Under which of the following conditions does an enzyme work at its maximum theoretical velocity? the reaction rate is first-order. the substrate concentration is equal to Km. the substrate concentration is much lower than Km. all of the enzyme exists as the enzyme-substrate complex. the temperature is close to 100 °C.

The protein catalase catalyzes the reaction 2H,0,(aq) – 2 H,0O(0) + 0,(g) and has a Michaelis–Menten constant of KM = 25 mM and a turnover number of 4.0 × 107 s-1. The total enzyme concentration is 0.016 µM and the initial substrate concentration is 6.58 µM. Catalase has a single active site. Calculate the value of Rmax (often written as Vmax ) for this enzyme. Rmax mM-s- Calculate the initial rate, R (often written as Vo), of this reaction. R = mM-s-

2. Imagine you run a reaction whose free energy of activation (AG*) is 18 kcal/mol. If it were a first order reaction, its’ half life (the time for half the starting material to go away) or half the product to form (assuming a 100% yield) would by In2/k where k is the rate constant. How long will it take for the reaction to be done if you run the reaction in an ice-salt bath at room temperature (20 °C) if the rate constant were 1 x 104/sec?

Chlorine dioxide, ClO2, is a reddish-yellow gas that is soluble in water. In basic solution it gives Cl03 and C102 ions. 2C102 (aq) + 20H (aq) → C1O3 (aq) + ClO2 (aq) + H2O(1) To obtain the rate law for this reaction, the following experiments were run and, for each, the initial rate of reaction of C1O2 was determined. Initial Concentration Initial Concentration Initial Rate of Cl02 (mol/L) of OH (mol/L) (mol/(L·s)) Exp. 1 3.9 x 10-2 6.0 x 10-2 2.09 x 10-2 Exp. 2 1.3 x 10-2 6.0 × 10–2 2.32 x 10-3 Exp. 3 1.3 x 10-2 0.18 6.97 x 10-3