Report

Basic Chemical Practicum Report - TERMOKIMIA AND LAW HESS

EXPERIMENT IV

I.                   PRACTICAL TITLE
TERMOKIMIA AND LEGAL HESS

II.                DATE AND TIME
MONDAY,  13 March 2017

III.             OBJECTIVES OF EXPERIMENT
1. Measuring the heat of reaction with a simple tool.
2. Collecting and analyzing thermochemical data.
3. Apply the Hess law.

IV.             PRAKTEK QUESTIONS
1. Provide an understanding of: a) enthalpy; B) isolated system; C) open system; D) closed system; E) the environment; F) calorimeter; G) exothermic

Answer:
A. Enthalpy is the sum of the energy of all forms of energy possessed by the substance whose quantity can not be measured.
B. Isolated system is a system with its environment can not exchange both energy and material.
C. Open systems are systems and environments can exchange energy and material.
D. Closed system is a system which enables the transfer of energy (heat) to the environment, but can not transfer mass.
E. The environment is anything outside the system that affects and limits the system.
F. The calorimeter is the process of measuring the heat of the reaction / measuring the temperature change of a certain amount of water over the solution. As a result of a chemical reaction in an isolated container.

2. What is the difference in enthalpy with energy in (ΔE)?

Answer:
· Entalpi is the sum total of all forms of energy possessed by the substance whose quantity can not be measured, whereas
· Deep energy is the total amount of total potential energy and kinetic energy of substances contained in a system.

V.                THEORETICAL BASIS
Thermodynamics explains the relationship between heat and other forms of energy. Its development, which was an important scientific achievement in the 19th century, was due to the efforts of physicists and engineers who wanted to achieve high efficiency in a heat engine. The interest in improving the engine once again becomes important because of the need to use fossil fuels effectively. However, in the last 75 years, the important application of thermodynamics is in the field of chemistry. The law of thermodynamics is an important tool for studying chemical reactions. Thermochemistry is the influence of the heat that accompanies the chemical reaction. The second law of thermodynamics is primarily the basis for deriving the equilibrium constant of the properties of thermodynamic properties, in the third law of thermodynamics will be unveiled the starting point for looking at the properties of experimental thermodynamic properties (Petrucci.1987: 225).

Each system has energy because the material particles (solid, liquid, or gas) always move randomly and diverse. There is translational motion, rotation, and ubrasi (vibrate). In addition, there may be a shift in the energy levels of electrons in atoms or molecules. Each movement, influenced by many factors and can change shape when colliding with each other. As a result, the energy of a particle's ganga will be different from the others. The total energy of all the particles will be different from the others. The total energy of all the particles in the system is called the inner energy (U). Therefore, the absolute value of U can not be calculated.
The first law of thermodynamics discusses the energy changes that accompany the event, and is useful for calculating the incoming or outgoing heat of the system. By equation: q = aU - w. The second law, to be discussed about spontaneous and non-spontaneous changes. The second law of thermodynamics sounds the natural process of adding natural entropy or entropy as it increases, and the third thermodynamic law sounds a pure element or compound in the form of perfect crystals having zero entropy at 0 ° C (Syukri.1999: 74).

The application of the first law of thermodynamics to chemical events is called thermodynamics, which deals with the heat that accompanies chemical reactions. Chemical reactions include isothermal processes, and when done in the open air then the reaction calories qp = ΔH. Consequently, the heat can be calculated from the enthalpy change of the reaction q = ΔH = Result H-reaction. So that there should be a uniformity of standard conditions, ie temperature 25 ° C and pressure 1 Atm. Thus, thermodynamic calculations are based on standard conditions (Shukri.1999: 84).

