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ENGLISH Final: Part 1

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ustaad
21 Nov 2021
School
University of Kashmir
Department
Humanities
Course
ENGLISH
Professor

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Document Summary

R-c-al basic u uuclespila) upplies e pai e ho, r h, ha all omc (taame maallic cemp@llmds. Cabocati on may 0aauge ezpavsi bn waiy take plad. Y kng meo stalle caulo ecatissu easiy me d uote hodl loe ma 9oacm. The ala e sn 9aac han tat s sn i r. Sa sabgth imibrlai x ctotieu takeo plo e2. bit imigohaiou io meo thau60. Reactamt &aiu sn ae (2 diasteskasamc ( racomic mi sctuea siie o haae. ) r: r/s, sn hausima r-canthe psceduct (2)s. Woe dees ymot qine s"2 ax l hall do edzul hallda. Viy eh,= chex beuzy of 9ives sn" 9caction mgood yioldlgenasall) Bhhzgl alide (rchafc ( m bo td ,2 ) halide tfk. 3 coplde &2 padial o bevds stuctwro cp (meluding an. Compto bend a l c a sn rea. Ne 9c0a9angemoib ditact ly tlack eu q ustata ot la0b caimd qeaupllildooes m iwseiou. Lo l i oo lod" aldan gnsosiou keasde atback eactty, csgct.

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Related Questions

For each complex indicate the geometry, coordination number, oxidation state of the metal, dn count, and total electron count.

For an ideal solution, we discussed in class that the chemical potential is given by: μ=μ_0+RT "ln" (c/c_0 ) where the units of µ written in this form are Joules/mole. The concentration c_0 is a reference concentration, which we will take to be 1 Molar for this problem. Suppose solution 1 with solute concentration of 1 mM (i.e., 1 x 10-3 Molar) is in contact with solution 2 with solute concentration of 0.1 mM (i.e., 1 x 10-4 Molar). If a differentially small amount of solute, dn, moves from the higher concentration to the lower concentration, the differential change in free energy is calculated as follows: dG = [-((∂G_1)/dn)_(P,T) dn + ((∂G_2)/dn)_(P,T) dn] where G_1 is the free energy of solution 1 (the higher concentration solution), G_2 is the free energy of solution 2 (the lower concentration solution), and the relative signs indicate that the number of moles of solute in solution 1 is decreased by dn, while the number of moles of solute in solution 2 is increased by dn. Recalling the definition of chemical potential, we can write: dG = [-μ_1 dn + μ_2 dn] = [μ_2-μ_1 ] dn

2a. Show that for the case described above dG = [RT "ln" (c_2/c_1 )] dn and insert the values of the high (c_1) and low (c_2) concentrations to confirm that, indeed, dG < 0 for transfer of solute from a high concentration solution to a low concentration solution. This result confirms that such transfer will be a spontaneous process if the solutions are placed in contact with one another.

2b. Without worrying about how it might be done, at T = 298 K, by what amount (in kJ/mole) would the chemical potential of the solute in the low concentration solution need to be increased in order for the system to be at equilibrium, even though the two solutions remain at their different concentrations of c_1=1 "mM" and c_2=0.1 "mM" ? (Note: In the case where the solute is an ion, and therefore charged, the most common way of changing the chemical potential of one solution versus the other is to apply a different electrical potential (in Volts) to one solution versus the other. Depending on the signs of the charge on the ion and of the electrical potential, the free energy of a charged solute can be either increased or decreased. This situation of maintaining different ionic concentrations in equilibrium is common in electrochemistry and in living cells.)