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

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