Thermochemistry is a part of thermodynamics that studies heat changes that follow chemical reactions. The amount of heat that arises or is required in a chemical reaction is called reaction heat. The heat of the reaction at P remains the same as the change in the ental strength, and the heat of reaction in U remains the same as the change in power.
The magnitude of the reaction pans depends on the type of reaction, the phase state of the substances in the reaction, the amount of the reacting agent, and the reaction temperature. In thermodynamic equations, the amount of substances in the reaction is expressed in moles while the heat is expressed in Kilocalories (Sukardjo 1990: 192).

·         · Change of enthalpy (ΔH) and internal energy changes (ΔU) in chemical reactions.
The reaction at constant pressure ΔH and heat of reaction at constant volume, ΔU is connected through the equation:
ΔU = ΔH-PΔU
If the heat of reaction is carved under constant pressure conditions at a constant temperature of 298 K, it is obtained -566.0 KJ, indicating that the 566.0 KJ energy has left the system as heat ΔH = -566.0 KJ. To evaluate the pressure-volume work
PΔU = P (Vt -Vi)
Then we can use the ideal gas equation. The kinetic equation
PΔV = RT (nf-ni)
Nf is the mole of gas in this product (2 moles CO2) is the number of moles of reactant solid gas (2 moles CO + 1 mole O2), so
PΔV = 0,0083145 KJ / mol K-1. 298 K × [2 (2 + 1)] mol
         = -2.5 KJ
Internal energy change is
ΔU = ΔH-PΔV
      = -566,0KJ - (- 2,5KJ)
      = -563.5KJ

·         The enthalpy change (ΔH) that accompanies the change of material form.
When the liquid comes into contact with the atmosphere, energized molecules on the liquid surface can overcome the attraction with each other and enter the form of gas or steam.

·         Indirect Determination ΔH: Hess's law of equilibrium of enthalpy changes
ΔH is an extensive nature
Hess's law concerning the sum of the constant heat (Petrucci.1992: 239-244).

G.H.Hess a Swiss chemist in 1840 investigated whether a chemical reaction taken through several streets would affect the heat of his reaction.
To answer the problem Hess did some experiments. The following presented data from experiments investigating the heat of sulfur formation reaction.
S (P) + 11 / 2O2 → SO3 (Q)
This SO3 formation reaction can be done through several ways. From each experiment will be investigated the amount of reaction heat generated.

1.      Direct way
S (S) + 3 / 2O2 (g) → SO3 (g) ΔH = -395.73

2.      Indirect way
S (S) + O2 (g) → SO2 (g)                                     ΔH = -296.83
SO2 (g) + 1/2 O2 (g) → SO3                               ΔH = -98.9
S (S) + 3 / 2O2 (g) → SO3 (g)                  ΔH = -395.73

Standard enthalpy change (ΔHo)
Some types of standard enthalpy changes, namely:

·         Change of enthalpy of standard formation (ΔHfo)
It is an enthalpy change that occurs in the formation of 1 mole of a compound of the most stable elements in the satandar state.

·         Change of standard decomposition enthalpy (ΔHdo)
It is an enthalpy change that occurs on the decomposition of 1 mole of a compound into its most stable elements in the standard state.

·         Change of standard burning enthalpy (ΔHoc)
Is the enthalpy change that occurs in the burning of 1 mole of a substance perfectly. Burning is the reaction of a substance with oxygen, thus when a substance is completely burned and it contains:
- C à CO2
- H à H2O
- S à SO2 (Susanto.2003: 46).

VI.             TOOLS AND MATERIALS

A. Tool
Ø Measuring cups
Ø Calorimeter
Ø Mixer
Ø Cup of trophies

B. material
Ø 40 mL distilled water
Ø 40 mL HCl 1 M
Ø 40 mL NaOH 1M
Ø 1M acetic acid
Ø Sodium Hydroxide 1M
Ø Sodium Acetate 1M
Ø 1M Nitrate Acid
Ø Ammonia 1M

VII.          WORK PROCEDURES

A. Determination of calorimeter constants

40 ml of distilled water

- Measured with measuring cups
- Poured chalorimeter
The completed calorimeter & stirrer tool is closed
- Recorded its contents

