Question

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Consider the indicated step in the elimination-addition reaction depicted below. Given that the pKaHA of NH3 is 35, and assuming that the pKabH of the negatively charged aniline species is 41.3, what is Keq for this step? To make your Keg Value feasible to enter into Canvas, take the log (base 10) of your calculated Keq value and enter this result to the nearest tenth. NH2 NH2 : N The free energy (AG) for a reaction under a set of initial conditions/concentrations is -11.9 kcal/mol. The standard free energy (AG°) of this reaction is -7.8 kcal/mol. Given this information, what will be the value of the ratio of the reaction quotient (Q) to the equilibrium constant (Kea)? In other words, what will Q/Keq equal? Assume a temperature of 25 °C. The value of the gas constant is 1.987 × 10-3 kcal/K·mol. To make your answer feasible to enter in Canvas, take the log (base 10) of your Q/Keg value and enter this result to the nearest tenth

Answer 1

guesham- -35 -PKo The reaction of Proton Telease of NH is as forlos- S0, the e xpression of, Xact NH3 is, Kow LNHrtH+3 INlH,)10-35 where concentradin PKa of aniine is 413 So, Ko of anilinhe o The Yeaction of Proton reltose of oniline is as oow -41.3 NH N H2 叶 So, the exPeion of , Klamine) NHL Now, he reactisn is NH2 呠曰N 1H The etPiesslon ơf equil,'bridm consent ofthe Yea chal , NHL e9 Mulpi mumerator ond 41310

Therebre he Key of the reachion is 0 G. 3 lo neares+ ten) Free energy ơf the reaction) Δ(x 11-5 kcal/mol The Standard free energy of the女reaction ,on: -7, 8 Kcad/moL TemPerature, T225+ 273) 298 K Guas constant, R , 987 xo-3 Kcal/Kmof eKnow the expsession of free enersd of a reachion as, where K is the esu tibrium comstant ond g is the reactian guten 24 Jn (3//RT Da/RT -11. g KCal èmol-ı // 1'98구 X10_ 3 KCah Kryma-1 x298 K O1 e9 -20 2-0 6x10- ThereforeノーKe, Now, oo9 0618710