Thermodynamics of the Solubility of Potassium Nitrate PDF

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17th May 2017

Caitlin Bettenay

Thermodynamics of the Solubility of Potassium Nitrate

ABSTRACT: Throughout this experiment, the solubility of

KN O 3

was

determined as a function of temperature, and this data was used in order to find the Ksp for the dissolution of the

∆ G,

∆ H , and

KN O 3 . This was used to calculate

∆ S for the dissolution process. It was conducted

through finding the temperature using a temperature probe at various points when adding

KN O 3

to water. The

−128913.061 , while the ΔH was

−21597 J /mol KJ and finally, the ΔS was

33.4057 J/Kmol. The percent error of the

∆ G was calculated to be

∆ H was found to be 42.67% and

∆ S was found to be 0.93%. Based on these results, the reaction was

found to be spontaneous and an exothermic reaction.

INTRODUCTION: The solubility product constant (Ksp) indicates how soluble a compound is in a solution (generally water). The higher Ksp (>1), the more soluble the compound. When Ksp = 1, the concentrations of compound and ions are equal.

Equilibrium will be established in a saturated

KN O 3

solution in water as

can be seen from Equation 1 below: −¿ (aq) +¿ ( aq ) +N O 3¿ ¿ KN O 3 ( s ) ⇌ K

The Ksp of this reversible reaction can be defined by Equation 2 below as the concentrations of the solids do not change, so only the aqueous and gaseous phases ae considered in the Ksp equation. + ¿¿ K ¿ −¿ ¿ N O3 Ksp=¿

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17th May 2017

Caitlin Bettenay

Where

+¿¿ K is the molar concentration of ¿

is the molar concentration of As the solubility of

KN O 3

+¿(mol/ L) and where ¿ K

−¿ ¿ N O3 ¿

−¿(mol/ L) ¿ NO 3

changes when the Temperature (T) changes,

the Ksp is a function of the temperature.

Thermodynamics deals with the relations between heat and other forms of energy. Three important thermodynamic parameters Energy),

∆ H (enthalpy change) and

∆ G (Gibbs Free

∆ S (entropy change) can be used

to obtain a better understanding of the dissolving process of G can be used to determine the spontaneity of process: negative

KN O 3

KN O 3 .

∆

dissolving

∆ G means the process is spontaneous while positive

∆ G means the process is non-spontaneous.

positive if heat needs to be provided for

∆ H for

KN O 3

to dissolve

(endothermic), and negative if heat is released for water (exothermic). Finally,

∆ S for

KN O 3

KN O 3 would be

KN O 3 dissolving in

dissolving in water is always

positive as the randomness of the system increases. Table 1 below shows the relationship between these three important thermodynamic parameters and the direction of the reaction associated.

Table 1: Relationship of the signs of

∆ G,

∆ H,

∆S

and direction of

the Reaction

Several important equations will be used in order to determine these three thermodynamic parameters:

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17th May 2017

Caitlin Bettenay

Gibs’ Free Energy equation, which is a mathematical expression that relates changes in free energy, enthalpy and tropy, can be seen from Equation 3 below: ∆ G =∆ H−T ∆ S This equation can also be expressed in terms of Ksp (Equation 4): ∆ G=−RTlnKsp Through combing both Equation 3 and 4 its possible to derive an equation that relates Ksp and the temperature (K) to the values associated with ∆ H and

∆ S which can be seen through Equation 5 below:

( ∆RH )(T1 )+ ( ∆RS )

lnKsp=−

Through plotting

( ∆RH )

− of

lnKsp Vs.

( 1T )

and the y-intercept is −1

8.314 J ∙ mol ∙ K

−1

it produces a line where the slope is

( ∆RS )

where R is the ideal gas constant

. The enthalpy and entropy of dissolution can be

determined by evaluating the temperature dependence of Ksp.

Throughout this experiment, the solubility of

KN O 3

will be determined

as a function of temperature, and this data will be used in order to find the Ksp for the dissolution of calculate the

∆ G,

KN O 3

∆ H , and

and will furthermore be used to ∆ S for the dissolution process. This will

be done through finding the temperature using a temperature probe at various points when adding

KN O 3

to water.