40 ml of distilled water

- Measured again with measuring cups
- Poured into dry cup
- Heated 60oc-70oc
- Measured hot water temperature appropriately
- Rapidly displaced kekalorimeter
- Recorded every 15 seconds while stirring
The temperature of the maximum solution and slowly decreases, recorded every 1 minute until constant

B. Determination of ΔH of neutralization for acid-base

The calorimeter is dried
40 ml of 1M NaOH solution

- Measured and incorporated kekalorimeter

40 ml 1M HCl

- Measured in 150ml cup glass
- Close near the calorimeter
- Temperature of acid and base solutions is measured
- The temperature of the two solutions should not be at odds of 0.5oc, when different temperatures are equalized
- When the temperature is the same, put it quickly into the calorimeter

Observation result

VIII.       OBSERVATION DATA

A. Determination of calorimeter constants

     

	Deuteronomy 1	Deuteronomy 2	Average
Hot water temperature, ° C	60 ° C	57 ° C	58.5 ° C
Cold water temperature, ° C	28 ° C	26 ° C	27 ° C
Mixed temperature, ° C	39 ° C	38 ° C		38.5 ° C

B. Determination of ΔH of neutralization for acid-base

·               NaOH + HCl 1M

	Deuteronomy 1
Temperature of acid solution Temperature of base solution Mixed temperature	29,5°C 30°C 39°C

·               KOH 1M +HNO₃ 1M

	Deuteronomy 1
Temperature of acid solution Temperature of base solution Mixed temperature	23°C 23°C 25°C

·               NH₃ 1M + HCl 1M

	Deuteronomy 1
Temperature of acid solution Temperature of base solution Mixed temperature	32°C 31°C 31°C

·               NH₃ + HNO₃ 1M

	Deuteronomy 1
Temperature of acid solution Temperature of base solution Mixed temperature	23°C 23°C 25°C

·               CH₃COOH + NaOH

	Deuteronomy 1
Temperature of acid solution Temperature of base solution Mixed temperature	26°C 30°C 27°C

IX.             DISCUSSION

A.    The calorimeter constant

In this experiment the calorimeter is cleaned and dried. Enter 20 mL of distilled water into the calorimeter, record the weight and measure the temperature, then take 40 mL of distilled water with a measuring cup, heat and record the hot water weight. Combine hot water and cold water into the calorimeter, note the temperature. Result of experiment of determination of calorimeter constant, by formula:

                                         C.Mp (Tp-Tm)      =   C.Md(Tm-Td) + W(Tm-Td)

                     4,184 J/g°C.20(60°C-39°C)     =   4,184 J/g°C.20(39°C-28°C) + W(39°C-28°C)

                                    4,184 J/g°C (21°C)    =   4,148 J/g°C (11°C) + W (11°C)

                                                     87,864 J    =   46,024 J + 11 W

                                                          11 W   =   87,864 J – 46,024 J

                                                          11 W   =   41,84

                                                               W   =   3,803 J°C

B.     Determination of ΔH is neutralized for acid-base

In this experiment we made observations of different mixtures of acid-base solutions. After conducting an experiment in accordance with work procedures, the data obtained are:

Work, then the data obtained are:

1.      The temperature of the acid solution (HCl) 1M           = 29.5 ° C

The temperature of the basic solution (NaOH)            = 30 ° C

Mixed temperature                                                       = 39 °

And get it:

Qreaksi = C. M. (Tf-Ti) + W (Tf-Ti)
               = 4.184.80 (30 -29.75) + (39 -29.75) .3,80
               = 334.72. (9,25) + 35,17
               = 3131.33 J
Qreaksi = - Qperiphery
-Q periphery = 3131.33 J

The equation of the reaction
HCl + NaOH → NaCl + H2O

Then ΔH reaction = -Qperiphery / The reacting mole
                                = 3131.33 J / 0.02 mol
                                = 156566,5J / mol

