A student conducts an experiment to determine the enthalpy of neutralization for phosphoric acid
reacting with sodium hydroxide. The student combines 25.0 mL of equimolar solutions of acid and
base, both having an initial temperature of 22.5 °C, in an open polystyrene calorimeter. Each
solution has a density of 1.00 g/mL and a specific heat of 4.18
J
g • °C
. The student collects data until
a maximum temperature of 26.4 °C is reached.
(a) What assumption is the student allowed to make?
(b) Explain how the student knows data collection is complete.
(c) Calculate the qrxn for the neutralization reaction.
(d) Is the reaction exothermic or endothermic? Justify your answer.
(e) Calculate the ΔHneut
if the initial molarities of both solutions is equal to 0.80 M. Be sure to
report your answer with the correct sign and units.
a) no loss of heat to the surroundings.
b) when the temperature of the solution stops rising.
c) qrxn = mass 50 mL H2O x specific heat x (Tfinal-Tinitial)
d) T rose so rxn must be exothermic.
e) Determine the limiting reagent. DHneut = q/mols
efw
(a) The student is allowed to assume that there is no heat transfer to or from the surroundings during the reaction.
(b) The student knows data collection is complete when the maximum temperature of 26.4 °C is reached and there is no further increase.
(c) To calculate qrxn, we need to use the equation:
qrxn = mcΔT
Where:
m = mass of solution = volume of solution x density = 25 mL x 1.00 g/mL = 25.0 g
c = specific heat of solution = 4.18 J/g • °C
ΔT = change in temperature = final temperature - initial temperature = 26.4 °C - 22.5 °C = 3.9 °C
qrxn = (25.0 g)(4.18 J/g • °C)(3.9 °C) = 396.15 J
(d) The reaction is exothermic because the increase in temperature indicates the release of heat.
(e) To calculate ΔHneut, we use the equation:
ΔHneut = qrxn / moles
The moles of acid and base can be calculated using the equation:
moles = concentration x volume
Since the initial molarities of both solutions are equal to 0.80 M, the volume of each solution is 25 mL or 0.025 L.
moles = 0.80 M x 0.025 L = 0.020 mol
ΔHneut = 396.15 J / 0.020 mol = 19807.5 J/mol
To convert J/mol to kJ/mol, we divide by 1000:
ΔHneut = 19807.5 J/mol / 1000 = 19.8 kJ/mol
Therefore, ΔHneut is equal to 19.8 kJ/mol.
(a) The assumption the student is allowed to make is that the specific heat capacity and density of the solution remain constant throughout the reaction.
(b) The student knows data collection is complete when the maximum temperature is reached and there is no further increase in temperature for a significant period of time.
(c) To calculate the qrxn for the neutralization reaction, we can use the equation:
qrxn = m * c * ΔT
Where:
qrxn is the heat released or absorbed by the reaction,
m is the mass of the solution (which can be calculated using density and volume),
c is the specific heat capacity of the solution, and
ΔT is the change in temperature.
First, calculate the mass of the solution:
Mass = Volume * Density
Mass = 25.0 mL * 1.00 g/mL
Mass = 25.0 g
Calculating ΔT:
ΔT = Final temperature - Initial temperature
ΔT = 26.4 °C - 22.5 °C
ΔT = 3.9 °C
Substituting the values into the equation:
qrxn = 25.0 g * 4.18 J/g • °C * 3.9 °C
qrxn = 387.15 J
The qrxn for the neutralization reaction is 387.15 J.
(d) To determine if the reaction is exothermic or endothermic, we need to look at the sign of qrxn. Positive values indicate endothermic reactions (heat is absorbed), while negative values indicate exothermic reactions (heat is released).
In this case, qrxn is positive, so the reaction is endothermic.
(e) To calculate ΔHneut, we need to use the equation:
ΔHneut = qrxn / moles of limiting reactant
We need to determine the moles of limiting reactant first. Since both solutions have equal initial molarities, the moles of limiting reactant can be calculated using the volume of the solution and its molarity:
Moles of limiting reactant = Volume * Molarity
Moles of limiting reactant = 25.0 mL * 0.80 mol/L
Moles of limiting reactant = 0.020 mol
Substituting the values into the equation:
ΔHneut = 387.15 J / 0.020 mol
ΔHneut = 19357.5 J/mol
The ΔHneut for the neutralization reaction is 19357.5 J/mol.
(a) The student is allowed to assume that the heat loss to the surroundings is negligible during the experiment. This assumption is reasonable since the calorimeter is open and there is no insulation or any other means to prevent heat loss.
(b) The student knows data collection is complete when a maximum temperature is reached and the temperature starts to stabilize or decrease slightly. This indicates that the reaction has completed, and no further heat is being absorbed or released.
(c) To calculate the qrxn, we can use the formula:
qrxn = m * C * ΔT
Where:
- qrxn is the heat absorbed or released by the reaction (in joules),
- m is the mass of the solution (in grams),
- C is the specific heat capacity of the solution (in J/g • °C),
- ΔT is the change in temperature (final temperature - initial temperature) in °C.
In this case, the mass of the solution is equal to its volume multiplied by its density:
Mass of solution = volume * density
= 25.0 mL * 1.00 g/mL
= 25.0 g
The change in temperature (ΔT) is the final temperature minus the initial temperature:
ΔT = 26.4 °C - 22.5 °C
= 3.9 °C
Now, we can substitute the values into the formula:
qrxn = 25.0 g * 4.18 J/g • °C * 3.9 °C
= 384.15 J
Therefore, the qrxn for the neutralization reaction is 384.15 J.
(d) The reaction is exothermic. This can be justified by the fact that the temperature increased during the reaction. In an exothermic reaction, heat is released to the surroundings, resulting in a temperature increase.
(e) To calculate ΔHneut, we can use the equation:
ΔHneut = qrxn / moles of limiting reactant
Since both the acid and base are equimolar, the limiting reactant will be the reactant that is completely consumed. In this case, the acid and base have the same initial molarities of 0.80 M.
The moles of the limiting reactant can be calculated using the equation:
moles = molarity * volume (in liters)
Since the volume is given in milliliters, we need to convert it to liters:
Volume = 25.0 mL * (1 L / 1000 mL)
= 0.025 L
Now we can calculate the moles of the limiting reactant:
moles = 0.80 M * 0.025 L
= 0.02 moles
Finally, we can substitute the values into the equation:
ΔHneut = 384.15 J / 0.02 moles
= 19207.5 J/mol
Therefore, ΔHneut is equal to 19207.5 J/mol (with the correct sign indicating exothermicity).