1.- An enzyme acts as a catalyst for a biology reaction and accelerates the A R reaction, the values of the kinetic parameters are: Vmax = 5.7 mM I-1 min-1 a Km of 6X10-2 mM. It was tested with a substrate concentration (Ca0) of 3.1 M. Using flow reactors it is desired to have a substrate conversion of 95% given a feed of 3.5 l / min. Use the following formula to solve the problem -ra = Vmax. Ca / Km + Ca;
Remember that the integral is τ = Km / Vmax ln (Ca0 / Ca) + (Ca0-Ca) / Vmax
a) Calculate the volume of a piston flow reactor to obtain the desired conversion.
b) Calculate the volume of a mixed flow reactor to obtain the desired conversion
c) Calculate the volume of the reactors if Km changes according to the following table
Km (Mm) = 2.90X10-2; 290X10-1; 3.6; 40; 100 VFP VFM
d) Explain the reason for the change in the volume of the FM and FP reactors.
Dear user
Please upload question in relevant discipline as this is not the part of chemical engineering section.
Thank you
Chegg SME
Wish you all the best buddy ??
1.- An enzyme acts as a catalyst for a biology reaction and accelerates the A R...
CHEM3250 Assignment-Enzyme Inhibition Consider the data below for an enzyme catalyzed reaction. The rate of the reaction has been determined with and without an inhibitor. A total concentration of enzyme of 20 uM was used in the experiment. SHOW WORK AND UNITS!!! Without Inhibitor With Inhibitor [substrate] (mM)Rate of formation of te of formation of product product (mM/min) mM/min) 6.67 5.25 0.49 7.04 38.91 1.0 2.2 6.9 41.8 44.0 1.5 3.5 1 a) On the same graph, plot the data...
Question 1: Design of isothermal reactors 30 Marks The irreversible, gas-phase reaction A+B D is to be carried out in an isotherma °C) plug-flow reactor (PFR) at 5.0 atm. The mole fractions of the feed streams are A 0 B 0.50, and inerts 0.30. The activation energy for the above reaction is 80 000 cal/mol. the pressure drop due to fluid friction in the reactor is so small that it can be ignored, perform the following tasks: 2T a s...
An enzyme catalyzes the reaction M N. The enzyme is present at a concentration of 3.5 nM, and the Vmax is 2.9 PM s-. The Km for substrate M is 6.5 MM. Calculate kcat kcat = 1 What values of Vmax and Km would be observed in the presence of sufficient amounts of an uncompetitive inhibitor to generate an a' of 1.3? apparent Vmax = UM S-1 apparent Km = UM
An enzyme catalyzes a reaction with a Km of 9.00 mM and a Vmax of 3.95 mM·s–1. Calculate the reaction velocity, v0, for the following substrate concentrations: A. 1 mM B. 9 mM C. 11 mM
An enzyme catalyzes a reaction with a Km of 7.00 mM and a Vmax of 4.00 mM·s–1. Calculate the reaction velocity, v0, for the following substrate concentrations. A) 1.25 mM B) 7.00 mM C) 10.0 mM
An enzyme catalyzes a reaction with a Km of 9.50 and a Vmax of
1.75
An enzyme catalyzes a reaction with a K_m of 9.50 mM and a of 1.75 mM- s^-1. Calculate the reaction velocity, V_0, for the following substrate concentrations. 2.50 mM 9.50 mM 12.0 mM
1. The turnover number for an enzyme is known to be 5000 min. From the following set of data, calculate the Km, Vmax and the amount of enzyme present in this experiment. Use excel to obtain the lineweaver burk plot. Substrate concentration (MM) 1 Initial velocity (umol/min) 167 250 334 376 498 499 100 1,000
Score: 3318/3/00 15 of 37 > An enzyme catalyzes a reaction with a km of 8.50 mM and a Vmax of 2.45 mM.s. Calculate the reaction velocity, to, for each substrate concentration. S = 3.75 mm mMs-1 S = 8.50 mM mM.s-1 [S] = 10.5 mM about us Careers Privacy policy terms of use contact us help
An enzyme catalyzes a reaction with a Km of 5.00 mM and a Vmax of 1.85 mM s-1. Calculate the reaction velocity, vo. for the following substrate concentrations. a) 3.00 mM Number mM.s b) 5.00 mM Number mM.s c) 10.5 mM Number
An enzyme catalyzes a reaction with a Km of 8.00 mM and a Vmax of 4.45 mM s-1. Calculate the reaction velocity, Vo, for the following substrate concentrations a) 1.00 mM Number mM.s-1 b) 8.00 mM Number mM s-1 c) 11.0 mM Number mM.s-1