2.      Temperature of CH3COOH 1M                                  = 26 ° C
The temperature of 1M NaOH base solution is            = 30 ° C
Mixed temperature                                                       = 27 ° C

To get Qreaksi used formula:

Q reaction         = C. M. (Tf-Ti) + W (Tf-Ti)

                          = 4,184J/g°C.80gr(27°C-28°C) + (27°C-28°C ) . 3,803 J/°C

                          = -334,72 J/°C . (-1°C) +(-3,803 J)

                          = -338,523 J

-Qperiphery      = Q reaction

                          = -338,523 J

The reacting mole
CH3COOH + NaOH → NaCH3COO + H2O

Then ΔH reaction = -Qperiphery/ The reacting mole
                                = 338,523 J/0.02 mol
                                = 16926.15 J/mol

3.      Temperature of 1M (NaCH₃COO) acid solution                     = 23 ° C
The aqueous solution temperature of (HCl) is 1M                   = 23 ° C
Mixed temperature                                                                   = 25 ° C
To get Qreaksi used formula:

Qreaction    = C . M . (Tf-Ti) + W (Tf-Ti)

                    = 4,184J/g°C.80gr(25°C-23°C) + (25°C-23°C ). 3,803 J/°C

                    = 334,72 J/°C . (2°C) +3,803 J/°C (2°C)

                    = 677,046 J

-Qperiphery            = Qreaction

                    = 677,046 J

NaCH₃COO + HCl → NaCl + CH₃COOH

Then ΔH reaction         = -Qperiphery/ The reacting mole
                                    = 677,046 J/0.02 mol
                                    = 33852,3 J/mol

4.      Temperature of 1M HNO3 acid solution                     = 31 ° C
The aqueous solution temperature of NaOH is 1M      = 32 ° C
Mixed temperature                                                       = 31 ° C
To get Qreaksi used formula:

Qreaction    = C . M . (Tf-Ti) + W (Tf-Ti)

                    = 4,184J/g°C.80gr(31°C-31,5°C) + (31°C-31,5°C ). 3,803 J/°C

                    = 334,72 J/°C . (-0,5°C) +(-1,9015 J)

                    = -169,2615 J

-Qperiphery= Qreaction

                    = -169,2615 J

Mol that reacts → mol        = MV

                                              = 1 mol / V. 0.021 V

  = 0.021 mol

Then ΔH reaction         = -Qperiphery/ The reacting mole
                                    = -169,2615  J/0.02 mol
                                    = 8403,075 J/mol

5.      Temperature of 1M (HCL) acid solution                                 = 31 ° C
The aqueous solution temperature of NH­₃) is 1M                    = 27 ° C
Mixed temperature                                                                   = 29 ° C
To get Qreaksi used formula:

Qreaction    = C . M . (Tf-Ti) + W (Tf-Ti)

                    = 4,184J/g°C.80gr(29°C-29°C) + (29°C-29°C )                    

                    = 0 J

-Qperiphery            = Qreaction

                    = 0 J

HCl + NH₃ → NH₄Cl

Then ΔH reaction         = -Qperiphery/ The reacting mole
                                    = 0 J/mol

X.                DISCUSSION

A.    DETERMINATION OF CALORYMETER STIPULATION

A calorimeter is a tool used to measure the heat of a reaction. While the heat absorbed by the calorimeter is called the calorimeter constant. The calorimeter constant can be determined by experimental temperature measurements on cold water, hot water, and a mixture of cold and hot water. Then we put the water into the calorimeter alternately to measure the temperature. After getting the temperature respectively, then we can determine the calorimeter constant that has been discussed in the discussion. From the discussion obtained the calorimeter constant (W) 3.803 J / ° C.
Our group has determined from our experimental results that the calorimeter constant is 3.803J / ° C. This calorimeter constant is also used to determine ΔH neutralization. According to the theory of the determination of the calorimeter is 0 or the smaller the value the better the calorimeter used.
So our experiments got results greater than zero. This experiment has been performed in accordance with work procedures, it's just possible tool error at the time of lab work, other things can also happen to praktikan itself.

