Question

The interconversion of L-α-lysine and L-β-lysine, for which at Kc=7.20 at 333 K, is catalyzed by the enzyme lysine 2,3-aminomutase. At 333 K, a solution of L-α-lysine at a concentration of 3.90×10−3 M is placed in contact with lysine 2,3-aminomutase. (Figure 1) Part A What are the equilibrium concentrations of L-α-lysine? Express your answer to two significant figures and include the appropriate units. [L- α α lysine] = Previous AnswersRequest Answer Incorrect; Try Again; 5 attempts remaining Part B What are the equilibrium concentrations of L- β lysine? Express your answer to three significant figures and include the appropriate units. [L- β β lysine] = Previous AnswersRequest Answer Incorrect; Try Again; 5 attempts remaining Provide Feedback Figure1 of 1

Answer 1

The reaction is

L-$\alpha$- lysine $\rightleftharpoons$ L-$\beta$-lysine.

ICE table is

	[L-$\alpha$-lysine]	[L-$\beta$-lysine]
Initial	3.90*10-3	0
Change	-x	+x
Equilibrium	(3.90*10-3 -x)	x

Now, Equilibrium constant Kc = [L-$\beta$-lysine]/[L-$\alpha$-lysine]

Using the concentration at equilibrium

Kc = (x)/(3.90*10^-3 -x)

Given , value of Kc = 7.20 ( at 333K)

Then,

7.20 = (x)/ (0.0039-x)

Or, x = 7.20*(0.0039 -x)

Or, x = 0.02808 - 7.20x

Or, 8.20 x = 0.02808

Or, x = 0.00342 = 3.42*10^-3 M.

Then, equilibrium concentration of L-$\alpha$-lysine

= (0.0039 - 0.00342) = 0.000480 M = 4.80 *10^-4 M.

And, equilibrium concentration of L-$\beta$-lysine = 3.42*10^-3 M.

Now, calculated Kc = (3.42*10^-3)/(4.80*10^-4) = 7.125

As calculated Kc is close to given Kc value , hence answers are valid .