Chapter 7: Problem 146
The rate law for the reaction \(\mathrm{RCl}+\mathrm{NaOH} \rightarrow \mathrm{ROH}+\mathrm{NaCl}\) is given by Rate \(=\mathrm{k}(\mathrm{RCl})\). The rate of the reaction is a. Halved by reducing the concentration of \(\mathrm{RCl}\) by one half. b. Increased by increasing the temperature of the reaction. c. Remains same by change in temperature. d. Doubled by doubling the concentration of \(\mathrm{NaOH}\).
Short Answer
Step by step solution
Identify the Rate Law
Analyze Option (a)
Analyze Option (b)
Analyze Option (c)
Analyze Option (d)
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
First-Order Reaction
- \(k\) is the rate constant, and
- \([A]\) is the concentration of the reactant.
Effect of Temperature on Reaction Rate
- Molecules move faster.
- There is a greater frequency of collisions between molecules.
- More molecules have enough energy to surpass the activation energy barrier of a reaction.
- \(k\) is the rate constant,
- \(A\) is the frequency factor,
- \(E_a\) is the activation energy,
- \(R\) is the gas constant, and
- \(T\) is the temperature in Kelvin.
Concentration Dependence
- When the concentration of \(\text{RCl}\) is increased, the reaction rate increases proportionally.
- Conversely, decreasing the concentration of \(\text{RCl}\) results in a reduced rate.
Activation Energy
- Molecules must collide with sufficient energy and correct orientation.
- Reactions with lower activation energy proceed faster than those requiring higher amounts of energy.