B.     DETERMINATION ΔH TO NETRALIZATION OF BASIC ACID
From this experiment, we get the ΔH neutralization value of acids and bases of:

1. ΔH neutralization between CH3COOH + NaOH             = 16926,15 J / mol
2. ΔH neutralization between NaCH3COOH + HCl             = 33852,3 J / mol
3. ΔH neutralization between HNO3 + NaOH                      = 8403,075 J / mol
4. ΔH of neutralization between HCl + NaOH                      = 156566,5J / mol
5. ΔH neutralization between HCl + NH3                             = 0 J / mol
Thus, from these experiments we obtained a different ΔH of neutralization between the respective base acids of the experiments.

XI.             POST-FACTORY QUESTIONS

1.      For acid-base reaction in procedure B, how much is OH neutralization if you wrongly assume that the caloric received calorimeter is zero? Show by count using your data!
Answer:
Q = C.M (Tf - Ti) + W (Tf - Ti)
      = 4.184 x 80 (39-29.75) + 3.803 (39-29.75)
      = 334.72 (9,25) + 3,803 (9,25)
      = 3131.34 J

2.      What is the effect on ΔH of neutralization when reaction with 1 M NaOH is HCl with concentration more than 1 M?
Answer:
No effect, since the reaction ΔH used is that the reacting mole is 0.04 mole of NaOH.
Thus, ΔH reaction with 1M HCl, ΔH HCl reaction> 1M.

3.      Show that the reaction calorie of the following pair:
A. NaOh + HCl
B. NaOH + HOAC
C. HCl + NaOH
Can be used to describe Hess's law. Collect data from other practitioners, and compute the reaction ΔH for the reaction (a). Do the results match the experimental results?
Answer:
NaOH + CH3COOH → CH3COONa + H2O    ΔH = 432.3 J
HCl + CH3COONa → NaCl + CH3COOH        ΔH = 397,718 J
NaOH + HCl → NaCl + H2O                              ΔH = 830,018 J

4.      Conclude the ΔH price of neutralization for acids and bases with different strengths!
Answer:
The stronger the acid and base react, the greater the enthalpy (ΔH) of the reaction. It can be seen that the reaction of HCl with NaOH yields the greatest reaction ΔH compared to the others.

XII.          CONCLUSION

1.      In measuring the reaction can use a simple calorimeter.

2.      To determine the calorimeter constant, it takes data in the form of weight, heat type, and temperature of the substance. To determine the calorimeter constant based on the black principle, the equation is:
C.MP. (Tp-Tm) = c.Md (Tm-Td) + W (Tm-Td)
information:
C = Water heat type 4,148 J / g ° C
Mp = Weight of hot water
Md = Cold water weight
Tp = Hot water temperature
Td = Cold water temperature
Tm = mixed temperature
W = The calorimeter constant J / g

3.      For ΔH the reaction is used equation
Where, Qrex = Qreaksi
                                   = Q solution + Q calorimeter
                                   = C. Mtot (Tf-Ti) + W (Tf-Ti)

4.      The calculation of heat of reaction can be done by using:
A. Hess's Law
B. Standard enthalpy entries
C. Average reactant energy

XIII.       BIBLIOGRAPHY

Agus. 2009.Kimia Dasar Universitas.Jakarta:Erlangga.

Petrucci, Raip H. 1992. Kimia Dasar. Jakarta: Erlangga

Sukardjo. 1990. Kimia Dasar. Jakarta : Erlangga

Susanto. 2003. Panduan Belajar Sukses SPMBPTN. Yogyakarta: UGM

Syukri. 1999. Kimia Dasar I. Bandung: ITB