Nahco3 And Nahco3

OBJECTIVE
To determine the change of standard Gibbs free energy for the decomposition of sodium hydrogen carbonate (NaHCO3) from the change of standard enthalpy and the change of standard entropy.

THEORY
Free energy is a state function that expresses the spontaneity of a chemical process in term of enthalpy and entropy change of a system under conditions of constant temperature and volume (Hemholtz free energy) and constant temperature and pressure (Gibbs free energy). It is a quantity of non-pressure-and-volume work that a system can perform.

Gibbs free energy also known as Gibbs function.
G = H – TS , where H, S and T denote enthalpy, entropy and temperature respectively.

Gibbs function for a chemical reaction system is based
on differences in standard free energies formation of products and reactants, in which, ∆G < 0 means reaction proceeds to the right, ∆G > 0 means reaction proceeds to the left and ∆G = 0 reaction is at equilibrium. Gibbs function also expresses the “escaping tendency” of a reaction component in terms of pressure of a gas, phase change and concentration of a solute.

When the state of the system changes at constant temperature : dG = dH – TdS (if the change at constant infinitesimal) ∆G = ∆H – T∆S (if t is measurable) and at standard conditions of 25˚C and 1 atm ∆Gᶿ = ∆Hᶿ - T∆Sᶿ
By determining the quantity of ∆Hᶿ experimentally according to Hess’s Law and extraction of ∆Sᶿ from the data ∆Gᶿ of the reaction can be estimated easily.
In this experiment, the heat released/absorbed by decomposition of NaHCO3 can not be measured directly and need to be carried out in two separate experiments in determining the quantities ∆Hᶿ for the reactions of NaHCO3 and Na2CO3.

The reaction are as follow:

2NaHCO3 (s) + H2SO4 (aq)  Na2SO4 (aq) + 2CO2 (g) + 2H2O (l) [1] ∆Hᶿ = Y kJmol-1
Na2CO3 (s) + H2SO4 (aq)  Na2SO4 (aq) +H2O (l) + CO2 (g) [2] ∆Hᶿ = Z kJmol-1
_____________________________________________________________________________________________________
[1] – [2’]
2NaHCO3 (s)  Na2CO3 (s) +H2O (l) + CO2 (g) [3] ∆Hᶿ = X kJmol-1

The enthalpy, H is defined as : H = U + PV
A change in enthalpy s equal to the heat supplied at constant pressure to a system in the casewhere the system does no additional work : dH = dq
For a measurable change, ∆H = qp . The heat released/absorbed by each reaction is calculated by using formula : qp = mCp∆T.

CHEMICALS
Sodium hydrogencarbonate powder, NaHCO3
Sodium carbonate powder, Na2CO3
1M Sulphuric acid, H2SO4

APPARATUS
50mL pipette, thermos, thermometer, 100mL beaker and stopwatch
PROCEDURE 4.4318g NaHCO3 was weighed and recorded. 50mL of 1M H2SO4 was pipetted out and transferred into a thermos. The cork with thermometer was used to cover the thermos. The stopwatch was started and the temperature of H2SO4 was measured and recorded every 1 min for 4 min. At 5th min, the weighed NaHCO3 was poured into the thermos and covered with cork and thermometer. The temperature was measured and recorded every 10s for the next 4 min. The steps above were repeated with 3.3222g Na2CO3.

RESULTS AND DISCUSSION
Mass of NaHCO3 : 4.4318g
Mass of Na2CO3 : 3.3222g

Table 1 : Temperature of the H2SO4 every 1 min for 4 min
Time (min)
Temperature of H2SO4 for NaHCO3 (˚C)
Temperature of H2SO4 for Na2CO3 (˚C)
1
33.0
34.0
2
33.0
34.0
3
33.0
34.0
4
33.0
34.0

Table 2 : Temperature after NaHCO3 and Na2CO3 were transferred into H2SO4 every 10s for 4 min respectively.
Time (s)
Temperature after adding NaHCO3 (˚C)
Temperature after adding Na2CO3 (˚C)
10
31.0
34.0
20
31.0
34.5
30
30.0
35.0
40
29.5
35.0
50
29.5
35.0
60
29.0
35.5
70
29.0
35.5
80
29.0
35.5
90
29.0
35.5
100
28.0
35.5
110
28.0
35.5
120
28.0
35.5
130
28.0
35.5
140
28.0
35.5
150
28.0
35.5
160
28.0
35.5
170
28.0
35.5
180
28.0
35.5
190
28.0 35.0
200
28.0
35.0
210
28.0
35.0
220
27.0
35.0
230
27.0
35.0
240
27.0
35.0

The heat change for the reactions of :
a.
NaHCO3 with H2SO4
2NaHCO3 (s) + H2SO4 (aq)  Na2SO4 (aq) + 2CO2 (g) + 2H2O (l)

qp = mCp∆T
= (50mL x 1cm3/mL x 1.84g/cm3 x 1kg/1000g) x (1.38kJkg-1K-1) x (28.0-33.0) = + 0.6348 kJ
∆Hᶿ = qp / n
= 0.6348 kJ / (4.4318 g / 84.008 gmol-1)
= 0.6348 kJ / 0.05275 mol
= + 12.034 kJmol-1

b. Na2CO3 with H2SO4
Na2CO3 (s) + H2SO4 (aq)  Na2SO4 (aq) +H2O (l) + CO2 (g) qp = mCp∆T
= (50mL x 1cm3/mL x 1.84g/cm3 x 1kg/1000g) x (1.38kJkg-1K 1) x (35.5-34.0) = - 0.1904 kJ
∆Hᶿ = qp / n
= 0.1904 kJ / (3.3222 g / 105.99 gmol-1)
= 0.1904 kJ / 0.03134 mol
= - 6.077 kJmol-1

