Given that a reaction is a second order process, determine amount remaining after 12.84 minutes when 2.85 M of a material to react if the rate constant is 0.2712/min

The hydrolysis of the sugar sucrose to the sugars glucose and fructose, C12H22O11+H2OC6H12O6+C6H12O6 follows a first-order rate equation for the disappearance of sucrose: Rate =k[C12H22O11] (The products of the reaction, glucose and fructose, have the same molecular formulas but differ in the arrangement of the atoms in their molecules.) (a) In neutral solution, k=2.11011s1 at 27 C and 8.51011s1 at 37 C. Determine the activation energy, the frequency factor, and the rate constant for this equation at 47 C (assuming the kinetics remain consistent with the Arrhenius equation at this temperature). (b) When a solution of sucrose with an initial concentration of 0.150 M reaches equilibrium, the concentration of sucrose is 1.65107M . How long will it take the solution to reach equilibrium at 27 C in the absence of a catalyst? Because the concentration of sucrose at equilibrium is so low, assume that the reaction is irreversible. (c) Why does assuming that the reaction is irreversible simplify the calculation in pan (b)?

Many biochemical reactions are catalyzed by acids. A typical mechanism consistent with the experimental results (in which HA is the acid and X is the reactant) is Step 1: Step 2: Step 3: Derive the rate law from this mechanism. Determine the order of reaction with respect to HA. Determine how doubling the concentration of HA would affect the rate of the reaction.

Consider the first-order decomposition reaction A+BCwhere A (circles) decomposes to B (triangles) and C (squares). Given the following boxes representing the reaction at t=2 minutes, fill in the boxes with products at the indicated time. Estimate the half-life of the reaction.

The initial rate for a reaction is equal to the slope of the tangent line at t 0 in a plot of [A] versus time. From calculus, initial rate = d[A]dt . Therefore. the differential rate law for a reaction is Rate = d[A]dt=k[A]n. Assuming you have some calculus in your background, derive the zero-, first-, and second-order integrated rate laws using the differential rate law.

Make a graph of [A] versus time for zero-, first-, and second-order reactions. From these graphs, compare successive half-lives.

Apply the method of initial rates to determine the order of a chemical reaction with respect to Reactant X. Create a set of hypothetical experimental data that would lead you to conclude that the reaction is second order in X.

The hydrolysis of the sugar sucrose to the sugars glucose and fructose follows a first-order rate equation for the disappearance of sucrose. C12H22O11(aq)+H2O(l)C6H12O6(aq)+C6H12O6(aq) Rate =k[C12H22O11] In neutral solution, k=2.11011 at 27 C. (As indicated by the rate constant, this is a very slow reaction. In the human body, the rate of this reaction is sped up by a type of catalyst called an enzyme.) (Note: That is not a mistake in the equation—the products of the reaction, glucose and fructose, have the same molecular formulas, C6H12O6, but differ in the arrangement of the atoms in their molecules). The equilibrium constant for the reaction is 1.36105 at 27 C. What are the concentrations of glucose, fructose, and sucrose after a 0.150 M aqueous solution of sucrose has reached equilibrium? Remember that the activity of a solvent (the effective concentration) is 1.

If you know some calculus, derive the integrated second-order rate law for the reaction 2Aproducts Follow the same steps listed in Question 121.

Consider the following reaction in aqueous solution: 5Br(aq)+BrO3(aq)+6H+(aq)3Br2(aq)+3H2O(l) If the rate of disappearance of Br(aq) at a particular moment during the reaction is 3.5104Ms1, what is the rate of appearance of 3Br2(aq) at that moment?