The percentage of calcium carbonate that reacted is 2.5%. The reaction in question allows us to determine the equilibrium Kp: Kp = the partial pressure of carbon dioxide, since the other components are solids. We'll apply the ICE table to the provided equilibrium. At the start, we have 0.2 for calcium carbonate with no initial moles of other substances. As the reaction progresses, we set the changes to be -x for calcium carbonate, +x for carbon dioxide, and +x for the other product, leading us to an equilibrium of 0.2-x for calcium carbonate while both other products are at x. Using Kp = Kc(RT)ⁿ, where n represents the mole difference of gaseous products and reactants, we find n to equal 1 for this reaction. With R as the gas constant (8.314 J/mol K) and the temperature at 800 °C (1073 K), we substitute the values accordingly. Upon calculation, we find x = 0.005, which indicates the amount of calcium carbonate that dissociated or reacted, leading us to the reacted percentage.
Answer:
B) Hyperbolic curve; substrate saturation
Explanation:
Enzymatic kinetics examines the rates of reactions catalyzed by enzymes. These studies offer insights into the mechanism of the catalytic reaction and enzyme specificity. Determining the reaction rate facilitated by an enzyme is generally straightforward, as purification or isolation of the enzyme is frequently unnecessary. Measurements are taken under optimal conditions for pH, temperature, and the presence of cofactors, utilizing saturating substrate concentrations. Under these circumstances, the observed reaction rate is the maximum velocity (Vmax). The rate can be measured by monitoring either product formation or substrate consumption.
Following the rate of product formation (or substrate consumption) over time yields the so-called reaction progress curve, or merely, reaction kinetics. This reacts as a hyperbolic curve
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