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- 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'?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 ________________________subunit G.H.I please
- Give clear handwritten answer- Derive an expression for the time dependence of the degree of polymerization (the average number of monomer residues per polymer molecule) for a condensation polymerization in which the reaction is known to have rate law d[M]/dt = -k[M] 3 , where [M] represents the concentration of the monomer!The aqueous-phase decarboxylation of CH2NO2CO2H, CH2NO2CO2H CH3NO2(aq) + CO2(g), is studied at a certain temperature, giving the following data: Time (s) [CH2NO2CO2H] (M) 0.0e+00 0.100 1.0e-01 0.0466 2.0e-01 0.0217 3.0e-01 0.0101 4.0e-01 0.00473 Figure out whether this reaction is first-order or second-order with respect to the concentration of CH2NO2CO2H. Calculate the value of the rate constant. Pick the choice from below which gives the correct reaction order and value of the rate constant for this reaction.Given the following data, determine the rate law and calculate K Experiment [NO] (M) [Cl2] (M) Rate (M/s) 1 0.0300 0.0100 3.4 x 10-4 2 0.0150 0.0100 8.5 x 10-5 3 0.0150 0.0400 3.4 x 10-4 the units on K are M-2s-1. You should enter the answer without units to 2 sig figs.
- The degradation of the antibiotic clindamycin stored at 343 K in aqueous solution at pH 4 is found to be first order with a rate constant of 2.49 x 10−7 s −1. Over the temperature range 320 K to 360 K theactivation energy was found to be 123.3 kJ mol−1. (a) Calculate the rate constant at 325 K.(b) The threshold for product safety is 1% degradation. At 295 K the time taken for 1% of the antibiotic to degrade is found to be close to 0.01/ k. Comment on the shelf-life of the drug.A dye degradation experiment was carried out and the rate law was determined to be: rate=k[A]2. The initial absorbance, [A]0, was 1.00 and the absorbance at t=60.0 s, [A]t, was 0.00125. What is the value of the rate constant, k for the experiment?The rate constant for the fi rst-order decomposition of N2O5 in the reaction 2 N2 O5(g) → 4 NO2(g) + O2(g) with v = kr[N2O5] is kr = 3.38 x 10-5 s-1 at 25 oC. What is the ha lf- life of N2O5? What w ill be the total pressure, init ial ly 78.4 kPa for the pure N2O5 vapour, (a) 5.0 s, (b) 5.0 min after init iation of the reaction?
- The rate of photodecomposition of the herbicide piclo- ram in aqueous systems was determined by exposure to sunlight for a number of days. One such experiment produced the following results. (Data from R.T. Hedlun and C.R. Youngson, “The Rates of Photodecomposition of Picloram in Aqueous Systems," Fate of Organic Pesticides in tbe Aquatic Environment, Advances in Chemistry Series, #111, American Chemical Society (1972), 159—172.) Exposure Time, t (days) [Pidoram] (mol L_1) 0 4.14 X 10-6 7 3.70 X 10-6 14 3.31 X 10-6 21 2.94 X 10~6 28 2.61 X 10~6 35 2.30 X 10-6 42 2.05 X 10-6 49 1.82 X 10"6 56 1.65 X 10-6 Determine the order of reaction, the rate constant, and the half-life for the photodecomposition of picloram.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. 75Ff.22. With explanation...