An enzyme-catalyzed reaction has a Km of 1 mM and a Vmax of 4 nmole/L sec-1. The reaction velocity (nmole/L sec-1) when the substrate concentration is 0.25 mM is: A. 1.25 B. 10.0 C. 5.0 D. 0.50 E. 1.0 F. 100 G. 150 H. 55 I. 75
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- An enzyme-catalyzed reaction has a Km of 1 mM and a Vmax of 4 nmole/L sec-1. The reaction velocity (nmole/L sec-1) when the substrate concentration is 0.25 mM is:
A. 1.25
B. 10.0
C. 5.0
D. 0.50
E. 1.0
F. 100
G. 150
H. 55
I. 75
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- The color change accompanying the reaction of phenolphthalein with strong base is illustrated below. The change in concentration of the dye can be followed by spectrophotometry (Section 4.9), and some data collected by that approach are given below. The initial concentrations were [phenolphthalein] = 0.0050 mol/L and [OH] = 0.61 mol/L. (Data are taken from review materials for kinetics at chemed.chem.purdue.edu.) (For more details on this reaction see L Nicholson, Journal of Chemical Education, Vol. 66, p. 725, 1989.) (a) Plot the data above as [phenolphthalein] versus time, and determine the average rate from t = 0 to t = 15 seconds and from t = 100 seconds to t = 125 seconds. Does the rate change? If so, why? (b) Use a graphical method to determine the order of the reaction with respect to phenolphthalein. Write the rate law, and determine the rate constant. (c) What is the half-life for the reaction?Instantaneous rates for the reaction of hydroxide ion with Cv+ can be determined from the slope of the curve in Figure 11.3 at various concentrations. They are (1) At 4.0 105 mol/L, rate = 12.3 107 mol L1 s1 (2) At 3.0 105 mol/L, rate = 9.25 107 mol L1 s1 (3) At 2.0 105 mol/L, rate = 6.16 107 mol L1 s1 (4) At 1.5 105 mol/L, rate = 4.60 107 mol L1 s1 (5) At 1.0 105 mol/L, rate = 3.09 107 mol L1 s1 (a) What is the relationship between the rates in (1) and (3)? Between (2) and (4)? Between (3) and (5)? (b) What is the relationship between the concentrations in each of these cases? (c) Is the rate of the reaction proportional to the concentration of Cv+? Explain your answer.An enzyme-catalyzed reaction has a Km of 1.4 mM and a Vmax of 7 nM/s. What is the initial velocity when the substrate concentration is 0.6 mM?
- An enzyme-catalyzed reaction is studied in the absence and presence of an inhibitor and the following data was obtained. [S] in mmoles/L Velocity in mmoles/L/min-1 No inhibitor With inhibitor 1.25 1.72 0.98 1.67 2.04 1.17 2.50 2.63 1.47 5.00 3.33 1.96 10.00 4.17 2.38 Calculate the Km of the enzyme in the reaction without inhibitor ________________________ Km’ of the enzyme in the reaction with inhibitor ________________________ Vmax of the uninhibited reaction ________________________ Vmax’ of the inhibited reaction ________________________The enzyme-catalysed conversion of a substrate at 25 oC has a Michaelis constant of 0.015 mol dm-3 . and a max-imum velocity of 4.25 x 10-4 mol dm-3 s-1 when the enzymeconcentration is 3.60 x 10-9 mol dm-3. Calculate koat and the catalytic efficiency η. Is the enzyme 'catalytica lly perfect'?The kinetics of an enzyme are measured as a function of [S] in the presence and absence of 2 mM I. Compute Km and Vmax in the absence and presence of I S] (uM) V (uM/min) without I with I 3 10.4 4.1 5 14.5 6.4 10 22.5 11.3 30 33.8 22.6 90 40.5 33.8
- Benzene on graphite follows the Langmuir isotherm. At a pressure of 1 atm, theThe volume of benzene adsorbed on a sample of graphite was 4.2 mm3 atconditions T = 0°C and 1 atm. At 3 atm it was 8.5 mm3. Suppose 1 molecule ofbenzene occupies 30Å and estimate the surface area of graphite.(a) Most commercial heterogeneous catalysts are extremelyfinely divided solid materials. Why is particle sizeimportant? (b) What role does adsorption play in the actionof a heterogeneous catalyst?An enzyme kinetics experiment is carried out by adding 1.00mg of a 50.0kDa enzyme into a total volume of 50uL of buffer, and measuring the initial rates while adding increasing amounts of substrate (1 Da = 1 g/mol). The calculated Vmax from this series of experiments was 30.5mM/s. Calculate the turnover number for the enzyme.
- A An enzyme that follows Michaelis-Menten kinetics has a KM value of 16.0 uM and a kcat value of 181 s-1. At an initial enzyme concentration of 0.0100 uM, the initial reaction velocity was found to be 1.07 x 10- uM/s. What was the initial concentration of the substrate, [S], used in the reaction ? Express your answer in micromolar to three significant figuresIn the fluorination reaction of butadiene, the butadiene is adsorbed onto a catalyst C.To test the efficiency of the adsorption, the adsorption isotherm was determined.The volume of butadiene per gram of C varied as follows at constant temperature:p( kPa)- 13.3; 26.7 ; 40.0 ; 53.3 ; 66.7 ; 80.0V(cm3)-17.9 ; 33.0 ; 47.0 ; 60.8 ;75.3 ; 91.3At the temperature used, p* for butadiene is 200 kPa. Find a suitable isotherm for the data and determine Vmon. please assist.For the 2° reaction H2(g) + I2(g) → 2HI(g) at 650 K, the collisional cross section, sigma H2-I2, is 0.36 nm2, the reducedmass is 3.32 × 10−27 kg, and the activation energy is 171 kJ mol−1. Use collision theory to calculate the rateconstant. If the actual value of the rate constant is 8.4 × 10−6 M-1s-1, determine the steric factor for this reaction.