∆Hᶿ for the decomposition of NaHCO3 :
2NaHCO3(s)+H2SO4(aq)Na2SO4(aq)+2CO2(g)+2H2O(l) [1] ∆Hᶿ=+12.034 kJmol-1
Na2CO3 (s)+H2SO4(aq)Na2SO4(aq)+H2O(l)+CO2(g) [2] ∆Hᶿ =-6.077 kJmol-1
_____________________________________________________________________________________________________
[1] – [2’]
2NaHCO3 (s)  Na2CO3 (s) +H2O (l) + CO2 (g) [3] ∆Hᶿ = X kJmol-1

∆Hᶿ = X kJmol-1 = [1] – [2’]
= [+12.034kJmol-1] – [+6.077 kJmol-1]
= +5.957
kJmol-1

∆Sᶿ for the decomposition of NaHCO3 :
∆Sᶿ = S[products] – S[reactants] = 135.98 JK-1mol-1 – 102.09 JK-1mol-1
( obtained from Appendix 3 Thermodynamic Data at std condition) = 33.89JK-1mol-1

∆Gᶿ for the decomposition of NaHCO3 :
∆Gᶿexp = ∆Hᶿ - T∆Sᶿ = 5957 Jmol-1 – [(33.0+34.0)/2) + 273.15 K] x [ 33.89 JK-1mol-1] = 5957 Jmol-1 – [306.65 K x 33.89 JK-1mol-1] = 5957 Jmol-1 – 10393.37 Jmol-1 = - 4435.37 Jmol-1

∆Gᶿtheory = G[products] – G[reactants] = [-1047.67 Jmol-1] – [-851.86 Jmol-1] = -195.81 Jmol-1
Thus, ∆Gᶿexp < ∆Gᶿtheory
Since ∆Hᶿ > 0, ∆Sᶿ > 0 and ∆Gᶿexp < 0, the decomposition of NaHCO3 can be considered as spontaneous and forward reaction which more stable at higher temperature.

The major sources of error during this experiment are while when the cork with thermometer was not really suited with the thermos. This shows that the thermos was not really isolated. There might be some heat change between the system and the surrounding. Besides, while transferring NaHCO3 and Na2CO3 into the thermos respectively, some of them were stacked to the mouth of the thermos and split out onto due to strong wind causing some losses of mass which affected the accuracy while calculating the ∆Hᶿ. Other than that, there was a problem while taking temperature reading on every 10s for 4 min. It was quite rushed to take temperature reading of the first 10s due to limited time to cover the thermos and to focus on stopwatch as well as the temperature reading. This caused the delay of taking the actual temperature reading for few seconds which affect the accuracy and precision.

CONCLUSION
∆Gᶿexp (4435.37 Jmol-1) < ∆Gᶿtheory. (-195.81 Jmol-1)
While ∆Hᶿ > 0, ∆Sᶿ > 0 and ∆Gᶿexp < 0, the decomposition of NaHCO3 can be considered as spontaneous and forward reaction which more stable at higher temperature.

REFERENCES
Atkins, P, Paula, JE. (2006). Physical Chemistry. Ninth Ed. : Oxford University Press.
Lower, S. (2010). Free energy: the gibbs function (gibbs energy). Retrieved on March 27
2013 from http://www.chem1.com

QUESTIONS
1. Ti = 22.6 ˚C Tf = 26.3 ˚C Mass of HCl = 50mL x 1.04gmL-1 = 52g Mass of NaOH = 75mL x 1.04gmL-1 = 78g qp = mHCl CHCl (Tf-Ti) + mNaOH CNaOH (Tf-Ti) = (52g) (4.186Jg-1 ᵒC-1) (26.3-22.6ᵒC ) + (78g) ( 4.186Jg-1 ᵒC-1) (26.3-22.6ᵒC ) = (217.672 JᵒC-1) (3.7ᵒC)+ (326.508 JᵒC-1) ( 3.7ᵒC) = 805.39 J + 1208.08 J = - 2013.47 J
This is an exothermic process due to rise in temperature during chemical reactions.

2. H2 (g) + ½ O2 (g)  H2O (l) (b) Mg (s) + 2HCl (aq)  MgCl2 (aq) + H2 (g) (c) MgO (s) + 2HCl (aq)  MgCl2 (aq) + H2O (l) (d)
(b) – (d’)
H2 (g) + ½ O2 (g)  H2O (l) (b)
MgCl2 (aq) + H2O (l)  MgO (s) + 2HCl (aq) (d’)
H2 (g) + ½ O2 (g) + MgCl2 (aq)  MgO (s) + H2O (l)(e) Mg (s) + 2HCl (aq)  MgCl2 (aq) + H2 (g) (c)
(e) – (c)
Mg (s) + ½ O2 (g)  MgO (s)