Tag: rates of reaction

Questions Related to rates of reaction

The rate law for a reaction between the substances $A$ and $B$ is given by rate$=k{ \left[ A \right]  }^{ n }{ \left[ B \right]  }^{ m }$. On doubling the concentration of $A$ and having the concentration of $B$ halved, the ratio of the new rate to the earlier rate of the reaction will be as:

  1. $\cfrac { 1 }{ { 2 }^{ m+n } } $

  2. $(m+n)$

  3. $(n-m)$

  4. ${2}^{(n-m)}$


Correct Option: D
Explanation:

Given that

                    $R=K[A]^n[B]^m$
after doubling the concentration of $A$ and concentration of $B$ is halfed 
$R^1=K[2A]^n[\dfrac{B}{2}]^m$
$R^1 =(2)^{n-m} R$
$ \dfrac{R^1}{R}= \dfrac{2^{n-m}}{1}$

The rate equation for the reaction $2A+B \rightarrow C$ is found to be rate = $k[A] [B]$. The correct statement in relation to this reaction is that the :

  1. units of $k$ must be$\ mol^{-1} L$ $s^{-1}$.

  2. $t _{1/2}$ is constant

  3. rate of formation of C is twice the rate of disappearance of A

  4. value of $k$ is independent of the initial concentration of A and B


Correct Option: A
Explanation:

The given reaction is $2A+B\longrightarrow C$

The given rate equation is:-
$rate=K[A] [B]$

The unit of rate is $mol L^{-1} s^{-1}$
Unit of $[A]= mol L^{-1}$
Unit of $[B]= mol L^{-1}$

Unit of $K$=$\cfrac {mol L^{-1} s^{-1}}{mol L^{-1} mol L^{-1}}$
$=mol^{-1} L$ $s^{-1}$.

The reaction $A(g)+2B(g)\rightarrow C(g)+D(g)$ is an elementary process. In an experiment in volving this reaction. The initial pressure of A and B are $P _A=0.6$ atm $P _B=0.8$atm respectively when $P _C=0.2$ atm, the rate of reaction relative to the initial rate is:

  1. $\displaystyle\frac{1}{6}$

  2. $\displaystyle\frac{1}{12}$

  3. $\displaystyle\frac{1}{36}$

  4. $\displaystyle\frac{1}{18}$


Correct Option: A
Explanation:
        $A(g) + 2(B) \rightarrow C(g) + D(g)$
 t = 0  0.6         0.8              0     0
 at t     0.6-x      0.8-x          x       x
since this the elementary reaction
rate,r = $K[B]^2 [A]$
now $r _i = k (0.6)(0.8)^2 = 0.38K$
when $P _i = x - 0.2$ atm
when $P _A= 0.6-x =0.4$ atm
when $P _B= 0.8 - 2x =0.4$ atm
$r _f = K(0.4) (0.4)^2 = 0.064K$
$r _1/r _2 = 0.064/0.384 = 1/6$

In the reaction A + 2B $\longrightarrow $ 2C + D. if the concentration of A is increased four times and B is decreased to half of its initial concentration then the rate becomes:

  1. twice

  2. half

  3. unchanged

  4. one fourth of the rate


Correct Option: C
Explanation:

The given reaction is $A+2B\longrightarrow2C+O$

Rate law is given by:-
$Rate=[A][B]^2$             $- (i)$

Now, if the concentration of $A$ is increased $4$ times & concentration of $B$ is increased $1/2$ of the initial concentration. Then,

$(Rate) _{New}=[4A][B/2]^2$
$=4[A] \cfrac {[B]^2}{4}$
$\Rightarrow (Rate) _{New}= [A] [B]^{2}$       $- (ii)$

$(i)$ & $(ii)\Rightarrow$  Rate is unchanged

If $n _A$ and $n _B$ are the number of moles at any instant in the reaction : $2A _{(g)} \rightarrow 3B _(g)$ carried out in a vessel of $V\ L$, the rate of the reaction at that instant is given by ?

  1. $- \frac{1}{2} \frac{dn _A}{dt} = \frac{1}{3} \frac{dn _B}{dt}$

  2. $- \frac{1}{V} \frac{dn _A}{dt} = \frac{1}{V} \frac{dn _B}{dt}$

  3. $- \frac{1}{2V} \frac{dn _A}{dt} = \frac{1}{3V} \frac{dn _B}{dt}$

  4. $- \frac{1}{V} \frac{n _A}{t} = \frac{1}{V} \frac{n _B}{t}$


Correct Option: C
Explanation:
If $n _A$ and $n _B$ are the number of moles at any instant in the reaction : $2A _{(g)} \rightarrow 3B _(g)$ carried out in a vessel of VL, the rate of the reaction at that instant is given by

