4. You characterize an enzyme that catalyzes the following reaction: By measurin

Question

4. You characterize an enzyme that catalyzes the following reaction: By measuring equilibrium concentrations, you determine that Keq values for the reaction are 326 M, 378 M and 437 M at 25°C, 30°C, and 37°C, respectively. Using Excel or similar program, tabulate the data, make a van't Hoff plot with trendline and its equation, and use it to determine 1°, and S°, for the reaction. a. b. what is G, for this reaction at 37 °C? c. Is this reaction driven by entropy, enthalpy, or both? Explain. d. What is the Keq for the reaction at 42 °C?

Explanation / Answer

Van;t Hoff equation is

d/dT( lnKeq)= deltaH/RT2

when integrated, lnKeq=-deltaH/RT+C, since deltaG =-RTlnKeq a, ln Keq= -deltaG/RT and deltaG= deltaH-T*deltaS

lnKeq=-deltaH/RT+deltaS/R

so a plot of lnKeq vs 1/T ( T in K, T (K)= t(deg.c+273) gives straight line whose slope is –deltaH/R and intercept is deltaS/R

T in K and R= 8.314 J/mole.K

So a plot of ln K vs 1/T is prepared

from the equation of best fit, -deltaH/R= -2238 and deltaH= -2238*8.314 J/mole= -18607 J/mole=-18.607 Kj/mole

intercept is deltaS/R= 13.3, deltaS= 13.3*8.314= 110.57 J/mole.K

at 37 deg.c= 37+273= 310K

deltaG =deltaH-T*deltaS= -18607-110.57*8.314 =-18607-919.3=-19526.3 J/mole= -19.526 KJ/mole

since deltaG is -ve, the reaction is feasible and is driven, both deltaH and deltaS contribute to deltaG. since deltaH is -ve and deltaS is +ve, the reaction is driven by both.

at t=42 deg.c, T= 42+273= 315K, from the equation of best fit, lnK=-2238/315+13.3, K= 490

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