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Enthalpy of Neutralization
Introduction
Energy changes always accompany chemical reactions. If energy, in the form of heat, is liberated the reaction is exothermic and if energy is absorbed the reaction is endothermic. Thermochemistry is concerned with the measurement of the amount of heat evolved or absorbed. The heat (or enthalpy) of neutralization (∆H) is the heat evolved when an acid and a base react to form a salt plus water.
Eq. 1 HNO2(aq) + NAOH(aq) → NaNO2(aq) + H 2 O(l) + Q
Q in the above equation is -∆H and is expressed in kJ/mol of water. Neutralization reactions are generally exothermic and thus ∆H is negative.
Heat measurements are performed by carrying out the reaction in a special container called a calorimeter. The heat (Q) given off by the neutralization reaction is absorbed by the reaction solution and the calorimeter. Both the solution and calorimeter increase in temperature due to the absorbed heat and this increase can be measured with a thermometer. ∆H is negative if heat is evolved and positive if heat is absorbed.
Eq. 2 - ∆Hneutralization = QSolution + QCalorimeter
To evaluate the calorimeter constant (also known as its heat capacity) in J/oC, one adds a known mass of hot water to a known mass of cold water which is in the calorimeter. Heat (Q) is lost by the hot water and is absorbed by the cold water and the calorimeter. Thus the heat absorbed by the calorimeter is the heat lost by the hot water minus the heat gained by the cold water.
QHot water = QCold water + QCalorimeter
Eq. 3 Qcalorimeter = QHot water - QCold water
It should be noted that we assume that the temperature of the calorimeter is the same as the solution inside it at all times. Q for both the hot and cold water is given by:
Eq. 4 Q = (4.184J/g-oC)(Mass in g)(∆t)
∆t is found by plotting temperature versus time for the system in the calorimeter and extrapolating the results to find ∆t at the instant of mixing (in this experiment, 5 minutes). A typical graph is shown in Figure 1.
Hot Water
∆t hot water Temperature, oC Mixture
∆t cold water and calorimeter
Cold Water
Time in Minutes Figure 1
The heat gained by the calorimeter is the difference between the heat lost by the hot water and the heat gained by the cold water. The calorimeter constant is this difference divided by the temperature change of the calorimeter (temperature change of the cold water)
Eq. 5 Calorimeter constant = QCalorimeter/∆tCold water
The ∆H of neutralization is found by mixing known quantities (moles) of an acid and a base (both initially at the same temperature) in a calorimeter and measuring ∆t of the mixture and the calorimeter. A typical graph for neutralization is shown in Figure 2.
Salt Solution
Temperature, oC
∆t of salt solution and calorimeter
Acid and Base
Time in Minutes Figure 2
Procedure
Acid-Base Neutralization
Each student will be assigned an acid and a base from the following list.
Acids: 2.00 M HCl Bases: 2.00 M NaOH 2.00 M HNO 3 2.00 M KOH 1.00 M H 2 SO 4 2.00 M NH 4 OH 0.667 M H 3 PO 4
Pipet 50.00 mL of your assigned acid into the clean dry calorimeter. Suspend the 50 oC thermometer in the acid. Pipet 50.00 mL of your assigned base into a 100 mL beaker. Since both the acid and base have been at room temperature for several hours, we can safely assume that both are at the same temperature. Record the acid (and base) temperature every one half minutes for 4.5 minutes. At the 5 minute mark quickly add the base to the acid in the calorimeter. Record the solution temperature every minute for 10 minutes giving a total of 15 minutes of readings. Record all temperatures to the nearest 0.01oC.
Clean and dry the beaker and calorimeter and repeat the above procedure for a second trial.
Plot graphs of temperature vs time for the two neutralization reaction trials. Extrapolate the two lines beyond the data points and determine the temperature changes (∆t) at the 5 minute mark as shown in Figure 2. Evaluate the average ∆H for the reaction on page 10.
Write a balanced equation for you assigned acid-base mixture, be sure to neutralize all of the acidic hydrogens of your acid. Determine the moles of water formed in your system from the molarity and volumes of acid and base used and correct your ∆H for the formation of one mole of water.
Using standard thermodynamic tables calculate the theoretical enthalpy change in kJ/1 mole water and compare to you experimentally determined value.
