H2O produced. You should confirm all of those estimates.
Then kJ/mol = 1.1/0.02 = ? kJ/mol.
Note: Where did the 0.02 mols H2O produced come fromj?
You had mols H2SO4 = M x L = about 56.
You had M x L NaOH = about 20 so NaOH is the limiting reagent. Go through the equation to confirm that.
To calculate the enthalpy change (ΔH) for the reaction per mole of H2O produced, we can use the equation:
ΔH = q / n
where:
- ΔH is the enthalpy change (in kilojoules per mole of H2O)
- q is the heat released or absorbed by the reaction (in joules)
- n is the number of moles of H2O produced.
To find q, we need to calculate the heat absorbed or released in the reaction, which can be determined using the equation:
q = m * c * ΔT
where:
- q is the heat absorbed or released (in joules)
- m is the mass of the solution (in grams)
- c is the specific heat capacity of water (in joules per gram and Kelvin)
- ΔT is the change in temperature (in Kelvin).
In this case, the volume of H2SO4 and NaOH are equal, so we don't need to consider the volume change.
First, let's calculate the mass of the solution:
mass = volume * density
Since both volumes are 60.0 mL and the density of water is 1.00 g/mL:
mass = 60.0 mL * 1.00 g/mL = 60.0 g
Next, let's determine the change in temperature:
ΔT = final temperature - initial temperature
ΔT = 24.86 °C - 22.68 °C = 2.18 °C
Now we can calculate the heat (q):
q = mass * c * ΔT
q = 60.0 g * 4.184 J/g·K * 2.18 K = 543.144 J
Since we want the answer in kilojoules per mole of H2O, we need to convert the heat from joules to kilojoules and divide it by the number of moles of H2O.
To find the number of moles of H2O produced, we need to calculate the number of moles of limiting reactant (either H2SO4 or NaOH) and multiply it by the stoichiometric coefficient of water.
Let's calculate the number of moles of H2SO4:
moles H2SO4 = volume * molarity
moles H2SO4 = 60.0 mL * 0.930 mol/L = 55.8 mmol
Let's calculate the number of moles of NaOH:
moles NaOH = volume * molarity
moles NaOH = 60.0 mL * 0.320 mol/L = 19.2 mmol
Since NaOH is the limiting reactant (as it is less), we will use its moles to calculate the moles of water. Based on the balanced chemical equation, the stoichiometric coefficient of water is 2 (2 moles of NaOH produce 2 moles of H2O).
moles H2O = 2 * moles NaOH
moles H2O = 2 * 19.2 mmol = 38.4 mmol
Now we can convert mmol to moles:
moles H2O = 38.4 mmol / 1000 = 0.0384 mol
Finally, let's calculate the enthalpy change (ΔH):
ΔH = q / n
ΔH = 543.144 J / 0.0384 mol = 14133.75 J/mol
Converting J to kJ:
ΔH = 14133.75 J / 1000 = 14.134 kJ/mol
Therefore, the enthalpy change (ΔH) for this reaction per mole of H2O produced is 14.134 kJ/mol.