Question

Data Sheet Collecting Data mass of KNO; weighed out 20.2g only one value needed since the grams of KNO3 do not change during the experiment Trials 2 3 4 5 1 6 25.0 29.2 33.4 37.6 41.8 46.0 total volume, ml temperature °C when crystals form in solution 67 56 47 40 37 33 Determining Ksp only one value needed since the moles of KNO, do not change during the experiment # of moles of KNO3 concentration, M=[K] = [NO3] K sp Determining AG T, K In Kg AG (J/mol) Determining AH 1/T, KP slope of the line defined by In K versus 1/T ΔΗ, J/mol Determining AS AS, J/(K mol) average AS

Fill in the data sheet and please show your work for the calculations

Thermodynamics of Potassium nitrate dissolving in water

KNO₃(s)     ↔   K⁺ (aq)   +   NO₃^-(aq

Determine the thermodynamic variables, ΔG, ΔH, and ΔS, for the dissolving of KNO₃ in water. You will find these

variables by creating a Van’T Plot, which is a graph of lnK_(sp) vs 1/T.

A set amount of KNO₃ is weighed and dissolved in 25mL of water and heated until all the KNO₃ is dissolved.
The solution is allowed to cool until crystals form, meaning it has reached the saturation point (equilibrium)
for that temperature. The temperature is recorded. The concentration of the K⁺ and NO₃ is changed by adding
5mL of water to the solution. Again, the solution is heated until all crystals dissolve and allowed to cool. When
crystals form it has reached the saturation point for that concentration, and the temperature is recorded. This
process continues for 6 trials.  

Extra information to help with the data table:

Determining K_(sp)

Find the moles of KNO₃ from the mass in the data table

Find the concentration (Molarity) of K⁺ and NO₃ using moles of KNO₃ and the volume of water for each trial.
The moles of your ions are going to be the same as KNO₃ since the reaction is a 1:1:1 ratio.

Find K_(sp). From equation 1 you can see the equilibrium expression is equal to the concentration of [K⁺]x [NO₃]
each ion concentration is the same as the concentration of KNO₃. Do this for each trial.

Determining ΔG

1. Convert the Celsius temperature to Kelvin for each trial.
2. Take the natural log of the K_(sp) above for each trial. (This data is the Y-axis for the graph
3. Calculate ΔG with the following equation. ΔG = -RT lnK_{sp ,}where R is the constant 8.314 J/(K∙ mol)
   and T is temperature in Kelvins, for each trial.

Determining ΔH

1. Find 1/T for the Kelvin temperature for each trial. (This data is the x-axis for the graph)
2. Plot the graph of lnK_(sp) vs 1/T on graph paper or in excel. Find the slope of the line and enter it in
   the data table. Use slope (m) = y₂ – y₁ / x₂ – x₁ if you plotted the graph by hand. If using excel the
   equation for the line will give the slope y = mx +b.
3. Calculate ΔH using ΔH = -R x slope, where R is the constant 8.314 J/(K∙ mol). Note there is only one
   value of ΔH for all the trials.

Determining ΔS

1. Find ΔS for each trial using ΔS = (ΔH – ΔG) / T or use ΔS = y-intercept x R, and enter on the data sheet.
2. Calculate the average ΔS for each trial and enter on the data sheet.

Answer 1

The solution is given below

determination of ksp

• molar mass value is taken as per literature value.

+ NO kt KNO3 2002 g taken Mass of KNO₃ (m) (M) こ Vol.1032 glomo) KNO3 Molar mass of = KNO3 of my moles 2 mo) 20.2 iolo1032 mol o. 199796 Note :- (*) Concentration molarity Moles of solute Volume of solution (noj [KNO3 ] = [k+ ] - Thus i kop Total (me) Temperature concentration of KNO3 (mol/L) Trials c) Volume 63.8695 67°C 7.99184 25 ml 6.84233 46-8175 56 2 29.2 ml 35.7834 47 5.98192 3 33.4 ml 40 28.2356 37.6 ml 5.31372 4.77981 37 2208466 41.8 ml U] 33 4.343391 188650 6 ml 46.0

Note that ksp [kt] [no; and Since [no] (KNC 103] ksp [KNO₃] 2 Thus Ksp for each trial has been given in table
determination of delta G value :-

• Temperature in kelvin= temperature in degree celsius +273.15 k

DG Temperature in kop AG= -RT in Kop w/mol) (in Kelvin) 1.) 340 15 401568 - 11 755.46 2) -10525 .60 329.15 3.8463 95 22.33 3.) 32015 3.5775 3:3406 8 697035 313:15 4) 3: 1288 806 7.88 5 310:15 3.9373 7476-40 6) 306 15

detwrmination of delta H :-

AH!- Temperature th in kop (T) (in Kelvin) OOO029399 340-15 4. 1568 0.0030381 3.84 63 2) 329:15 ♡ 0031235 3.5775 3.) 320-15 4.) 0031934 303406 313-15 0.0032242 310.15 31288 0.0032 664 2.9373 306.15
graph is given below,

the graph is explained in the following image

plot between In(ksp) and 1/T 4.5 y=-3665.3x + 14.974 3. 5 3 A 2.5 In (ksp)------> 2 1.5 1 - 0.5 0 0.0029 0.003 0.0031 0.0032 0.0033 1/T (in kelvin)-

determination of delta S -

3665.3 n + 14.974 = 1- E Quation Сс and У. where Ksp in co - 3665 - 3 ( as general equation to) so plope intercept y zmatb 14.974 -R ΔΗ x slope n Z - R X - 36653 AH 3047303042 Il mol

(by each trial) Trial AG AS (J/kmol) - 1175 5046 124o 147 2. 10525.GO 1240 560 3 1 24 0928 1250086 4 5 95 22 33 8697035 80670 88 7476.40 12.40 26 6 6 123 958 average AS 124.49.1 J/k mol

AS (by graph) A Sz y-intercept XR 14.974 x 8.314 Ilmol os ८ 124.494 Ilk mol