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Introduction to rate of reaction - class-X

Description: introduction to rate of reaction
Number of Questions: 44
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Tags: chemistry how far? how fast? rates of reaction types of chemical reaction redox reactions and rate of chemical reactions chemical reactions and equations
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For a chemical reaction, $A \rightarrow products$, the rate of reaction doubles when the concentration of A is increased by a factor of 4, the order of reaction is :

  1. 2

  2. 0.5

  3. 4

  4. 1


Correct Option: B
Explanation:
let x be order of reaction and K be rate of constant.
* A$\rightarrow $ products.
$ \cfrac { K\cfrac { { \left( A \right)  }^{ x } }{ time }  }{ K\cfrac { { \left( 4A \right)  }^{ x } }{ time }  } or,\quad \cfrac { 1 }{ 2 } = { \left( \cfrac { A }{ 4A }  \right)  }^{ x }\\ or,\quad \cfrac { 1 }{ 2 } ={ \left( \cfrac { 1 }{ 4 }  \right)  }^{ x }$      (reciprocating)
$\\ or,\quad 2={ \left( 4 \right)  }^{ x }\\ or,\quad 2={ { 2 }^{ 2 } }^{ x }\\ or,\quad 2x=1\\ or,\quad x=\cfrac { 1 }{ 2 } \\ x=0.5$

The term $-\dfrac{dc}{dt}$ in a rate equation refers to:

  1. the concentration of a reactant

  2. the decrease in concentration of the reactant with time

  3. the velocity constant of reaction

  4. none of the above


Correct Option: B
Explanation:

The term $-\dfrac{dc}{dt}$ in a rate expression indicates the decrease in concentration of the reactant with time. It is the minus sign which is used to show the decrease in concentration of the reactant.

For a first order reaction, A$\rightarrow$ products, the concentration of A changes from $0.1$M to $0.025$ M $80$ minutes. The rate of reaction when the concentration of A is $0.01$M, is:

  1. $1.73\times 10^{-5}$M/min

  2. $3.47\times 10^{-4}$M/min

  3. $3.47\times 10^{-5}$M/min

  4. $1.73\times 10^{-4}$M/min


Correct Option: D
Explanation:

K$= \frac { 2.303 }{ t } \log { \left[ \dfrac { { A } _{ 0 } }{ A }  \right]  } $


$\quad =\dfrac { 2.303 }{ 80 } \log [{ \dfrac { 0.1 }{ 0.025 }  }]$

$k= 0.0173\quad { min }^{ -1 }$

$\therefore $ Rate$= K\left[ A \right]$

$\quad \quad \quad =0.0173\left[ 0.01 \right]$

$\quad \quad \quad = 1 .73 \times { 10 }^{ -4 } $

$\quad \quad \quad =1.73\times{ 10 }^{ -4 }M/min$

Hence, the correct option is $(D)$

Which of the following is (are) true for first order reaction?

  1. Rate of reaction is fastest at the beginning of reaction.

  2. Rate of reaction is fastest when (reactants)=(products)

  3. Rate of reaction increases with temperature

  4. Ea decreases considerably as temperature as temperature increases, hence the reaction becomes faster


Correct Option: A,C
Explanation:

A] True. for first order reactions, products are formed at the fastest rate during the  beginning of the reaction.

B] False. When reactants>Products, the rate of reaction is fast.
C] True. Increasing the temperature increases the rate of reaction.
D] False. Activating energy does not decrease, but increasing the temperature allows crossing of energy barrier.

If enthalpy of $B$ is greater than that of $A$, the reaction $A\rightarrow B$ will be:

  1. instantaneous

  2. exothermic

  3. endothermic

  4. spontaneous


Correct Option: C
Explanation:

In an endothermic reaction, the enthalpy of the product is greater than the enthalpy of the reactant. 

Thus, the given reaction is an endothermic reaction.

For the reaction $2A + B + C \rightarrow 2D$. The observed rate law is Rate=$K[A]{ [B] }^{ 2 }$. Correct statements are
a) An increase of cone .of C does not affect the rate
b)Doubling the conc of A doubles the rate
c)Tripling the conc of B increases the rate by 9 times
d)Doubling the conc of C, doubling the rate

  1. a,b,c

  2. b,c,d

  3. d

  4. c,d


Correct Option: A
Explanation:

Given  rate is $k[A]{ [B] }^{ 2 }$

 It is overall ${ 3 }^{ rd }$ order  reaction with a following  first order  and B following second order.
 (a) True  C is not affecting the rate of reaction.
 (b) True $rate _{ 1 }\quad =k[A]{ [B] }^{ 2 }\ rate _{ 2 }\quad =k[2A]{ [B] }^{ 2 }=2k[A]{ [B] }^{ 2 }\ =2\times rate _{ 1 }$
 (c) True $rate _{ 1 }\quad =k[A]{ [B] }^{ 2 }\ rate _{ 3 }=k[A]{ [3B] }^{ 2 }=9Kk[A]{ [B] }^{ 2 }\ =9\times rate _{ 1 }$
(d)False C does not affected  the rate in any manner.

