Tag: chemistry

Questions Related to chemistry

For a particular $A+B \rightarrow C$ was studied at $25^{\circ}C$. The following results are obtained.


              [A]              [B]           [C]
    (mole/lit)       (moles/lit)  (mole  lit $^{-1} sec^{-2}$)  
$9 \times 10^{-5}$ $1.5 \times 10^{-2}$           $0.06$
$9 \times 10^{-5}$ $3 \times 10^{-3}$            $0.012$
$3 \times 10^{-5}$ $3 \times 10^{-3}$            $0.004$
$6 \times 10^{-5}$            x           $0.024$


Then the value of x is :

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

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

  3. $4.5 \times 10^{-3} moleslitre^{-1}$

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


Correct Option: D
Explanation:
$A+B\rightarrow C$

$ rate=k\left[ A \right] \left[ B \right] $

$Experiment \  3\& 2 \  chosen \  for \  value \  of \  k \  as\left[ B \right] is \  same \  in \  both$ 

$\dfrac { { r } _{ 3 } }{ { r } _{ 2 } } =\dfrac { 0.004 }{ 0.012 } =k\dfrac { \left[ { 3\times 10 }^{ -5 } \right] \left[ { 3\times 10 }^{ -3 } \right]  }{ \left[ { 9\times 10 }^{ -5 } \right] \left[ { 3\times 10 }^{ -3 } \right]  } $

$k=1 \ using \  this \  rate \  constant \  value \  in \  finding \  x\\$
$ \dfrac { { r } _{ 4 } }{ { r } _{ 3 } } =\dfrac { 0.024 }{ 0.004 } =k\dfrac { \left[ { 6\times 10 }^{ -5 } \right] \left[ x \right]  }{ \left[ { 3\times 10 }^{ -5 } \right] \left[ { 3\times 10 }^{ -3 } \right]  } $

$\\ \left[ x \right] ={ 9\times 10 }^{ -3 }\\ $

Compound $A$ and $B$ react to form $C$ and $D$ in a reaction that was found to be second-order over all and second-order in $A$. The rate constant -at ${ 30 }^{ 0 }C$ is $0.622$ L ${ mol }^{ -1 }{ min }^{ -1 }$. What is the half-life of A when $4.10\times { 10 }^{ -2 }$ M of A is mixed with excess $B$?

  1. $40$ min

  2. $39.21$ min

  3. $28.59$ min

  4. None of these


Correct Option: B
Explanation:

$A+B\longrightarrow C+D$


 rate$=k{ [A] }^{ 2 }$ (given)

$ =0.622{ [4.10\times { 10 }^{ -2 }] }^{ 2 }$

$ =0.001$  is the rate of reaction initially

 Half-life$={ t } _{ 1/2 }=\cfrac { 1 }{ K[A] } =\cfrac { 1 }{ 0.622\times [4.1\times { 10 }^{ -2 }] } \\ =0.3921\times { 10 }^{ 2 }\\ =39.21\quad minutes.$

The decomposition of dimethyl ether leads to the formation of $CH _4, H _2$ and CO and the reaction rate is given by $Rate=k[CH _3OCH _3]^{\frac {3}{2}}$
The rate of reaction is followed by increase in pressure in a closed vessel, so the rate can also be expressed in terms of the partial pressure of dimethyl ether, i.e., $Rate=k(P _{CH _3OCH _3})^{\frac {3}{2}}$
If the pressure is measured in bar and time in minutes, then the unit of rate constants is:

  1. $bar^{\frac {1}{2}} min$

  2. $bar^{\frac {3}{2}} min^{-1}$

  3. $bar^{-\frac {1}{2}} min^{-1}$

  4. $bar min^{-1}$


Correct Option: C
Explanation:

As $Rate=k(P _{CH _3OCH _3})^{\frac {3}{2}}$
$bar/min=k(bar)^{\frac {3}{2}}$
$\therefore$ unit of k$=bar^{-\frac {1}{2}}min^{-1}$

Taking the reaction, $A + 2B\rightarrow Products$, to be of the second order, which of the following may be the correct rate law expressions?

  1. $\frac {dx}{dt}=k[A][B]$

  2. $\frac {dx}{dt}=k[A][B]^2$

  3. $\frac {dx}{dt}=k[A]^2$

  4. $\frac {dx}{dt}=k _1[A]+k _2[B]^2$


Correct Option: A,C
Explanation:

option A and C are correct as the sum of their exponents equals to 2

The rate constant of a second order reaction is $10^{-2} lit.mole ^{-1}.sec^{-1}$. The rate constant when expressed as $cc. \ molecule^{-1} .\ min^{-1}$ is:

  1. $9.96\times 10^{-22}$

  2. $9.96\times 10^{-23}$

  3. $9.96\times 10^{-21}$

  4. $9.96\times 10^{-24}$


Correct Option: A
Explanation:

The rate constant of a second order reaction is $10^{-2} lit.mole ^{-1}.sec^{-1}$.
$1L=1000cc$


$1 mole = 6.023\times 10^{23}$molecules
$1min=60sec$
Hence, rate constant $=10^{-2} lit.mole ^{-1} sec^{-1}\times \dfrac {1000cc}{1L} \times \dfrac {1mole} {6.023\times 10^{23}molecules} \times \dfrac {60 sec} {1 min}$
$=9.96\times 10^{-22}cc\ molecule^{-1} \ min^{-1}$

In a certain reaction, 10% of the reactant decomposes in one hour, 20% in two hours, 30% in three hours and so on. Dimension of the velocity constant are:

  1. hour$^{-1}$

  2. mole litre$^{-1}$ hour$^{-1}$

  3. litre mol$^{-1}$ hour$^{-1}$

  4. mole sec$^{-1}$


Correct Option: B
Explanation:

For the zero-order reaction, the time taken for the decomposition of the reactant is independent of initial concentration which is the case here. 


