Problem 4(10pts) Two batch experiments have been conducted to determine the values of ,B, and k in the rate model rA=k(CA)(CBB) for the liquid-phase reaction A+2B2C+D. In these experiments, the concentration of A has been measured as a function of time, and the numerical derivative dCA/dt has been estimated. The derivative has been plotted versus concentration on log-log plots for the two different experiments, as shown below. For the data on the left, the initial concentration of B was 20mol/m3, and for the data on the right, the initial concentration of B was 2mol/m3. For both sets of data, the initial concentration of A was 1 mol/m3. The units on concentration and time in the plots are mol/m3 and min, respectively. Using the plotted data fits, estimate the values of ,, and k. A. For a reaction 2A+B2C, with the rate equation: Rate=k[A]2[B] (a) the order with respect to A is 1 and the order overall is 1 . (b) the order with respect to A is 2 and the order overall is 2 . (c) the order with respect to A is 2 and the order overall is 3. (d) the order with respect to B is 2 and the order overall is 2 . (e) the order with respect to B is 2 and the order overall is 3 . B. When the concentration of reactant molecules is increased, the rate of reaction increases. The best explanation is: As the reactant concentration increases, (a) the average kinetic energy of molecules increases. (b) the frequency of molecular collisions increases. (c) the rate constant increases. (d) the activation energy increases. (e) the order of reaction increases. C. Suppose the reaction: A+2BAB2 occurs by the following mechanism: Step I A+BAB slow Step 2AB+BAB2 fast Overall A+2BAB2 The rate law expression must be Rate = (a) k[A] (b) k[B] (c) k[A][B] (d) k[B]2 (e) k[A][B]2 D. A catalyst: (a) actually, participates in the reaction. (b) changes the equilibrium concentration of the products. (c) does not affect a reaction energy path. (d) always decreases the rate for a reaction. (e) always increases the activation energy for a reaction