DATA: 1. Concentration: Concentration=

mass ( g ) volume ( L ) 1 mol 2.007 g × Concentration= 0.00289 L 101.11 ( molar mass KN O 3) ThereforeConcentration =6.85 M

2. Ksp Value:

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17th May 2017

Caitlin Bettenay −¿ ( aq ) ¿ +¿ ( aq ) +N O 3 KN O 3 ( s ) ⇌ K ¿ +¿ ¿ K ¿ −¿ ¿ KNO 3 Ksp=¿

2

Concentration ¿ Ksp=¿

2

6.85 M ¿ So , Ksp=¿ 3.

∴ Ksp=46.9225

∆ G:

∆ G=−RTlnKsp Where R=8.314 J ∙ mol−1 ∙ K −1∧T =Temperature ∈ Kelvin So , ∆G =−8.314 × 330.45 K × 46.9225 4.

∆ H: Figure 1 shows that the slope (m) value can be used to

determine the

∆ H. This is an exothermic reaction, it can be seen that

∆ H is negative.

∆ H=m × R

5.

∆ G=−128913.061

∆ H=(−0.2598 ×1000)(8.314 )

∆ S: From Figure 1, the

∆ H =−21597 J /mol

∆ S can be determined through the c

value in y=mx + c ∆ S=b × R

∆ S=4.018 ×8.314

∆ S=33.4057 J/Kmol

6. Percent Error Calculations (absolute value): ∆ H Percent Error=

theoretical− experimental ×% theoretical

∆ H Percent Error=

(−494.6 )− (−2 1597 ) (−494.6 )

∆ H Percent Error=¿ 42.67%

∆ S Percent Error=

theoretical− experimental ×% theoretical

∆ S Percent Error=

(−494.6 )− (−33.4097 ) (− 494.6 )

∆ S Percent Error=¿ 0.93%

Table 2: 4

17th May 2017

Caitlin Bettenay Temp (K): 330.45

Ksp:

lnKsp:

¿ mols=

2.007 g =0.0198 101.11 g /mol

[ KN O 3] =

0.0198 mol =6.85 2.89 ×10−3

ln ( Ksp)

46.9225 ¿ ln ¿ ¿ 3.8485

2

316.95 310.65 306.35 303.35

6.85 ¿ =46.9225 2 Concentration¿ =¿ ¿ 30.3601 24.2064 19.6249 15.9201

3.4131 3.1866 2.9768 2.7676

Figure 1: Line Graph Depicting the 1/T Vs. lnKsp

DISCUSSION: Throughout this experiment, the solubility of

KN O 3

was

determined as a function of temperature, and this data was used to find the Ksp for the dissolution of the

∆ G,

∆ H , and

KN O 3 , which was then used to calculate

∆ S for the dissolution process. This was done

through finding the temperature using a temperature probe at various points when adding

KN O 3 to water. As can be seen from the Data

section of this report, the was

∆ G was found to be

−21597 J /mol , and the

∆ S was 33.4057

−128913.061 , the

∆ H

J/Kmol. From this, the 5

17th May 2017

Caitlin Bettenay percent error of

∆ H was found to be 42.67% and the

∆ S percent error

was found to be 0.93%. According to these values and to Table 1 found in the introduction of this report, the dissolution of reaction at all temperatures. This is due to the meant it was a spontaneous process, and meant that heat was released when

KN O 3

(exothermic reaction) while finally, the value, and the higher the

KN O 3 is a spontaneous

∆ G being negative which

∆ H was also negative which was dissolved in water

∆ S was found to be a positive

∆ S value, the more the randomness of the

reaction.

The relatively high percent error could have been due to eye-sight methods such as reading the liquid volume from the graduated cylinder and recording the data when first crystals appear, thus the resulted values would not be precise. In future testing it would seem appropriate to take the greatest care when measuring so as to ensure the most accurate readings are gained as possible. It would also seem appropriate to do more testing so that averages can be gathered so as to minimize the overall percent error. Although various errors were recorded during the experiment, they did not affect the trend lines of the data thus, the method was sound and the purpose was achieved.

QUESTIONS: 1. Yes the reaction was expected to be spontaneous, as the found to be negative, the the

∆ G was

∆ H was also found to be negative, and

∆ S was found to be positive which is affluent with table 2

found in the introduction of this report and agrees with the points raised in the discussion. 2. This process was exothermic which does match the calculated enthalpy value as the

∆ H was a negative value which coincides

with an exothermic reaction 3. This process was expected to result in an increase in disorder as there was a positive

∆ S and the reaction was spontaneous.

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Caitlin Bettenay

17th May 2017

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