Tag: order of reactions

Questions Related to order of reactions

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}$
$-\frac{dc}{dt}=-\frac{1}{V}\frac{dn}{dt}=-\frac{1}{RT}\frac{dP}{dt}$

The rate constant for the reaction is $2 10^{-4} s^{-1}.$ The reaction is :

  1. First order

  2. Second order

  3. Third order

  4. Zero order


Correct Option: A

For a second-order reaction of the type rate=$k[A]^2$ the plot of $\dfrac{1}{[A] _1}$ versus t is linear with a: 

  1. positive slope and zero intercept

  2. positive slope and non-zero intercept

  3. negative slope and zero intercept

  4. negative slope and non-zero intercept


Correct Option: B
Explanation:

Given that for a second order reaction

$Rate= K[A]^2$

We know that rate of the reaction is decrease in the concentration of the reactant with respect to time.

$Rate= \cfrac {-d[A]}{dt}$
$\Rightarrow$ $\cfrac {-d[A]}{dt}=K[A]^2$
$\Rightarrow \cfrac {-d[A]}{[A]^2}=+Kdt$

Integrating both sides we have:-
$\Rightarrow -\int \cfrac {d[A]}{[A]^2}=+\int Kdt$
$\Rightarrow - \left[\cfrac {-1}{[A]}\right]=+Kt+C$
$\Rightarrow Kt+C= \cfrac {1}{[A]}$      $- (i)$

Now, from the equation $(i)$ we can see that graph of $\cfrac {1}{[A]}$ $V/s$ $t$ has a positive slope $K$ and non-zero intercept $C$.