The reaction of 1,3-butadiene with HCI produces two addition products. (See below) H-CI CH CH CH-CH3 CH, CH=CH-CH3 + CHCH-CH=CH2 I II The relative amounts of the products that form is strongly temperature dependent. Here are the results: At 0°C, the following product composition is observed: product I (71%), product II (29%). At 42°C, the following product composition is observed: product I (15%), product II (85%). ANSWER THE FOLLOWING QUESTION: Why does the addition product I predominate at the lower temperature (0°C)? Select one: Product I predominates at lower temperature because it has more t а. electron delocalization than product II. O b. At lower temperature, the reaction is equilibrium controlled: product II is more stable than product I, so product Il predominates. Product I is less sterically hindered, so it predominates. С. d. At lower temperature, the reaction is kinetically controlled: product I forms faster, so product I predominates.

The reaction of 1,3-butadiene with HCI produces two addition products. (See below) H-CI CH CH CH-CH3 CH, CH=CH-CH3 + CHCH-CH=CH2 I II The relative amounts of the products that form is strongly temperature dependent. Here are the results: At 0°C, the following product composition is observed: product I (71%), product II (29%). At 42°C, the following product composition is observed: product I (15%), product II (85%). ANSWER THE FOLLOWING QUESTION: Why does the addition product I predominate at the lower temperature (0°C)? Select one: Product I predominates at lower temperature because it has more t а. electron delocalization than product II. O b. At lower temperature, the reaction is equilibrium controlled: product II is more stable than product I, so product Il predominates. Product I is less sterically hindered, so it predominates. С. d. At lower temperature, the reaction is kinetically controlled: product I forms faster, so product I predominates.

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

Standard molar entropies, S°, in J/K•mol, are given below each reactant and product in the reaction shown below. The standard entropy of reaction, AS°, for this reaction is J. CH4(g) + 2 O2(g) CO2(g) + 2 H2O() 213.6 186.2 _205.0 69.9 Express your answer using four significant figures.
Which of the following molecules would be expected to have the highest molar entropy? ethane O heptane O 2-methylbutane 2,2-dimethylbutane
Part A Use the data below to determine which of the following compounds are thermodynamically stable with respect to their constituent elements at 25 °C. Substance and State AHᵒf (kJ/mol) AG°f (kJ/mol) S° [J/(K-mol)] PC13 (1) - 272.3 217.1 CH4 (g) N₂O(g) HI(g) Check all that apply. PC13 (1) N₂O(g) HI(g) CH4 (g) Submit Request Answer -319.7 - 74.8 82.0 26.5 - 50.8 104.2 1.7 186.2 219.7 206.5
Calculate the G⁰reaction value for the following reaction running at 298K.CO (g) + ½ O2 (g) ↔ CO2 (g)Assuming that the ∆H⁰reaction value is constant in the temperature range operated, calculate the G⁰reaction value at 500K.
We will now assess the spontaneity of this runaway decomposition of ammonium nitrate using three methods: Method 1: Entropy change of the universe (AS univ) AS univ=AS Rxn + AS surroundings Method 2: Change in Gibb's Free Energy (AG) AG=AH-TAS° Method 3: Change in Gibb's Free Energy from free energy of formation (AG) AGRxn = EnAGf, products - EnAG₁°, reactants 6. Define spontaneity. What is the sign on AS univ for a spontaneous reaction? What is the sign on AG for a spontaneous reaction? 7. Determine AS (in J K¹ mol¹ ) for the universe at 300°C. 8. Determine AG° (in kJ mol-¹) at 300°C using Method 2. 9. Determine AG (in kJ mol-¹) using Method 3.
8) Consider the following reaction:2 Br1-(aq) + Cl2(g) ↔ Br2(l) + 2Cl1-(aq) ∆H° = -92.6 kJ/mol at 25°C.Given the following standard entropies at 25°C:Substance S°(J/molK)Br1- (aq) 80.7Br2 (l) 152.3Cl1- (aq) 56.5Cl2 (g) 223.0Calculate the equilibrium constant K for this reaction at 80°C, assuming that ∆H° and ∆S° remain constant over this temperature range
Do polymerization reactions tend to have △Sorxn values that are greater than zero or less than zero? Why?
In some reaction mechanism there is competition between enthalpy and entropy of reaction.some reaction are favored by entropy but not by enthalpy and vice versa. justify it with three different examples in which reaction is favored by one factor but slowed down by other.
Do polymerization reactions tend to have △Sorxn values that are greater than zero or less than zero? Why?
Which is the major organic product formed by this reaction? Studying the summary chart in Lesson VI.3.5 will help to correctly determining the products. NaBH4 ? CH3OH © 2018 ProtonGuru.com A) OH C) Et OCH3 OCH3 B) OH D) Et Et H E) No reaction or none of these is a major product
Consider the potential energy diagram. Which of the following is true for the reverse reaction? catalysed ∆Hrev= +50kJ Earev=+150kJ catalysed ∆Hrev= -50kJ Ea=+50kJ uncatalysed ∆Hrev= +50kJ Ea=+150kJ uncatalysed ∆H= -50kJ Ea=+150kJ