$  \displaystyle  - \frac{1}{2} \frac{d[A]}{dt} =+ \frac{1}{3} \frac{d[B]}{dt}$

$ \displaystyle  - \frac{1}{2V} \frac{dn _A}{dt} =+ \frac{1}{3V} \frac{dn _B}{dt}$

Note: 
$  \displaystyle  [A]= \frac{n _A}{V} $
$  \displaystyle  [B]= \frac{n _B}{V} $

The decomposition of ${N} _{2}{O} _{5}$ in ${CCl} _{4}$ solution at 320 K takes place as ${2N} _{2}{O} _{5}\rightarrow{4NO} _{2}+{O} _{2}$; On the bases of given data order and the rate constant of the reaction is :
$\begin{matrix}Time\ in\ mitues&10&15&20&25&\infty\Valume of {O} _{2}&6.30&8.95&11.40&13.50&34.75\end{matrix}$
evolved (in mL)

  1. $1,0.198$ ${min}^{-1}$

  2. $3/2, 0.0198$ ${M}^{-1/2}$ ${min}^{-1}$

  3. $0, 0.0198$ $ {M}$ $ {min}^{-1}$

  4. $1, 0.0198$ $ {min}^{-1}$


Correct Option: A

Consider the reaction  : 
$2H _2(g) + 2NO(g) \rightarrow\  N _2(g) + 2H _2O(g)$
The rate law for this reaction is :
$Rate = k[H _2][NO]^2$
Under what conditions could these steps represent the mechanism?
Step 1 : $2NO(g) \rightleftharpoons  N _2O _2(g)$
Step 2 : $N _2O _2  + H _2 \rightarrow\ N _2O + H _2O$
Step 3 : $N _2O + H _2 \rightarrow\ H _2O + N _2$

  1. These steps can never satisfy the rate law

  2. Step 1 should be the slowest step

  3. Step 2 should be the slowest step

  4. Step 3 should be the slowest step


Correct Option: A
Explanation:

The given reaction is:-

$2H _2(g)+2NO(g)\longrightarrow N _2(g)+2H _2O(g)$

The given rate law is:-
$Rate=K [H _2][NO]^2$

The rate of the chemical reaction is determined by the slowest step. So, in the slowest step we should have $2$ molecules of $NO$ and $1$ molecule of $H _2$ because the rate of the reaction is determined by that.

So, I. $2NO(g)+H _2(g)\longrightarrow N _2(g)+H _2O _2$ (slow)
      II. $H _2O _2+H _2(g)\longrightarrow 2H _2O(g)$ (fast)

This could be the mechanism of the reaction as given by rate law.

How many years it would take to spend Avogadro's number of rupees at the rate of $1$ million rupees in one second?

  1. $19.098\times 10^{19} years$

  2. $19.098\ years$

  3. $19.098\times 10^{9} years$

  4. None of these


Correct Option: C
Explanation:
Rate of sending rupee = 1 million/ sec.
$=1\times 10^{6}/sec$
Total time = $\dfrac{6.022\times 10^{23}}{1\times 10^{6}}$ second
$=6.022\times 10^{17}$ sec = $\dfrac{6.022\times 10^{17}}{3600\times 24\times 365}$ years
$=19.098\times 10^{9}$ year
Option C

In a first order reaction, the concentration of reactant, decrease from 0.8 M to 0.4 M in 15 minutes. The time taken for concentration to change from 0.1 M to 0.025 M is:

  1. 7.5 minutes

  2. 15 minutes

  3. 30 minutes

  4. 60 minutes


Correct Option: C
Explanation:

Its a 1st order reaction,


$k = \dfrac{2.303}{t} log \dfrac{[A]}{[A - x]}$

So,
$k = \dfrac{2.303}{15} log \dfrac{[0.8]}{[0.4]}$

In the 2nd Case,
$k = \dfrac{2.303}{{t}^{1}} log \dfrac{[0.1]}{[0.025]}$

On substituting the value of k, We get
$t = 30\space min$

The decomposition of $N _{2}O _{5}$ in $CCI _{4}$ solution at 320 K takes place as
$2N _{2}O _{5} \rightarrow 4NO _{2} + O _{2}$; On the bases of given data order and the rate constant of the reaction is :

Time in minutes 10 15 20 25 $\infty$
Volume of $O _{2}$ evolved (in mL) 6.30 8.95 11.40 13.50 34.75
  1. 1,0.198 $min^{-1}$

  2. 3/2, 0.0198 $M^{-1/2} min^{-1}$

  3. 0,0.198 $M^{-1/2} min^{-1}$

  4. 1,0.0198 $min^{-1}$


Correct Option: B