A biological molecule undergoes a reversible interconversion between its trans a

Question

A biological molecule undergoes a reversible interconversion between its trans and cis forms (trans ß à cis) with first- order reactions rate constants k1 (reverse). The following kinetic results were obtained starting with 10 mM of the pure trans form until equilibrium was achieved at t = infinity.

Time (t)    (s)      0                 200      600      1000    infinity

[trans]     (mM)    10.0              6.71     3.42     2.16     1.38

Determine k1, k-1, and the equilibrium constant Keq from these data.

Explanation / Answer

ANSWER:

(i) Value of K
Let
A = trans
B = cis
k1 = kf
k-1=kr
forward and reverse reactions are both first order:
A ->B; rf = -d[A]/dt = kf*[A]
B ->A; rr = -d[B]/dt = +d[A]/dt= kr*[B]
K = kf/kr = [B]eq/[A]eq

[A]0 = 10.0 mM
[A]eq = 1.38 mM
[B]eq = 10-1.38=8.62 mM

K = kf/kr = [B]eq/[A]eq = 8.62/1.38 = 6.25

Value of kf and kr
(ii)
dA = [A] - [A]eq
dB = [B] - [B]eq
dA = -dB
dA0 = [A]0 -[ A]eq = 10-1.38 = 8.62 mM

rdA = rr - rf = kr*[dB] - kf*[dA] = - (kf+kr)*[dA]
This follows first order rate law with the rate constant (kf+kr)

So, Log(dA0/dA) = (kf+kr)*t/2.303
Log(8.62/dA) = (kf+kr)*t/2.303

Time (t)    (s)                 0                 200      600      1000    infinity
[A]     (mM)                 10.0            6.71     3.42     2.16     1.38
[A]eq = 1.38 mM
dA =[A]-[A]eq (mM)     8.62          5.33     2.04    0.78       0

From first two data of time, t = 200s

Log(8.62/dA2) - Log(8.62/dA1) = (kf+kr)*200/2.303

Log(dA1/dA2) = (kf+kr)*200/2.303

(kf+kr) = 2.303/200*Log(dA1/dA2) = 2.303/200*Log(8.62/5.33) = 0.0024

kf = kr*6.25
(kf+kr) = = 0.0024
( kr*6.25+kr) = 0.0024
k-1 = kr = 0.0003 s-1
k1 = kf = kr*6.25 = 0.0021 s-1

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