Enthalpy of Neutralization

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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) + H2O(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.

1

Hot Water

∆t hot water

Temperature, oC

Mixture
∆t cold water and calorimeter
Cold Water
5.0
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
5.0
Time in Minutes
Figure 2

2

The heat given off by the neutralization reaction, ∆H, is the sum of the heat absorbed by the solution and calorimeter.
Eq. 6 -∆H = + Qsolution + Qcalorimeter
Eq. 7 Qsolution = (Sp. Ht.)(Volume)(Density)(∆t)
Eq. 8 Qcalorimeter = (Calorimeter Constant)(∆t)
The specific heat (Sp. Ht.) and the density of the solution of the salt formed from your assigned acid and base. You must calculate the ∆H for one mole of water formed in your system in order to compare it to the theoretical value. ∆t is the same for the solution and the calorimeter.

Procedure
Calorimeter Constant
Weigh 50.0 grams of distilled water into your assigned calorimeter. Be sure it is clean and dry. This is the cold water. Suspend the supplied 50oC thermometer into the water from a paper clip on a ring stand. Be sure to note the thermometer can be read to 0.01oC.

Weigh 50.0 grams of distilled water into a 100 mL beaker. Heat this water to 15 – 20oC above the temperature of the cold water. Remove from the hot plate and place on the bench. Suspend a 110oC thermometer (in your locker) from a ring stand with a paper clip. Be sure to note the starting time...
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