(a)
Interpretation:
Considering the factor of charge stability, it is to be determined whether the given elementary step is reversible or irreversible.
Concept introduction:
The stability of charged species on both sides of a reaction decides if the products are more stable than the reactants or vice versa. A reaction is irreversible if it’s
(b)
Interpretation:
Considering the factor of charge stability, it is to be determined whether the given elementary step is reversible or irreversible.
Concept introduction:
The stability of charged species on both sides of a reaction decides if the products are more stable than the reactants or vice versa. A reaction is irreversible if it’s
(c)
Interpretation:
Considering the factor of charge stability, it is to be determined whether the given elementary step is reversible or irreversible.
Concept introduction:
The stability of charged species on both sides of a reaction decides if the products are more stable than the reactants or vice versa. A reaction is irreversible if it’s
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Organic Chemistry: Principles and Mechanisms (Second Edition)
- Real Time uy 12. A particular reaction requires 54 kJ mol to activate and is found to have a rate constant 1 -1 -1 -1 S S of 3.42 x 10-4 M¹s¹ at 155.85 °C. Predict the rate constant (in M¹ s¹ to two decimal places) if the temperature was increased to 460.85 °C. Approximately how much faster (to the nearest whole number) is the reaction predicted at 460.85 °C relative to 155.85 °C? A: 540 times fasterarrow_forwardThe below reaction follows second order rate kinetics and proceeds via a bimolecular mechanism. KCNCH3B1 -» KBr + CH3CN If the concentration of potassium bromide were doubled, how many times faster would the reaction go? If the concentration of CH3Br were doubled. How much faster would the reaction go? Write the rate law for the above reaction. If [KCN] 0.5 M, [CH3Br] 0.5 M, and the relative reaction rate is 0.25 M/s, what is the rate constant? Also, what are the units of the rate constant?arrow_forwardWhich step in the mechanism for free-radical bromination of propane is rate-determining? O Br-Br 2 Br O (CH3)2C-H+ Br→ (CH3)2C + HBr O (CH3)2C + Br-Br (CH3)2CBr + Br. O (CH3)2C + Br. → (CH3)2CBrarrow_forward
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- Organic Chemistry: A Guided InquiryChemistryISBN:9780618974122Author:Andrei StraumanisPublisher:Cengage Learning