Temperature Measurement-Heat of Dissolution-Neutralization, Chemistry tutorial

Introduction

Temperature is a compute of how hot or cold an object is. Whenever there is a temperature difference, there will be a spontaneous heat flow from the object at higher temperature to the object at lower temperature. Thermometer is the instrument utilized to compute temperatures.

Celsius and Kelvin scales are utilized for metric and SI units correspondingly while Fahrenheit scale is the choice for British units. Such 3 scales are related via the subsequent relationships: 

K = 0C + 273   0F = 9/5 (0C) + 32   0C = 5/9 (0F - 32)

Every transform, physical or chemical, is connected through a transform in energy, generally in the form of heat. The energy transform of a reaction that takes place under a steady pressure is described as the heat of the reaction or the enthalpy transform. If heat is growing during the transform, the procedure is exothermic, and if heat is absorbed during the transform, the procedure is considered to be endothermic. By convention, enthalpy transform for an exothermic procedure has a negative value while that of an endothermic procedure has a positive value. 

We are well-known through different forms of energy. Heat energy, light energy, electrical energy, nuclear energy, chemical energy of the bonds in a molecule, are just a few instances of different forms of energy. From the Law of Conservation of energy, during any physical or chemical transform: 

Energy lost = Energy gained 

In this experiment, we will become well-known through temperature measurements and record the temperature transforms that take place when ammonium chloride and calcium chloride are dissolved in water. From this data, we will be able to compute the heat energy given off or absorbed during this dissolution process (heat of dissolution).

Heat absorbed/ evolved = (mass) (specific heat) (temperature change)   

The SI unit for heat is joule (J) while a non-SI unit calorie (cal) is widely utilized in scientific measurements. The relationship between such 2 units is: 

1 cal = 4.184 J

Specific Heat is the amount of heat needed to increase the temperature of one gram of a substance via one degree Celsius. It can be expressed in cal / g.0C or Joules / kg Kelvin. 

Water has a moderately high specific heat of 1 cal /g.0C while metals generally have low specific heat. To compute the heat of dissolution in water, specific heat of the aqueous solution will be considered to be that of pure water, 1 cal / g.0C. 

Calorimeter is an instrument utilized to compute heat flow in and out of a system. In this experiment, the calorimeter will consist of 2 Styrofoam cups, one nesting in the other.

Procedure 

Materials: 

Beaker (100 mL)                    Thermometer Hot Plate              Graduated Cylinder (50 mL) 

Ring Stand                             Thermometer Clamp    Stirring Rod    Balance 

Spatula Styrofoam Cups (2)    Cardboard Square 

Sodium Chloride (NaCl)          Calcium Chloride (CaCl2)   Ammonium Chloride (NH4Cl)

Ice

A. Temperature measurement: 

1. Using thermometer, compute the temperature of 50 mL of water in a 100 mL beaker. Be sure that the bulb is stable during the measurement and not touching the glassware. The bulb requires being fully absorbed in the liquid. 

2. Situate a 100 mL beaker with 50 mL water on a hot plate. Situate a thermometer in the water by the help of a stand and clamp. Bring the water to a boil indicated via steady stream of bubble formation from within the liquid. Once water starts to boil temperature is going to be stable until all of the water boils off. Compute the boiling point of water. 

3. Make about 30 mL of an ice -water mixture in a 100 mL beaker. Stir the ice slush and determine the Temperature. 

4. Add 3 tea spoons full of table salt, sodium chloride, to the slush and stir. Compute the temperature of the mixture. 

B. Heat of dissolution 

1. Work in pairs for this section. 

2. Weigh out about 10 grams of CaCl2. Be sure to record the exact mass. Construct a calorimeter through nesting 2 Styrofoam cups, one inside the other. Add 50 mL of water to the calorimeter. Permit the water to stand for 5 minutes to attain a steady temperature. Situate a small piece of card board to cover the cup. Make a small hole at the center of the card board and insert the thermometer through the hole. Make sure the thermometer bulb is under water. Determine the temperature of water. This is the initial temperature (Ti). 