Name:_________________________________ Data Calorimeter Constant, Trial 1 Time, Minutes Hot Water Temp, oC Cold Water Temp, oC Mixture Temp, oC 0.0 ---------- -----------
0.5 ---------- ----------
5.0 Mix The Samples
6.0 ---------- ----------
Name:_________________________________ Data Acid-Base Neutralization, Trial 1 Time, Minutes Acid and Base Temperatures, oC Salt Solution Temperature, oC 0.0 ----------
0.5 ----------
Mix the Acid and Base
Name:_________________________________ Data Acid-Base Neutralization, Trial 2 Time, Minutes Acid and Base Temperatures, oC Salt Solution Temperature, oC
0.
Mix the Acid and Base
Name:_________________________________
Calculations and Results
Enthalpy of Neutralization
Assigned Acid and Base: ______________ _________________
Balanced Equation for Complete Neutralization
Density of Salt Solution (g/mL) ____________________________
Specific Heat of Salt Solution (J/g-oC) _______________________
Trial 1
- Volume of Salt Solution (mL) ________________________________
- Mass of Salt Solution (g) ____________________________________
- ∆t of Salt Solution (oC) (from graph) __________________________
- Heat Gained by Salt Solution (J) ______________________________
- Heat Gained by Calorimeter (J) (Cal. Const. x 3) _________________
- Total heat Evolved by Reaction (J) (4 + 5) ______________________
Trial 2
- Volume of Salt Solution (mL) _________________________________
- Mass of Salt Solution (g) _____________________________________
- ∆t of Salt Solution oC) (from graph) _____________________________
- Heat Gained by Salt Solution (J) ________________________________
- Heat Gained by Calorimeter (J) (Cal. Const. x 3) ___________________
- Total Heat Evolved by Reaction (J) (4 + 5) ______________________
Average Heat Evolved by Reaction (Q) (Trials 1&2) _______________________
Name:_________________________________ Calculations and Results
Moles of Acid Used in the Neutralization Reaction
Moles Base of Used in the Neutralization Reaction
Moles of Water Formed in the Neutralization Reaction
Enthalpy of Neutralization (∆H) (Average Heat Evolved in kJ/ Moles Water Formed )
Theoretical Enthalpy of Neutralization
From Standard Heat of Formation Data and calculate ∆H in kJ per one mole of H 2 O. Remember ∆Ho^ = Σn∆Hf Products – Σn∆Hf Reactants
Physical Properties of 1.00 M Salt Solutions
Prestudy Page 1
A student studied the enthalpy of neutralization of HClO 4 and NaOH by reacting 50.0 mL of 2.00 M HClO 4 with 50.0 mL of 2.00 M NaOH in a calorimeter. He measured the temperature versus time for 15 minutes. The first 4.5 minutes show the temperature of the acid and base before mixing. The solutions were mixed at the 5 minute mark and the temperature of the resulting salt solution measured for 10 minutes to give the following data.
Time, Minutes Solution Temperature, oC 0.0 (Acid/Base) 23. 0.5 (Acid/Base) 23. 1.0 (Acid/Base) 23. 1.5 (Acid/Base) 23. 2.0 (Acid/Base) 23. 3.0 (Acid/Base) 23. 4.0 (Acid/Base) 23. 4.5 (Acid/Base) 23. 5.0 Acid & Base Mixed 6.0 (Salt Solution) 33. 7.0 (Salt Solution) 33. 8.0 (Salt Solution) 33. 9.0 (Salt Solution) 33. 10.0 (Salt Solution) 33. 11.0 (Salt Solution) 33. 12.0 (Salt Solution) 33. 13.0 (Salt Solution) 32. 14.0 (Salt Solution) 32. 15.0 (Salt Solution) 32.
Plot a graph of temperature versus time and determine the ∆t at the time of mixing, 5. minutes. See Figure 2.
The density of the final solution is 1.07 g/mL and its specific heat is 3.85 J/g-oC. The calorimeter constant is 85.8 J/oC.
Write a balanced equation for the reaction and calculate the moles of water formed.
Name:_________________________________
Enthalpy of Neutralization
Prestudy Page 2
Calculate the enthalpy (∆H) of neutralization in kJ per 1.00 mole of water formed. Remember ∆H = heated absorbed by the final salt solution plus the heat absorbed by the calorimeter.
Calculate the theoretical (true) enthalpy of neutralization in kJ per 1.00 mole water from heats of formation data given on page 12. ∆H = Σn∆Hf products – Σn∆f reactants.
Calculate the % error between the two results. % Error = (Exptl Value – True Value) x 100% divided by True Value % Error can be + or – depending whether the experimental value is greater or less than the true value.
