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Chemistry 2710 Spring 2000 Quiz 8 solution

The best method at your disposal for solving this problem is an Eadie-Hofstee plot. We first calculate v/s ( s=[CO2]):

v/s (s-1) 0.0222 0.0200 0.0167 0.0083
v ( $\mathrm{mol\,L^{-1}s^{-1}}$) $2.78\times 10^{-5}$ $5.00\times 10^{-5}$ $8.33\times 10^{-5}$ $1.66\times 10^{-4}$
The graph has the following appearance:
The data fit the Michaelis-Menten rate law very well. Note however that the selection of experimental points leaves somewhat to be desired: The point at $s = 20.00\,\mathrm{mmol/L}$ is absurdly distant from the others.

From the slope and intercept, we get $K_M = 9.93\,\mathrm{mmol/L}$ and $v_\mathrm{max} =
2.49\times 10^{-4}\,\mathrm{mol\,L^{-1}s^{-1}}$. Therefore

\begin{displaymath}k_{-2} = \frac{v_\mathrm{max}}{e_0} = \frac{2.49\times
10^{-4...
...es 10^{-9}\,\mathrm{mol/L}}
= 1.07\times 10^5\,\mathrm{s}^{-1}.\end{displaymath}



Marc Roussel
2000-04-07