# Enthalpy Lab

Topics: Oxygen, Ethanol, Carbon dioxide Pages: 7 (1026 words) Published: April 22, 2013
LAB OF ENTHALPY CHANGE IN COMBUSTION

Objective:
Determine the Enthalpy change of combustion ΔHc of three different alcohols. Methanol, Ethanol and Isopropilic acid.

Procedure:

1. Fill the spirit micro burner with Ethanol and weight it 2. Pour 100 cm3 of water into the aluminum cup
3. Arrange the cup a short distance over the micro burner 4. Measure the temperature of water
5. When the temperature of the water has risen by 10°C, record the temperature. 6. Reweight the microburner. Record
7. Repeat steps 1 to 6 but now with Methanol
8. Repeat step 1 to 6 with Isopropilic acid.

Data and Processing
Alcohols | Initial mass of microburner fill with alcohol (g)± 0.01| Final mass of microburner fill with alcohol (g)± 0.01| Initial temperature of water(°C)± 0.1| Final temperature of water(°C)± 0.1| Volume of water in metallic calorimeter (cm3) ±0.5| Ethanol | 5.38| 5.08| 23.0| 33.0| 100.0|

Methanol | 5.33| 4.94| 24.0| 34.0| 100.0|
Isopropolic acid | 5.45| 5.20| 24.0| 34.0| 100.0|

Find the mass of water

ρ=mv

ρ (density) H2O = 1.0 g /cm3

Calculate % Uncertainty in mass of water

As the mass of water is the same in the 3 alcohols the %uncertainty is the same for all the alcohols

Absolute uncertainty of the measuring cylindermass of water × 100
Calculating Δ mass change (alcohol’s burned mass)

(initial mass ± 0.01 g)-(final mass ± 0.01 g)

Calculating percentage uncertainty in alcohol burned mass

Absolute uncertainty of alcohol's burned massalcohol's burned mass × 100

Calculate the percentage uncertainty of alcohol burned moles

percentage uncertainty of alcohol burned mass+percentage uncertainty of alcohol`s molar mass

Calculating ΔH (enthalpy change)

ΔH=-mass of water x specific heat of water x Δ T of water mol of alcohol

* The specific heat for water is 4.18

=100x4.184x 10=4,184 J or 4,184 KJ exothermic

Methanol=

* H2O = 100 ml
* mH2O= 100 mg

* t1 H2O= 23°C

Mass (i) methanol= 5.38g

* tf= H2O=33°C

Mass (f) methanol= 5.08 g

ΔT= TF-TI=
ΔT= 10°C

Calculating mass change

Δm=mi-mf=

5.38-5.30=0.30g

Δm=ΔmMr=0.3032.04=0.009 mol

ΔH=-4.1840.009=-464888.9jmol

% uncertainity(balance)=±0.020.30x 100=6.67 %

% uncertainity(thermometer )=±110x 100=10 %

%error=-726000-(-464888.9)-726000x 100=36%

Qualitative Observations

We could see from the burn of methanol that the flame owas of color orange red, moreover therewere not dirt in the bottle.

Ethanol

* H2O = 100 ml
* mH2O= 100 mg

* t1 H2O= 24°C

Mass (i) ethanol= 5.33 g

* tf= H2O=34°C

Mass (f) ethanol= 4.94 g

ΔT= TF-TI=
ΔT= 10°C

Calculating mass change

Δm=mi-mf= 0.39 g

5.33-4.94= 0.39 g

ethanol 0.3946.07 g/mol=0,008 mol

ΔH=-4.1840.008=-523,000jmol

% uncertainity(balance)=±0.020.39x 100=13 %

% uncertainity(thermometer )=±110x 100=10 %

%error=-1360000-(-523000.0)-1368000x 100=61.8%

Qualitative Observations

We can observe a lost of weight during the experiment, moreover the flame was orange blue but with a big strong orange , it didn’t burn complete therefore show dirt in the cup.

Isopropolic acid

* H2O = 100 ml
* mH2O= 100 mg

* t1 H2O= 24°C

Mass (i) = 5.45 g

* tf= H2O=34°C

Mass (f) ethanol= 5.20g

ΔT= TF-TI=10 c

Δm=mi-mf= 0.25 g

Isopropolic acid 0.25 60,1g/mol=0,004 mol

ΔH=-4.1840.04=-1,046,000jmol

% uncertainitybalance=±0.020.25x 100=8%

% uncertainity(thermometer )=±110x 100=10 %

%error=-2006.9-(-1046.0)-2006.9x 100=47.9%

* At last, the alcohol used was Isopropilic acid. The flame with this alcohol was the strongest flame, it was very strong, was very yellow at the top and blue at the bottom. * We could also notice that all the 3 alcohols produced Soot. (is a general term that refers to impure carbon particles resulting from the incomplete combustion)

Conclusion =

As we...