Question

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7. (18 points) Consider the reaction: Nia (aq) +H.000) = OH(aq) + HNla(aq) K = 6.25x101 @25°C SHOW ALL WORK Note: HNia weak acid Niacin (COHON, one of the B vitamins) AG = A. (6 points) What is AG"(KI) for this reaction ſhow your work. --RTink B. (12 points) What are the equilibrium concentrations (M) of each species in a solution that was prepared by mixing 0.157 moles of HNia and 0.315 moles of Nia- in a 1.000 liter volumetric flask and diluted to the mark with water. No other source of OH was added (ie, the initial concentration of OH is 0). -Your work MUST include the expression for K, the ICE table and a complete final check. Use the approximation method [x is small) You MUST include a statement of the approximation and demonstratie its validity. No quadratic equation answers will be accepted. Reaction: Initial Change Equilibrium Calculated Equilibrium Values FINAL ANSWERS

Answer 1

a)

AG = -RTIn(K) = -8.314(273+25) In (6.25 x 10-10) = -52.5 kJ/mol

b)

The ICE table will be

	[Nia-]	[H2O]	[OH-]	[Nia]
Initial	0.315	-	0	0.157
Change	-x		+x	+x
Equilibrium	0.315 - x		x	0.157+x
Calculated Equilibrium Values	approx (0.315)		1.254 * 10^(-9)	approx(0.157)

K = Products] [Reactants [Nia][OH-] Nia-1 (0.157 +r) (0.315 - 2) = 6.25 x 10-10

Assuming x is very small in comparison to the value of 0.157 and 0.315, we can write

r = 6.25 x 10-10 X 10" X 0.315 0.157 = 1.254 x 10-9

Hence the equilibrium value of x obtained is very small, as assumed before start of reaction.

Note - Post any doubts/queries in comments section.