8.15 Consider the following hypothetical reaction: A+B → C Calculate the average rate of the reaction...
Which statements listed below can explain why a chemical reaction does not take place? concentration of reactants is too low the temperature is too low not enough effective collisions are occurring some of these answers all of these answers For a chemical reaction the rate of the forward reaction is at a maximum when when the concentration of the reactants and products is the same a state of equilibrium is reached the reaction begins the concentration of products is greater...
Consider the hypothetical reaction: A + B +2C -------> 2D + 3E, where the rate law is: Rate = -delta[A] / delta t = k[A][B]^2. An experiment is carried out where [A]0 = 1.0*10^-2M, [B]0 = 3.0M, [C]0 = 2.0M. The reaction is started, and after 8.0 seconds, the concentration of A is 3.8*10^-3M. a) Calculate the value of k for the reaction. b) Calculate the half-life for this experiment. c) Calculate the concentration of A after 13.0 seconds d)...
Consider the hypothetical reaction: A + B + 2 + 2D + E where the rate law is Rate = -444 = k[A][B An experiment is carried out where [A] = 6.9 x 10 M, (B) = 2.4 M and (C) = 4.0 M. The reaction is started, and after 10.7 seconds, the concentration of A is 4.9 x 10 M. Calculate the value of k for this reaction in units of M's Answer:
Consider the following reaction: A + B + C +D The rate law for this reaction is as follows: Rate = k Alla B11/2 Suppose the rate of the reaction at certain initial concentrations of A, B, and C is 1.14x10-2 M/s. Part A What is the rate of the reaction if the concentrations of A and C are doubled and the concentration of B is tripled? VO AED A O O ? Rate 2 = M/s Submit Request Answer
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(A). For the hypothetical reaction A+B → C the following rate data (with and without catalyst) were obtained: Rate, M/s Rate, M/s (A), M [B, M (uncatalyzed) (catalyzed) 0.100 0.200 3.51 x 10 4 7.14 x 10-2 0.100 0.100 1.75 x 104 7.14 x 10-2 0.0500 0.200 1.76 x 104 3.57 x 10-2 0.0500 0.100 8.80 x 10-5 3.57 x 10-2 Based on these data: a. Determine the uncatalyzed and catalyzed rate laws for this reaction,...
A mixture initially contains A, B, and C in the following concentrations: [A] = 0.350 M , [B] = 0.700 M , and [C] = 0.400 M . The following reaction occurs and equilibrium is established: A+2B⇌C At equilibrium, [A] = 0.150 M and [C] = 0.600 M . Calculate the value of the equilibrium constant, Kc.
1. The initial reaction rate of a hypothetical reaction | A +B →G was measured at 25°C for several different starting concentrations of A and B. The results are shown below: Experiment 0.2 Initial reaction rate M/s 2.0 X 10-6 1.2 X 10-5 8.0 X 10-6 0.4 0.3 0.6 0.2. A) Use these data to determine the values of m and n and k in the following equation: Rate = k AmBn B) What will the reaction rate be when...
Part A: The rate constant for a certain reaction is k = 1.90×10−3 s−1 . If the initial reactant concentration was 0.150 M, what will the concentration be after 7.00 minutes? Part B: A zero-order reaction has a constant rate of 4.60×10−4 M/s. If after 30.0 seconds the concentration has dropped to 8.00×10−2 M, what was the initial concentration? Part C: A certain reaction has an activation energy of 60.0 kJ/mol and a frequency factor of A1 = 7.80×1012 M−1s−1...
Review Constants Periodic Table Consider the following reaction: A + B + C D The rate law for this reaction is as follows: Part A Rate = k., A][C] B1/2 What is the rate of the reaction if the concentrations of A and C are doubled and the concentration of B is tripled? Suppose the rate of the reaction at certain initial concentrations of A, B, and C is 1.19x10-2 M/s. 18 AU A O O ? Rate 2 =...
32. iven the following informa -a[S03]/At, t/min tion, calculate the average rate, between 10- and 40.0 minutes for the loss of so o.0 10.0 20·0 30.0 40.0 50.0 o.124 o.092 0.068 O.050 0.037 0.028 O.0 0.032 O 056 0.074 0.087 0.096 O.01 0-028 0.037 0 044 0.048 a. 2.18 x 10-4 d. 545 b. 1.8 x 10-3 e. 459 c.1.1l x 10-3 33. The following data were measured for ther eaction: 2C12 (g) SOCI2(g) + c120 (g) → so2 (g)...