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Chemistry 1060 Exam 2 Practice: Thermodynamics and Equilibria - Prof. Michael S. Sommer, Exams of Chemistry

Practice questions for exam 2 in chemistry 1060, spring 2009. The questions cover topics such as entropy changes during phase transitions, gibbs free energy, entropy change during isothermal mixing, energy levels and configurations of particles, and industrial chemical reactions. Students are required to use given data and equations to calculate answers.

Typology: Exams

Pre 2010

Uploaded on 08/18/2009

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Chemistry*1060*–*Spring*2009*
Exam%2,%PRACTICE%
1*of*3*pages* * mssommer*
Useful Information and Equations:
dG =dH d(TS)
ΔG=ΔHTΔS
K=
(aj,eq )
ν
j
j
prods
(ak,eq )
ν
k
k
reacts
=
(aj
ν
jx)
ν
j
j
prods
(ak
ν
kx)
ν
k
k
reacts
G=G˚+RT ln(a)
ΔG˚=ΔH˚TΔS˚=RT lnK
ΔG=ΔG˚+RT ln(Q)
S=kBlnW
ε
=wnet
qin
=qin +qout
qin
=Thigh Tlow
Thigh
dS d q /T
W=N!
(nj)!
j
ΔS=ntotalR X iln(Xi)
i
dS d q /T
T(K)=tC)+273.15
4.184J=1cal
1mL =1cm3
1atm =760torr(mmHg)
R=1.987 cal
mol K
R=8.314 J
molK
kB=R/NA=1.38 ×1023 J
K
R=0.082 Latm
molK
101.3J=1Latm
ΔHrxn =Hproducts Hreactants
1J=1kg m2/s2
Xi=ni
nj
j
ΔHrxn =biDi
i=react
bonds
breaking
bjDj
j=prod
bonds
forming
+ΔHphase
change
dH =dE +d(PV )
P
i=XiP
total
ΔHrxn =
ν
iΔH˚i,f
i=prod
ν
jΔH˚j,f
j=react
pf3

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Download Chemistry 1060 Exam 2 Practice: Thermodynamics and Equilibria - Prof. Michael S. Sommer and more Exams Chemistry in PDF only on Docsity!

Exam 2, PRACTICE Useful Information and Equations: € dG = dHd ( TS ) €

Δ G = Δ H − T Δ S

K =

( a (^) j , eq ) ν (^) j j prods

( ak , eq )^ ν^ k k reacts

( a (^) j − ν (^) j x ) ν (^) j j prods

( ak − ν (^) k x )^ ν^ k k reacts

G = G ˚+ RT ln( a ) €

Q =

( a (^) j ) ν (^) j j prods

( ak )^ ν^ k k reacts

Δ G ˚= Δ H ˚− T Δ S ˚= − RT ln K € Δ G = Δ G ˚+ RT ln( Q ) € S = kB ln W € ε = − wnet qin

qin + qout qin

ThighTlow ThighdSd q / T

W =

N!

( n (^) j )! j

Δ S = − ntotal R Xi ln( Xi ) i

dSd q / T

T ( K ) = tC ) + 273. € 4.184 J = 1 cal € 1 mL = 1 cm 3 € 1 atm = 760 torr ( mmHg ) € R = 1.987 (^) molcalKR = 8.314 (^) molJKkB = R / NA = 1.38 × 10 − (^23) J KR = 0.082 (^) molL^ • atmK € 101.3 J = 1 Latm € Δ Hrxn = HproductsHreactants € 1 J = 1 kgm^2 / s^2 € Xi = ni n (^) j

∑ j

Δ Hrxn = biDi i = react bonds breaking

∑ −^ b^ j Dj

j = prod bonds forming

∑ +^ Δ Hphase

changedH = dE + d ( PV ) € Pi = XiPtotal € Δ Hrxn = ν (^) i Δ H ˚ i , f i = prod

∑ −^ ν^ j Δ H ˚^ j , f

j = react

Exam 2, PRACTICE PART 1 P1. As long as the temperature of the system is uniform, any phase change is reversible. What is the entropy change for the freezing of 2.50 moles of water? NOTE : Δ H ˚ for the process H 2 O( l ) → H 2 O( s ) is – 1.436 kcal/mol. P2. What is ∆ G ˚ at T = 303 K for the reaction N 2 ( g ) + 3H 2 ( g ) → 2NH 3 ( g )? NOTE : The ∆ H ˚f,298 for NH 3 ( g ) is – 11.0 kcal/mol and you will need the following data: compound S ˚ 298 (cal/K mol) H 2 ( g ) 31. N 2 ( g ) 45. NH 3 ( g ) 46. P3. The entropy change for reversible, isothermal mixing is € Δ S = − ntotal R Xi ln( Xi ) i

Calculate Δ S for the mixing of 0.75 mol of Ne and 0.50 mol of H 2. P4. Consider a sample of N = 10 molecules distributed among four non-degenerate energy levels with energies of 0, ε 0 , 2 ε 0 , 3 ε 0. The available energy E is 5 ε 0. How many unique configurations are possible? What is the number of arrangements ( W ) for each configuration? Which is the most probable configuration? It will help to draw the configurations, being mindful of N and E. P5. Suppose we have obtained the box of particle-in-a-box fame. Further, we find that there are four particles in this box with exactly 34 ε of total energy, where ε is the energy of the ground state of the box. Recalling that the particle-in-a-box energy levels increase as n^2 (that is, ε , 4 ε , 9 ε , …), what is the number of arrangements that are possible for these four particles?