#12 -4-1The rate constant of a first-order reaction is 3.15 × 10 Sat 350. °C. If the activation energy is 107.constant is 8.65 × 10-4 - 1S Round your answer to 3 significant digits.°CNote: Reference the Fundamental constants table for additional information..x10XŚkJmol'calculate the temperature at which its rate

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-1 The rate constant of a first-order reaction is 3.15 x 10 S at 350. °C. If the activation energy is 107. -4 constant is 8.65 x 10 S . Round your answer to 3 significant digits. Note: Reference the Fundamental constants table for additional information. °C 0 X S kJ mol' calculate the temperature at which its rate

-1 The rate constant of a first-order reaction is 3.15 x 10 S at 350. °C. If the activation energy is 107. -4 constant is 8.65 x 10 S . Round your answer to 3 significant digits. Note: Reference the Fundamental constants table for additional information. °C 0 X S kJ mol' calculate the temperature at which its rate

Introductory Chemistry: A Foundation

9th Edition

ISBN:9781337399425

Author:Steven S. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Donald J. DeCoste

Chapter17: Equilibrium

Section: Chapter Questions

Problem 9CR

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11.35 For the reaction 2 NO(g) + 2 H?(g) — N,(g) + 2 H,O(g) at 1100°C, the following data have been obtained: [NOJ [HJ Rate = A(N2]/At (mol L~1) (mol L_1) (mol L-1 s_1) 5.0 X 10’1 0.32 0.012 1.0 X 10~’ 0.32 0.048 1.0 X 10"2 0.64 0.096 Derive a rate law for the reaction and determine the value of the rate constant.
Hydrogen iodide, HI, decomposes in the gas phase to produce hydrogen, H2, and iodine, I2. The value of the rate constant, k, fur the reaction was measured at several different temperatures and the data are shown here: Temperature (K) k (M -1 5-1) 555 6.23107 575 2.42106 645 1.44104 700 2.01103 What is the value of the activation energy (in kJ/mol) for this reaction?
The reaction H2SeO3(aq) + 6I-(aq) + 4H+(aq) Se(s) + 2I-3(aq) + 3H2O(l) was studied at 0C, and the following data were obtained: [H2SeO3]0 (mol/L) [H+]0 (mol/L) [I]0(mol/L) Initial Rate (mol/L s) 1.0 104 2.0 102 2.0 102 1.66 107 2.0 104 2.0 102 2.0 10-2 3.33 107 3.0 104 2.0 102 2.0 102 4.99 107 1.0 104 4.0 102 2.0 102 6.66 107 1.0 104 1.0 102 2.0 102 0.42 107 1.0 104 2.0 102 4.0 102 13.2 107 1.0 104 1.0 102 4.0 102 3.36 107 These relationships hold only if there is a very small amount of I3 present. What is the rate law and the value of the rate constant? (Assumethatrate=[H2SeO3]t)
Why does a catalyst increase the rate of a reaction? What is the difference between a homogeneous catalyst and a heterogeneous catalyst? Would a given reaction necessarily have the same rate law for both a catalyzed and an uncatalyzed pathway? Explain.
Use the data provided in a graphical method to determine the order and rate constant of the following reaction: 2PQ+W Time (s) 9.0 13.0 18.0 22.0 25.0 [P] (M) 1.077103 1.068103 1.055103 1.046103 1.039103
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A study of the rate of dimerization of C4H6 gave the data shown in the table: 2C4H6C8H12 (a) Determine the average rate of dimerization between 0 s and 1600 s, and between 1600 s and 3200 s. (b) Estimate the instantaneous rate of dimerization at 3200 s from a graph of time versus [C4H6]. What are the units of this rate? (c) Determine the average rate of formation of C8H12 at 1600 s and the instantaneous rate of formation at 3200 s from the rates found in parts (a) and (b).
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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)?