Tag: chemistry

Questions Related to chemistry

The rate constant of a first-order reaction is $3 \times 10^{-6}$ per second and initial concentration is 0.10 M. Then the initial rate of reaction is:

  1. $3 \times 10^{-6} Ms^{-1}$

  2. $3 \times 10^{-8} Ms^{-1}$

  3. $3 \times 10^{-7} Ms^{-1}$

  4. $3 \times 10^{-9} Ms^{-1}$


Correct Option: C
Explanation:

Rate = k [concentration] for first order reaction.
Given $k = 3 \times 10^{-6}/sec,$ [concentration] = 0.1 M
$\therefore rate = 3 \times 10^{-6} \times 0.1  = 3 \times 10^{-7} ms^{-1}$

What is the unit for the rate constant of a second order reaction?

  1. $\displaystyle { s }^{ -1 }$

  2. mol $\displaystyle { L }^{ -1 }$

  3. mol $\displaystyle { L }^{ -1 }{ s }^{ -1 }$

  4. L $\displaystyle { mol }^{ -1 }{ s }^{ -1 }$

  5. $\displaystyle { mol }^{ 2 }{ L }^{ -2 }{ s }^{ -2 }$


Correct Option: D
Explanation:

The unit for the rate constant of a second order reaction is $\displaystyle \displaystyle L { mol }^{ -1 }{ s }^{ -1 } $. 
For a second order reaction, $\displaystyle rate = k [A]^2 $
$\displaystyle mol { L }^{ -1 }{ s }^{ -1 }  = k (mol { L }^{ -1 })^2$
$\displaystyle k = L { mol }^{ -1 }{ s }^{ -1 }  $

Statement I : In a second order reaction doubling [A] quadruples the rate
Because
Statement II : The rate equation is $\displaystyle r={ k\left[ A \right]  }^{ 2 }$ for such a reaction

  1. Statement 1 and Statement 2 are correct and Statement 2 is the correct explanation of Statement 1 .

  2. Both the Statement 1 and Statement 2 are correct and Statement 2 is not the correct explanation of Statement 1.

  3. Statement 1 is correct but Statement 2 is not correct.

  4. Statement 1 is not correct but Statement 2 is correct.

  5. Both the Statement 1 and Statement 2 are not correct.


Correct Option: A
Explanation:

Secon order rate expression is given as $r$$=$$k[A]^2$ so if we double the concentration of a then rate increases by 4 times, hence both statements are correct and statement 2 is correct explanation of statement 1.

For the reaction $A + B \rightarrow C$, determine the order of the reaction with respect to $B$ from the information given below.

$\displaystyle { \left[ A \right]  } _{ \circ  }$ $\displaystyle { \left[ B \right]  } _{ \circ  }$ Initial rate (M/s)
1.00 1.00 2.0
1.00 2.00 8.1
2.00 2.00 15.9
  1. Zero order

  2. First order

  3. Second order

  4. Third order

  5. Fourth order


Correct Option: C
Explanation:

Analyzing first and second raw,when concentration of $B$ is doubled then initial rate increased by four times,so it is increasing by square of concentration.Hence it is a second order reaction.


$Rate$$=$$k[A]^x [B]^2$  Using this rate expression and keeping $[A]$ constant,if we double $[B]$ then rate increases four times. 

Statement 1: In a second-order reaction with respect to $A$, when you double [$A$], the rate is quadrupled.
Statement 2: The rate equation is $r = k[A]^2$ for such a reaction.

  1. Statement 1 and Statement 2 are correct and Statement 2 is the correct explanation of Statement 1.

  2. Both the Statement 1 and Statement 2 are correct, but Statement 2 is not the correct explanation of Statement 1.

  3. Statement 1 is correct, but Statement 2 is not correct.

  4. Statement 1 is not correct, but Statement 2 is correct.


Correct Option: A
Explanation:

$\bullet \quad $If rate = ${ K\left[ A \right]  }^{ 2 }$ for a reaction.

       then it is a second order reaction with respect to A.
$\bullet \quad $Let ${ \gamma  } _{ 1 }={ K\left[ A \right]  } _{ t }^{ 2 }$
       if ${ \left[ A \right]  } _{ t }$ is doubled.
             ${ \gamma  } _{ 2 }={ K\left( { 2\left[ A \right]  } _{ t } \right)  }^{ 2 }$
   $\Rightarrow \quad { \gamma  } _{ 2 }={ 4\gamma  } _{ 1 }$.
Hence, statement 1 and statement 2 are correct and statement 2 is the correct explanation of statement 1.

