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

from Van't Hoff equation

d(lnK)/dT= deltaH/RT2

when integrated, the equation

lnK= -deltaH/RT+C

so a plot of lnK vs 1/T( T in K) gives straight line whose slope is -deltaH/R.

the plot of lnK vs 1/T is drawn.

from the plot, the slope is -deltaH/R = -2238, deltaH= 2238*8.314 J/mole=18607J/mole= 18.607 Kj/mole. deltaH is postive.

since lnK= -deltaH/RT+C

since deltaG= deltaH-TdeltaS

deltaG= -RTlnK

-RTlnK= deltaH-T*deltaS

lnK= -deltaH/RT+ deltaS/R

hence the intercept is deltaS/R= 13.3, deltaS= 13.3* 8.314 J/mole.K=110.6 J/mole.K, since entropy is +ve, the reaction is driven by entropy.

the equation of best fitt is lnK=-2238*1/T+13.3

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

lnK= -2238/310+13.3=437.3

deltaG= -RT lnK= -8.314*310* 437.3=-1127100.

at 42deg.c= 42+273= 315K, lnK=-2238/315+13.3=490.5

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