The dimension of the velocity constant for the zero-order reaction is = mole litre$^{-1}$ hour$^{-1}$

Option B is correct.

When ethyl acetate was hydrolysed in presence of 0.1 N HCl, the rate constant was found to be $5.40\times 10^{-5}sec^{-1}$. But when 0.1 N $H _2SO _4$ was used for hydrolysis, the rate constant was found to be $6.25\times 10^{-5} sec^{-1}$. Thus, it may be concluded that:

  1. $H _2SO _4$ is stronger than HCl

  2. $H _2SO _4$ is weaker than HCl

  3. $H _2SO _4$ and HCl both have the same strength

  4. the data are not sufficient to compare the strength of $H _2SO _4$ and HCl


Correct Option: A
Explanation:

Option (A) is correct.  $H _2SO _4$ is stronger than $HCl$.
Relative strength in favour of $H _2SO _4$
$=\frac {\text {rate constant of reaction catalysted by }H _2SO _4}{\text {rate constant of reaction catalysted by HCl}}$.

The rate constant (K) for the reaction, $2A+B\rightarrow$ Product was found to be $2.5\times 10^{-5}$ litre $mol^{-1} sec^{-1}$ after 15 sec, $2.60\times 10^{-5} litre\  mol^{-1} sec^{-1}$ after 30 sec and $2.55\times 10^{-5} litre \ mol^{-1}sec^{-1}$ after 50 sec. The order of reaction is:

  1. $2$

  2. $3$

  3. $0$

  4. $1$


Correct Option: A
Explanation:

 The order of reaction is 2.


K does not change with time; thus its value remains unchanged during the reaction.

By seeing the equation, it can be said that reaction is a third order but the unit of K suggest it to be 2nd order.

Assertion : In a second-order reaction with respect to A, when you double [A], the rate is quadrupled.
Reason : The rate equation is $\displaystyle r={ k\left[ A \right]  }^{ 2 }$ for such a reaction.

  1. Both Assertion and Reason are true and Reason is the correct explanation of Assertion

  2. Both Assertion and Reason are true but Reason is not the correct explanation of Assertion

  3. Assertion is true but Reason is false

  4. Assertion is false but Reason is true

  5. Both Assertion and Reason are false


Correct Option: A
Explanation:

Rate equation is as follows:
$Older Rate = k [A]^{2}$                       equation -1
This shows that reaction is $2nd $ order.
So if we double the concentration of the A , then new reaction rate can be written as follows:
$New Rate = k [2A]^{2}$
$New Rate = k\times 4[A]^{2}$
$New Rate = 4 k [A]^{2}$                     equation -2
Now, comparing equation -1 and 2 we get :
New Rate = $4\times Older Rate$


For a gaseous reaction, $A\left( g \right) \longrightarrow $ Product, which one of the following is correct relation among $\dfrac { dP }{ dt } ,\dfrac { dn }{ dt }$ and $\dfrac { dc }{ dt } $?
($\dfrac { dP }{ dt } =$ Rate of reaction in $atm$ ${ sec }^{ -1 }$; $\dfrac { dc }{ dt } =$ Rate of reaction in molarity ${ sec }^{ -1 }$; $\dfrac { dn }{ dt } =$ Rate of reaction in $mol$ ${ sec }^{ -1 }$)

  1. $\dfrac { dc }{ dt } =\dfrac { dn }{ dt } =-\dfrac { dP }{ dt } $

  2. $-\dfrac { dc }{ dt } =-\dfrac { 1 }{ V } \dfrac { dn }{ dt } =-\dfrac {1}{RT}\dfrac{ dP }{ dt } $

  3. $\dfrac { dc }{ dt } =\dfrac { V }{ RT } \dfrac { dn }{ dt } =\dfrac { dP }{ dt } $

  4. None of the above


Correct Option: B
Explanation:
$PV=nRT$
$\frac{dP}{dt}.V=RT.\frac{dn}{dt}$=> $\frac{1}{V}\frac{dn}{dt}=\frac{1}{RT}\frac{dP}{dt}$-(1)
Now, PV=nRT or $P=\frac{n}{V}RT$
Concentration $c=\frac{n}{V}$
P=cRT
$\frac{dP}{dt}=\frac{dc}{dt}.RT$=> \frac{dc}{dt}=\frac{1}{RT}\frac{dP}{dt}$
$-\frac{dc}{dt}=-\frac{1}{V}\frac{dn}{dt}=-\frac{1}{RT}\frac{dP}{dt}$