The unit for the rate constant is calculated from the rate law.
For the given rate law, determine the units of the rate constant for rate $= k[A]^{2} [B]$.

  1. $s^{-1}M^{-3}$

  2. $s^{-1}M^{-2}$

  3. $s^{-1}M^{-1}$

  4. $s^{-1}$


Correct Option: B
Explanation:

Since rate constant is given by $k[A]^2[B]$.
Where 1+2=3, so it is probably third order reaction.
So rate constant should have units of $L^2 mol^{-2} s^{-1} \implies M^{-2}s^{-1}$

A graph of concentration versus time data for a second-order reaction gives a straight line in which of the following plots of the data?

  1. $[A] _{t} = -kt + [A] _{0}$

  2. $ln [A] _{t} = -kt + ln [A] _{0}$

  3. $\dfrac {1}{[A _{t}]} = kt + \dfrac {1}{[A _{0}]}$

  4. All of the above

  5. None of the above


Correct Option: A
Explanation:

Let the concentration at $t=0$ be $[{ A } _{ o }]$ & at $t=t$ be $ [{ A } _{ t }]$

 By graph,
$ [{ A } _{ t }]=Kt+[{ A } _{ o }]$
 For negative slop,
$ [{ A } _{ t }]=-Kt+[{ A } _{ o }]$

The rate of the reaction, $C{ Cl } _{ 3 }CHO+NO\longrightarrow CH{ Cl } _{ 3 }+NO+CO$, is given by the equation, rate $=k\left[ C{ Cl } _{ 3 }CHO \right] \left[ NO \right] $. If concentration is expressed in ${mol}/{litre}$, the unit of $k$ is:

  1. ${ mol }^{ -2 }{ L }^{ 2 }{ s }^{ -1 }$

  2. $mol$ ${ L }^{ -1 }{ s }^{ -1 }$

  3. $L$ ${ mol }^{ -1 }{ s }^{ -1 }$

  4. ${ s }^{ -1 }$


Correct Option: C
Explanation:

Since it is second order reaction as clear by rate equation so unit of $k$ is given by $M^{-1} s^{-1}$ or $L\space mol^{−1}s^{−1}$

Units of rate constants for first and zero order reactions in terms of molarity $M$ unit are respectively:

  1. ${ sec }^{ -1 },M\ { sec }^{ -1 }$

  2. ${ sec }^{ -1 },M$

  3. $M\ { sec }^{ -1 },{ sec }^{ -1 }$

  4. $M,{ sec }^{ -1 }$


Correct Option: A
Explanation:

$K=\frac{dx}{dt[A]} $ for Ist order = $sec^{−1}$
$ K = \frac{dx}{dt}$ for zero order = $mol \space litre^{−1}sec^{−1}$

Hence A is the correct answer.

Consider the reaction, $2A+B\longrightarrow $ Products. When the concentration of $B$ alone was doubled, the half-life did not change. When the concentration of $A$ alone was doubled, the rate increased by two times. The unit of the rate constant for this reaction is:

  1. ${ s }^{ -1 }$

  2. $L$ ${ mol }^{ -1 }{ s }^{ -1 }$

  3. Unitless

  4. $mol$ ${ L }^{ -1 }{ s }^{ -1 }$


Correct Option: B
Explanation:

Concentration change in '$B$' does not change half life, it means the reaction is first order with respect to $B$. When concentration of only '$A$' is doubled, the rate of reaction becomes double, thus order with respect to $A$ will also be one.
Overall order of reaction $=2$
Unit of rate constant $=L$ ${ mol }^{ -1 }{ s }^{ -1 }$