Equilibrium constants aren't changed if you add (orchange) a catalyst. Since at equilibrium forward and backward rates are equal: ... Adding a catalyst will affect both the forward reaction and the reverse reaction in the same way and will not have an effect on the equilibrium constant. The catalyst will speed up both reactions thereby increasing the speed at which equilibrium is reached.he equilibrium constant never depends on the type of catalyst used, no matter the nature of the reaction (homogeneous or heterogeneous). The equilibrium constant is a thermodynamic variable that is linked to the free Gibbs energy, and also varies with the temperature according to the Van't Hoff Equation. Catalysts only speed up the reaction rates; they don't have any effect on the equilibrium of the reactions.
However an adsorbent or a membrane may be able to selectively remove some of the reactor product and allow the reaction to continue to equilibrium on one side of the membrane and "beyond equilibrium" on the other side.
Explain the concept/meaning of kcat, KM and kcat/KM.
Question 15: Let n 〉 r 〉 1 . Prove that any n-vertex graph of minimum degree more than n -n/r contains Kr+1 without using Turán's Theorem.
Question 15: Let n 〉 r 〉 1 . Prove that any n-vertex graph of minimum degree more than n -n/r contains Kr+1 without using Turán's Theorem.
Below are the Kcat and Km values for 5 hypothetical enzymes. Which enzyme would have the largest specificity constant? Enzyme A - Kcat = 2.6 x 103 , Km = 4 x 10-4 Enzyme B - Kcat = 9 x 102 , Km = 6.1 x 10-6 Enzyme C - Kcat = 1.7 x 10-3 , Km = 3.2 x 105 Enzyme D - Kcat = 2.6 x 10-7 , Km = 4 x 104 Enzyme E - Kcat =...
Please calculate the kcat/Km and k (with correct units) for P1 and P2. Aldehyde Dehydrogenase P1 (person 1) P2 (person 2) kcat (s^-1) 6 0.3 Km (M) 0.0002 0.006 kcat/Km ? ? k ? ? [S] (M) 0.002 0.002
Please calculate the kcat/Km and k with correct units for person 1 (P1) and person 2 (P2). Use the Michaelis-Menten Equation Aldehyde Dehydrogenase Person 1 Person 2 kcat (S-1) 6.0 0.3 Km (M) 0.0002 0.006 Kcat/Km ? ? k ? ? [S] (M) 0.002 0.002
Km represents the.... formation of EP the back reaction of product to ES formation of ES kcat
Km represents the.... formation of EP the back reaction of product to ES formation of ES kcat
Q3 (Prove that P∞ k=1 1/kr < ∞ if r > 1) . Let
f : (0,∞) → R be a twice differentiable function with f ''(x) ≥ 0
for all x ∈ (0,∞).
(a) Show that f '(k) ≤ f(k + 1) − f(k) ≤ f '(k + 1) for all k ∈
N.
(b) Use (a), show that Xn−1 k=1 f '(k) ≤ f(n) − f(1) ≤ Xn k=2 f
'(k).
(c) Let r > 1. By finding...
In a certain reaction, a Kcat of 60 min-1 was observed, with a Vmax of 10 mMs-1. What is [E]?
Values for an Enzyme and a substrate are: Km=4 uM and kcat=20/min. For an experiment where [S]=6mM, it was found that Vo=480nM/min. What was the enzyme concentration? Give your answer in nM. Using the same kcat and Km as the previous question, if [Et]=0.5uM gives a Vo=5 uM/min, what wat the [S]? Give your answer in uM. reaction is run with the kcat=20/min and the Km=4uM. Use the enzyme concentration from question 1 above. A very strong inhibitor is added creating...
Does increasing enzyme concentration changes the value of Vmax, Km and Kcat? Why?