3. Holding the calorimeter stable, add all of the CaCl2 to water, situate the cover, and stir quickly through a thermometer. Be watchful through the bulb of the thermometer while stirring.

4. After mixing, time - temperature data should be recorded. One partner should record the temperature while bother reads the time and keeps the record. 

5. For 5 minutes, right from the start of mixing, take temperature at intervals of every 30 seconds. The highest temperature attained is the final temperature (Tf) of water. 

6. Print the temperature versus time plot using the graph paper offered in the lab book. 

7. After recording our data, wash contents of the cup down the sink through lots of water. 

8. Repeat steps 1-7 using approximately 10 grams of ammonium chloride. The minimum temperature reached in this case is the final temperature (Tf).

Description on Heat of Neutralization

Energy transforms always accompany chemical reactions. If energy, in the form of heat, is released the reaction is exothermic and if energy is absorbed the reaction is endothermic. Thermo chemistry is concerned through the measurement of the amount of heat developed or absorbed. The heat (or enthalpy) of neutralization (ΔH) is the heat evolved whenever an acid and a base react to form a salt plus water.

We had started a discussion in the last chapter on heat. We executed an experiment on heat of dissolution. We as well studied able enthalpy (heat flow) as different solutes were melted in different solutions. In this chapter we shall focus on another kind of heat flow known as heat of neutralization.

Theory

Heat of neutralization is described as; the constant quantity of heat released when one mole of strong acid is neutralized through one mole of strong base to create one mole of water. As an instance;

Fig: exothermic reaction                      

This indicates that the process is an exothermic reaction. Strictly speaking (aq) implies that the reaction is taking place in these dilute solution that the addition of additional solvent reasons no heat change, for example the heat of dilution is zero; this is exactly true, though, only at infinite dilution. The steady quantity of heat is equal to the heat of formation of one mole of undissociated water from combination of one mole of H+ and one mole of OH-, where

       ; ΔH = - 13.733 cal/mol.

The neutralization reactions of weak acid or weak base evolved not only heat of neutralization but as well heat of ionization. So the quantity of heat of reaction Q will be either lower or higher than 13.733 cal/mol, according to the endothermic or the exothermic ionization process.

Equipment

To find out the heat of neutralization of hydrochloric acid and sodium hydroxide we shall require the subsequent equipment:

Calorimeter 

Bekmann thermometer 

Graduated cylinder 

 Beaker (250 ml) 

Distilled water

 Burette  

 Pipette 

 Conical flask 

 Funnel 

Spatula 

Phenolphthalein -pH indicator.

Procedure

1. Prepare 500 ml of 5N HCI and 500 ml of NaOH, thermostat the 2 solutions at room      temperature T1.

2. Titrate 50 ml NaOH (in a flask) against 5N HCI via using phenolphthalein as pH indicator, determine V1 (HCl).

3. Weigh the empty calorimeter W1.

4. Place 50 ml of 5N NaOH in the calorimeter and record the temperature at steady state T2.

5. Introduce quickly V1 (HCl) and justly record the temperature T3.

6. Weigh the calorimeter by its contents W2.

7. The weight of NaCl solution, W3 = W2 - W1.

Recording Data

 Cc

CNaCl  

V1

W1

  W2 

W3

T1 

  T2 

T3

 

 

 

 

 

 

 

 

 

Calculations:

1. Calculate ΔT= [T3 - (T1  +T2 )/2] ; T1 may be = T2

2. Calculate the heat absorbed by 5N NaCl solution where,

ΔH1 = CNaCl   * W3 * ΔT

Consider ( CNaCl  = 3.895 cal)

3. The heat absorbed by the calorimeter;    

        ΔH2 = CC * ΔT

NB The heat capacity of the calorimeter; =

         CC = Q / ΔT 

Where Q the amount of heat absorbed through the calorimeter: the amount of heat lost via water

4. Compute heat of neutralization where;

      ΔH = ΔH1 + ΔH2

5. Compute the number of water produced from neutralization reaction (undissociated water) where; 

     n = (5 * 36.5 * V1)/1000

6.  Compute the molar heat of neutralization;

     ΔH' = ΔH/n

Results

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