The rate constant of a second order reaction is ${10^{ - 2}}lit.mol{e^{ - 1}}.{\sec ^{ - 1}}$ . The rate constant when expressed as $cc.{molecule^{ - 1}}.{\min ^{ - 1}}$ is:
A. $9.96 \times {10^{ - 22}}$
B.$9.96 \times {10^{ - 23}}$
C.$9.96 \times {10^{ - 21}}$
D.$9.96 \times {10^{ - 24}}$
Answer
Verified
459.6k+ views
Hint: We are given the rate constant of a second order reaction to be ${10^{ - 2}}lit.mol{e^{ - 1}}.{\sec ^{ - 1}}$. We will try to find the relation and convert moles into a number of molecules, litres into cubic centimetre and second into minutes. And finally we will put it in ${10^{ - 2}}lit.mol{e^{ - 1}}.{\sec ^{ - 1}}$ to find the required answer.
Complete step by step answer:
As the rate constant of a second order reaction is given ${10^{ - 2}}lit.mol{e^{ - 1}}.{\sec ^{ - 1}}$and we have to find the rate constant of the second order reaction in the unit $cc.{molecule^{ - 1}}.{\min ^{ - 1}}$ . We know that
$
1L = {10^3}c{m^3} \\
1mol = 6.022 \times {10^{23}} \\
1\sec = \dfrac{1}{{60}}\min \\
$
We will have to put the required units of volume , mole and time in ${10^{ - 2}}lit.mol{e^{ - 1}}.{\sec ^{ - 1}}$ to find the answer in $cc.{molecule^{ - 1}}.{\min ^{ - 1}}$, We will have to do it carefully so that we can avoid calculation error. So:
$
{10^{ - 2}}lit.mol{e^{ - 1}}.{\sec ^{ - 1}} \\
= \dfrac{{{{10}^{ - 2}}{{.10}^3}cc}}{{6.022 \times {{10}^{23}}molecule.\dfrac{1}{{60}}\min }} = 9.96 \times {10^{ - 22}}cc.{molecule^{ - 1}}.{\min ^{ - 1}} \\
$
From the above explanation and calculation it is clear to us that
The correct answer of the given question is option: A. $9.96 \times {10^{ - 22}}$
Additional information:
The Rate constant is a proportional constant of the equation which expresses the Relation between the rate of the chemical equation and the concentrations of the reactants of the chemical equation. One mole of a substance contains $6.022 \times {10^{23}}$particles. This is a special number and it is also known as the Avogadro’s number or Avogadro constant. It is named after the great scientist Amedeo Avogadro. Amedeo Avogadro was from Italy.
Note:
Always remember that one mole has $6.022 \times {10^{23}}$particles , one litre is equal to ${10^3}c{m^3}$. One minute has sixty seconds. Always try to avoid silly mistakes and calculation errors while solving the numerical. To convert a given value in it’s different unit, always use the concept of units and dimensions as used in the above question.
Complete step by step answer:
As the rate constant of a second order reaction is given ${10^{ - 2}}lit.mol{e^{ - 1}}.{\sec ^{ - 1}}$and we have to find the rate constant of the second order reaction in the unit $cc.{molecule^{ - 1}}.{\min ^{ - 1}}$ . We know that
$
1L = {10^3}c{m^3} \\
1mol = 6.022 \times {10^{23}} \\
1\sec = \dfrac{1}{{60}}\min \\
$
We will have to put the required units of volume , mole and time in ${10^{ - 2}}lit.mol{e^{ - 1}}.{\sec ^{ - 1}}$ to find the answer in $cc.{molecule^{ - 1}}.{\min ^{ - 1}}$, We will have to do it carefully so that we can avoid calculation error. So:
$
{10^{ - 2}}lit.mol{e^{ - 1}}.{\sec ^{ - 1}} \\
= \dfrac{{{{10}^{ - 2}}{{.10}^3}cc}}{{6.022 \times {{10}^{23}}molecule.\dfrac{1}{{60}}\min }} = 9.96 \times {10^{ - 22}}cc.{molecule^{ - 1}}.{\min ^{ - 1}} \\
$
From the above explanation and calculation it is clear to us that
The correct answer of the given question is option: A. $9.96 \times {10^{ - 22}}$
Additional information:
The Rate constant is a proportional constant of the equation which expresses the Relation between the rate of the chemical equation and the concentrations of the reactants of the chemical equation. One mole of a substance contains $6.022 \times {10^{23}}$particles. This is a special number and it is also known as the Avogadro’s number or Avogadro constant. It is named after the great scientist Amedeo Avogadro. Amedeo Avogadro was from Italy.
Note:
Always remember that one mole has $6.022 \times {10^{23}}$particles , one litre is equal to ${10^3}c{m^3}$. One minute has sixty seconds. Always try to avoid silly mistakes and calculation errors while solving the numerical. To convert a given value in it’s different unit, always use the concept of units and dimensions as used in the above question.
Recently Updated Pages
Class 12 Question and Answer - Your Ultimate Solutions Guide
Master Class 12 Social Science: Engaging Questions & Answers for Success
Master Class 12 Physics: Engaging Questions & Answers for Success
Master Class 12 Maths: Engaging Questions & Answers for Success
Master Class 12 English: Engaging Questions & Answers for Success
Master Class 12 Chemistry: Engaging Questions & Answers for Success
Trending doubts
Explain sex determination in humans with the help of class 12 biology CBSE
Give 10 examples of unisexual and bisexual flowers
How do you convert from joules to electron volts class 12 physics CBSE
Differentiate between internal fertilization and external class 12 biology CBSE
On what factors does the internal resistance of a cell class 12 physics CBSE
A 24 volt battery of internal resistance 4 ohm is connected class 12 physics CBSE