|MadSci Network: Chemistry|
Tony, You didnít describe how you were monitoring the catalase reaction, so Iím going to assume that you were judging the rate of the reaction by observing the formation of bubbles of oxygen gas. Catalase decomposes hydrogen peroxide into water and oxygen. Like any enzymatic reaction, the rate is described by a formula called the Michaelis-Menten equation. The rate of the catalase reaction with hydrogen peroxide would be given by this equation: reaction rate = (Vmax) x [Concentration of hydrogen peroxide] x [Enzyme concentration] ------------------------------------------------------------------------ Km + [Concentration of hydrogen peroxide] Vmax is the maximum reaction rate, and is a constant. Km is the Michaelis-Menten constant, and for catalase, Km = 25mM (25 millimoles/liter). This number gives a measure of how well an enzyme binds its substrate. The Michaelis-Menten equation reflects the fact that the reaction rate for an enzyme-catalyzed reaction exhibits saturation. That is, the rate reaches a maximum reaction rate (Vmax), and stays at that maximum rate regardless of how much additional hydrogen peroxide you add Ė since the concentration of hydrogen peroxide appears in both the numerator and the denominator. A good way to think about this is to remember that the enzyme is a catalyst, and is present in small concentrations. The concentration of the catalase is much less than that of the hydrogen peroxide. Once you have enough hydrogen peroxide present so that all of the enzyme molecules are rapidly binding hydrogen peroxide and decomposing it, adding more hydrogen peroxide will not make the reaction go any faster. This may have been what your teacher meant when she told you that having more peroxide wouldnít make the reaction go faster. However, the rate is also dependent on the concentration of enzyme present. In your experiment, you used 3 mL of catalase in 125 mL total volume, instead of 1 mL of catalase in 100 mL total volume. So, under your conditions the rate of the reaction WILL be higher, because the enzyme concentration is higher. In addition, since you have much more peroxide in your reaction mixture that you were supposed to, much more oxygen gas will be produced.
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