2612 Thermodynamics of KNO3 Solubility
- Page ID
- 440629
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)EXPERIMENT 2612 - THERMODYNAMICS OF KNO3 SOLUBILITY
1.0 INTRODUCTION
1.1 OBJECTIVES
- Students will interpret data on a graph.
- Students will use the data to make several calculations regarding the solubility of the salt.
- Students will graph their data to further determine the thermodynamics of the solubility of the salt.
1.2 BACKGROUND
Equilibrium constants are connected to the thermodynamics of a reaction through the equation:
ΔG0 = -RTlnK
The ΔG of the reaction, and the equilibrium constant, change with temperature.
ΔG = ΔH – TΔS
However, enthalpy and entropy change very little over a wide temperature range. We can connect the equilibrium constant, measured at different temperatures, to the enthalpy and entropy as follows:
-RTlnK = ΔG0 = ΔH0 – TΔS0
lnK = -ΔG0/RT = -ΔH0/RT+TΔS0/RT = -ΔH0/RT+ΔS0/R = (-ΔH0/R)1/T+ΔS0/R
y = m x + b
If we plot lnK for a reaction (y-axis) versus 1/T (x-axis), the slope of the line is equal to (-ΔH0/R) and the y-intercept is equal to ΔS0/R. We can then get ΔG0 either from the lnK at 298K or by solving ΔG0 = ΔH0 – TΔS0.
The reaction to be studied is the solubility of KNO3(s).
KNO3(s) ↔ K+(aq) + NO3-(aq)
Let s dissolve +s +s
K = Ksp = [K+][NO3-] = s2
The source of your data is the solubility graph below. (You will need to convert the solubility units given in the graph to molarity for the calculation of the equilibrium constant. Grams of solute per 100 grams water = grams solute per 100mLs water = grams solute per 0.100L. Just change grams of solute to moles of solute and complete the calculation.)
2.0 DIRECTIONS
- Select 5 temperatures from the graph over the range given.
- Read the solubility and convert it to appropriate units.
- Calculate Ksp, lnKsp, 1/T.
- Graph lnK versus 1/T.
- Use the slope and y-intercept to determine ΔH0, ΔS0, ΔG0.
- Answer the questions that follow.
SOLUBILITY OF KNO3 VERSUS TEMPERATURE (Data taken from IUPAC and NIST Tables)
3.0 DATA RECORDING TABLE
4.0 ANALYSIS OF DATA
Make a graph of lnK versus 1/T and paste it here. Use it to answer the questions which follow.
5.0 CONCLUSIONS
Show your work.
- What is the slope of the line?
- What is the ΔH0 of the reaction?
- What is the y-intercept of the line?
- What is the ΔS0 of the reaction?
- What is the ΔG0 of the reaction from the graph?
- What is the ΔG0 of the reaction from the ΔH0 and ΔS0?
6.0 POST-LAB QUESTION
Explain why the signs (+ or -) of the thermodynamic properties make sense.
ΔG0:
ΔH0:
ΔS0: