Question

Consider the following reaction at 298 ! Mg () + 2H+ (aq) + 2Cl(a) = Mg²+ (aq) + 2Cl(aq) + H2 (8) Table 3: Activity Coefficients and Concentrations, Table 3 shows the activity coefficients and the amounts of all species when the not at equilibrium [] equilibrium reaction is not at equilibrium and when the Mg2+ (og) 0.690 0.0055 M 0.0012 M reaction is at equilibrium. H ) 0.914 0.035 M 0.018 M Cl")_ 0.899 0.023 M 0.023 M H2 (c) 1.00 0.756 bar 0.634 bar . When not at equilibrium, which direction will the reaction proceed to reach equilibrium? - Is the reaction spontaneous or nonspontaneous in the forward direction? (Prove by calculation.) - Draw (by hand is fine) a plot of the Gibbs Free Energy as a function of composition of the reaction mixture. Indicate what the slope is equal to, when the slope is zero, positive, and negative, how the chemical potentials of the reactants and products compare for each sign of the slope, and how Q and K compare. Also, indicate the region of the curve the reaction is at for Part A.

Answer 1

Question is solved using basic concepts of reaction spontaneity.

Solution - - The given seaction equation is – Mgcs) + 24 tag) +2679) = mogle ing) Ezcccam) tuca) Henul section quotient can be easitter as be anger x lax am a z Ving+x {mg24] x H₂X [ H2CG) ] (as azymalality hox [At])² Lazy [consunday 0.69 0 X 00055 X X .756 (Mac)] z 3 do know PV z nRT na l [ RT Ibarzo. g8oth 0-756 Xo. 98 atm 0.0821X atualitey X 298 Keleon male keluth [109]] z o.o30rok lite Hence putting in the above we get- 0-690 XOOOSS X X 0.030 (0.914 80.035)

DATE: V PAGE NO.: 1.1388 109 (3.19x102) 2 1.138 x et 10-17X10% Q z 0:11189 Similary we can calculate equilibacium constant Kz Qat equilibelum Ka hught X [mph] X 42 X [429] - [harx [44]) ? [H229)] at equilibrium - Th2 (9)] P 0-634X0.98 after 0.0821 ath-litey x 298 keleita male-keleth [H₂ Col] = 0.0253 male liter 0.690 X0-0012 X 1X 0.0253 - 10:912X0.018)2___ K, 2.094 X10-5 2.099X1055 (GSX102) ² 2.728109 = 7.69167-544 27.698 102 KZ00769

DATE: 11 PAGE NO.: should go because K Hence the reaction backward face attaining equilibcune we can calculate AG for forward direction as o aciz - RTln keq ACY = -0.0824 X 298 X 2.303 logro keq AGO 2 -0.0821 x atm-liter x 298 kaloin 2-303.09.2007 mole-Keltoin atm-liten AG - -0.0821 X 298 X2:303X -4 116) atu mole AG = 62.88 atm-liten mate Ditve Hence in forward is not spantaneous . direction reaction

0 PAGE NO.: and reaction The graph between composition can be gibbe free energy drawn as free energy harkA) Q3K QCK_ pon reactant pori product equilibrium mixtury QzK, AGRO V A equilibrium Blope 18 wo before equilibrium elope 18 negative and after equilibrium slope is positive We know that - chemical potential - DAG z slope ani when Qck ie before equilibrium slope is negative that means composition of reactant decrease and product indwases Blously a 7 at Azk, composition of seadant z composition of product at this point slop zo that means chemical potential = 0 → at azk Composition of product jucuseases -