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Dupre

March 27, 2017

Abstract:

Entropy (∆Sº) and Enthalpy (∆Hº) are values used in thermodynamics which are then used to calculate Gibbs Free Energy (∆Gº). ∆Gº is a thermodynamic potential that can be used to calculate the maximum or reversible work that may be performed by a system at a constant pressure of 1 atm, and temperature of 25ºC. To determine the values of ∆Hº and ∆Sº a series of titrations using HCl and Borate samples at decreasing temperatures was performed. The data obtained from the titrations were used to solve for the molarities of and . With the help of stoichiometry we can obtain the equation: ln = ln(4) + 3 ln [] to calculate the equilibrium constant. After graphing ln K vs. 1/T, we can solve for ∆Hº and ∆Sº in the equation format of ln(K) = (∆Sº/R) - (∆Hº/RT). Through this formula, it was calculated that ∆Hº= 107.48 kJ and ∆Sº= 335.06 J/K.

Purpose:

This experiment determined ∆Hº and ∆Sº with varying temperatures by using the information gained from multiple titrations to graph lnK vs. 1/T. The equation in slope-intercept form was used along with the equation lnK= (∆Sº/R) – (∆Hº/RT), to solve for ∆Hº and ∆Sº.

Procedure:

The procedures from laboratory packet (D. Mead, J. Robinson, S. Summers. General Chemistry I Laboratory Experiments Packet, pg. 39-43) were followed without modification.

Data and Results:

Molarity of HCl: 0.4763 M

Table 1- Sample Data Table:

Test Tube #

1

2

3

4

5

Temp (°C)

63 ºC

54 ºC

45 ºC

35 ºC

25 ºC

Buret Reading

/ mL

24.82 mL

19.46 mL

22.35 mL

11.42 mL

6.86 mL

Buret Reading

/ mL

6.28 mL

0.78 mL

3.52 mL

4.99 mL

2.55 mL

Test Tube #

1

2

3

4

5

Temp (K)

336.15 K

327.15 K

318.15 K

308.15 K

298.15 K

1/T

0.002975

0.003057

0.003143

0.003245

0.003354

Volume HCl (L)

0.01854 L

0.01868 L

0.01883 L

0.00643 L

0.00431 L

Moles

()()

0.008831

0.008897

0.008969

0.003063

0.002053

Moles

(moles / 2)

0.004415

0.004449

0.004484

0.001531

0.001026

[]

(moles / 0.00500 L)

0.8830 M

0.8898 M

0.8968 M

0.3062 M

0.2052 M

ln []

-0.124430

-0.116759

-0.108922

-1.183517

-1.583770

3 ln []

-0.373290

-0.350276

-0.326767

-3.550550

-4.751310

ln

1.013004

1.036018

1.059527

-2.164256

-3.365016

Table 2- Calculations Table:

Graph: ln vs. 1/T

y= -12927x + 40.298

∆Hº = [-(-12927) x 8.3145] / 1000 = 107.48 kJ

∆Sº = 40.298 x 8.3145 = 335.06 J/K

Sample Calculations:

A. Temperature from ºC to K: (table 2, test tube # 1)

63 ºC + 273.15 = 336.15 K

B. 1/T

1/ 336.15 K = 0.002975

C. Volume HCl used:

Final – Initial = Total

24.82 mL – 6.28 mL = 18.54 mL

D. mL to L

18.54 mL x   1 L   = 0.01854 L

    1000 mL

E. Moles of

()()

(0.01854 L)(0.4763 M) = 0.008831 mol

  F. Moles of

    moles / 2

  (0.008831/2) = 0.004415 mol

    G.  []

    (moles / 0.00500 L)

    (0.004415 mol / 0.00500 L) = 0.883 M

  H. ln []

  ln [0.883M] = -0.124430

  I. 3 ln []

    3 x ln[0.883M] = -0.373290

  J. ln

    ln = ln(4) + 3 ln []

    (1.386294) + (-0.373290) = 1.013004

    

Discussions and Conclusions:

By using various temperatures of borate ions, ∆Hº and ∆Sº can be calculated. Five test tubes were prepared with water while a beaker full of 30.2 g of borax is heated with water to about 70ºC. After cooling each of the five test tubes from 70 ºC to 65 ºC, 55 ºC and so on the solution was added to an Erlenmeyer flask to start the process of titration with HCl. The indicator used in these titrations was Methyl Orange. Once the indicator methyl orange went from yellow to the end point color peach the final mL amount of HCl used was recorded and compared to the initial amount to find the total amount of HCl used during the titrating of the samples. Once completed the data was used to calculate 1/T, Volume of HCl, moles of H, moles of Borax, the concentration of Borax, and the natural logarithm of the concentration of Borax and the equilibrium constant. After creating a scatterplot of ln K vs. 1/T the question used to find ∆Hº and ∆Sº was obtained. ∆Hº was calculated to be 107.48 kJ and ∆Sº was found to be 335.06 J/K.

When conducting experiments chances of error are more than likely to pop up. The first systematic error that could have occurred could have been that not all the borax crystals were dissolved in heat before starting the titration of each sample. Clear crystals can be hard to see in clear liquid. This could have affected the end point during the titration. A human error could have occurred when misjudging the color of the end point. Some could think peach is a darker color or light. Going too far over the end point would result in turning your sample in the flask pink.

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