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Chapter 18: Problem 17

The rate law for the reaction below is given by the expression \(\mathrm{k}[\mathrm{A}]\) [B] \(\mathrm{A}+\mathrm{B} \rightarrow\) Product If the concentration of \(\mathrm{B}\) is increased from \(0.1\) to \(0.3 \mathrm{~mole}\), keeping the value of \(A\) at \(0.1\) mole, the rate constant will be: [Main Online April 10, 2016] (a) \(3 k\) (b) \(9 k\) (c) \(k / 3\) (d) \(k\)

Short Answer

Expert verified
(d) k

Step by step solution

01

Understand the Rate Law

The rate law for the reaction is given by \( ext{Rate} = k[A][B] \). Here, \( k \) is the rate constant, \([A]\) and \([B]\) are the concentrations of reactants A and B respectively.
02

Initial Rate Calculation

Initially, the concentrations are \([A] = 0.1 \) mole and \([B] = 0.1 \) mole. The initial rate is given by: \[ ext{Rate}_{1} = k imes 0.1 imes 0.1 = 0.01k\]
03

New Rate Calculation After Increase in [B]

The concentration of \(B\) is increased to \(0.3\) moles while \( A \) remains at \(0.1\) mole. The new rate is calculated as: \[ ext{Rate}_{2} = k imes 0.1 imes 0.3 = 0.03k\]
04

Determine Effect on Rate Constant

The rate constant \( k \) itself is independent of concentrations and depends only on temperature and the nature of the reactant. Therefore, changes in concentrations do not affect \( k \).
05

Conclusion

Since the rate constant \( k \) does not change with changes in concentration, the value of \( k \) remains the same despite the change in \([B]\). Hence, the correct answer is \(d\), which is \(k\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Kinetics
Chemical kinetics explores how chemical reactions occur and the factors affecting their speed. These reactions happen at different rates, depending on various conditions.
This study helps chemists understand reactions better and optimize them for industrial or laboratory purposes. It's like having a recipe and knowing exactly how long each step will take.
In chemical kinetics, the reaction rate is vital. It measures how fast reactants turn into products.
  • Temperature: Changes can increase or decrease reaction rates.
  • Concentration: More reactant particles usually result in faster reaction.
  • Catalysts: These substances speed up reactions without being consumed themselves.
All these elements define how a reaction proceeds, helping predict and control it effectively.
Rate Constant
The rate constant, denoted as \( k \), is a crucial part of the rate law and indicates the reaction's speed at a given temperature.
In mathematical terms, it is the proportionality factor in the rate equation. For instance, in a rate law like \( \text{Rate} = k[A][B] \), \( k \) links the concentration of reactants \([A]\) and \([B]\) to the reaction rate.
Here are some key characteristics of the rate constant:
  • Specific to a particular reaction and set of conditions.
  • Units depend on the overall order of reaction.
  • Unchanged by altering reactant concentrations.
A constant temperature means \( k \) remains unchanged, making it a reliable measurement for comparing reaction speeds.
Reaction Order
The reaction order is another critical aspect in understanding rate laws.
It describes how the rate is affected by the concentration of each reactant. This gives insight into the mechanism of the reaction and helps in predicting reaction behavior.
The reaction order isn't always the same as the stoichiometric coefficients in the balanced equation. It can be:
  • Zero order: Rate doesn't depend on reactant concentration.
  • First order: Rate directly proportional to one reactant's concentration.
  • Second order: Rate proportional to either the square of one reactant's concentration or the product of two reactants' concentrations.
To determine the overall reaction order, add the orders of all reactants in the rate law. This information is pivotal for predicting how a reaction will respond to changes in concentration or other conditions.

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Most popular questions from this chapter

Consider the chemical reaction, \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{~g}) .\) The rate of this reaction can be expressed in terms of time derivative of concentration of \(\mathrm{N}_{2}(\mathrm{~g}), \mathrm{H}_{2}(\mathrm{~g})\) or \(\mathrm{NH}_{3}(\mathrm{~g})\). Identify the correct relationship amongst the rate expressions. [2002S](a) Rate \(=-\mathrm{d}\left[\mathrm{N}_{2}\right] / \mathrm{d} t=-1 / 3 \mathrm{~d}\left[\mathrm{H}_{2}\right] / \mathrm{d} t=1 / 2 \mathrm{~d}\left[\mathrm{NH}_{3}\right] / \mathrm{d} t\) (b) Rate \(=-\mathrm{d}\left[\mathrm{N}_{2}\right] / \mathrm{d} t=-3 \mathrm{~d}\left[\mathrm{H}_{2}\right] / \mathrm{d} t=2 \mathrm{~d}\left[\mathrm{NH}_{3}\right] / \mathrm{d} t\) (c) Rate \(=\mathrm{d}\left[\mathrm{N}_{2}\right] / \mathrm{d} t=1 / 3 \mathrm{~d}\left[\mathrm{H}_{2}\right] / \mathrm{d} t=1 / 2 \mathrm{~d}\left[\mathrm{NH}_{3}\right] / \mathrm{d} t\) (d) Rate \(=-\mathrm{d}\left[\mathrm{N}_{2}\right] / \mathrm{d} t=-\mathrm{d}\left[\mathrm{H}_{2}\right] / \mathrm{d} t=\mathrm{d}\left[\mathrm{NH}_{3}\right] / \mathrm{d} t\)

Higher order \((>3)\) reactions are rare due to : [Main 2015] (a) shifting of equilibrium towards reactants due to elastic collisions (b) loss of active species on collision (c) low probability of simultaneous collision of all the reacting species (d) increase in entropy and activation energy as more molecules are involved

Decrease in atomic number is observed during (a) alpha emission (b) beta emission (c) positron emission (d) electron capture.

The gas phase decomposition of dimethyl ether follows first order kinetics. $$ \mathrm{CH}_{3}-\mathrm{O}-\mathrm{CH}_{3}(\mathrm{~g}) \rightarrow \mathrm{CH}_{4}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g})+\mathrm{CO}(\mathrm{g}) $$ The reaction is carried out in a constant volume container at \(500^{\circ} \mathrm{C}\) and has a half life of \(14.5\) minutes. Initially, only dimethyl ether is present at a pressure of \(0.40\) atmosphere. What is the total pressure of the system after 12 minutes? Assume ideal gas behaviour. [1993 - 4 Marks]

A flask contains a mixture of compounds \(\mathrm{A}\) and \(\mathrm{B}\). Both compounds decompose by first-order kinetics. The half-lives for \(\mathrm{A}\) and \(\mathrm{B}\) are 300 \(\mathrm{s}\) and \(180 \mathrm{~s}\), respectively. If the concentrations of \(\mathrm{A}\) and \(\mathrm{B}\) are equal initially, the time required for the concentration of \(A\) to be four times that of \(\mathrm{B}\) (in s) is: (Use \(\ln 2=0.693\) ) [Main Sep. \(\mathbf{0 5}, \mathbf{2 0 2 0}\) (I)](a) 180 (b) 900 (c) 300 (d) 120
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