${ SO } _{ 2 }$ react with ${ O } _{ 2 }$ as follows :


 ${ 2SO } _{ 2 }+{ O } _{ 2 }\rightarrow { 2SO } _{ 3}$ 


The rate of disappearance of ${ SO } _{ 2 }$ is $2.4\times { 10 }^{ -4 }$ mol ${ lit }^{ -1 }{ min}^{ -1 }$, then :

  1. Rate of reaction is $1.2\times { 10 }^{ -4 }\quad mole{ lit }^{ -1\quad }{ min }^{ -1 }$

  2. Rate of appearance of ${ SO } _{ 3 }$ is $2.4\times { 10 }^{ -4 } { mole\quad lit }^{ -1 }min^{ -1 }$

  3. Rate of disappearance of ${ O } _{ 2 }$ is $1.2\times { 10 }^{ -4 } { mole\quad lit }^{ -1 }min^{ -1 }$

  4. Rate of reaction is twice the rate of disappearance of ${ SO } _{ 2 }$


Correct Option: B,C,D

Which of the fallowing can enhance the rate of the reaction ?

  1. increasing the temperature

  2. increasing the concentration of products

  3. increasing the activation energy

  4. using a +Ve catalyst


Correct Option: A,D
Explanation:

Solution:- (A) increasing the temperature and (D) using a +ve catalyst
  • Temperature is a measure of the kinetic energy of a system, so higher temperature implies higher average kinetic energy of molecules and more collisions per unit time. Hence an increase in temperature is accompanied by an increase in the reaction rate. 
  • A higher concentration of reactants leads to more effective collisions per unit time, which leads to an increased reaction rate (except for zero order reactions). Similarly, a higher concentration of products tends to be associated with a lower reaction rate.
  • Catalysts (e.g., enzymes) lower the activation energy of a chemical reaction and increase the rate of a chemical reaction without being consumed in the process. 
  • Hence, using a positive catalyst and increasing the temperature, both can enhance the rate of reaction.

By the action of enzymes, the rate of biochemical reaction:

  1. does not change

  2. increases

  3. decreases

  4. Either $a$ or $c$


Correct Option: B
Explanation:

Enzymes are the very efficient catalyst for the enzymatic biochemical reaction and Enzymes speed up the reaction by providing lower activation energy.

Rate of formation of $SO _{3}$ according to the reaction $2SO _{2}+O _{2} \rightarrow 2SO _{3}$ is $1.6 \times 10^{-3}\ kg\ min^{-1}$ Hence rate at which $SO _{2}$ reacts is :-

  1. $1.6 \times 10^{-3}\ kg\ min^{-1}$

  2. $8.0 \times 10^{-4}\ kg\ min^{-1}$

  3. $3.2 \times 10^{-3}\ kg\ min^{-1}$

  4. $1.28\times 10^{-3}\ kg\ min^{-1}$


Correct Option: B

$C _{4}H _{8}\rightarrow 2C _{2}H _{4}$; rate constant $=2.303\times 10^{4}\sec^{-1}$, After what time the molar ratio of $\dfrac{C _{2}H _{4}}{C _{4}H _{8}}$ attain the value $1$

  1. $176\ sec$

  2. $3522\ sec$

  3. $1661\ sec$

  4. $1761\ sec$


Correct Option: A

In which of the following cases, the reaction goes farthest to completion?

  1. $A \rightleftharpoons B (K = 10^3)$

  2. $P \rightleftharpoons Q (K = 10^{-2})$

  3. $A + B \rightleftharpoons C + D (K = 10)$

  4. $X + Y \rightleftharpoons XY _2 (K = 10^{-1})$


Correct Option: A
Explanation:

Higher the equilibrium constant, faster the rate of reaction

$A+B \rightleftharpoons C+D$
$\Rightarrow K=\cfrac {[C][D]}{[A][B]}$
To complete the reaction fastest the numerator of $RHS$ will be higher.
$\therefore$ Option A 
$A\rightleftharpoons B(K=10^3)$ is correct, because equilibrium constant in option A is higher than remaining options.

On increasing the pressure three fold, the rate of reaction of ${ 2H } _{ 2 }{ S }$ + ${ O } _{ 2 }$ $\rightarrow $ products would increase

  1. 3 times

  2. 9 times

  3. 12 times

  4. 27 times


Correct Option: D
Explanation:

Rate$=$ ${K _{{p^2}}}{H _2}S \times P{O _2} = x$ $($rate$)$


On increasing the pressure three fold :


Rate$=$ $K{\left( {3{P _{{H _2}S}}} \right)^2}\left( {3P{O _2}} \right)$

$=$ $K \times 9{P^2} _{{H _2}S} \times 3P{O _2}$

$=$ $K \times 27 \times {P _{{H _2}S}} \times P{O _2} = 27$

the rate will increases $27$ times 

Hence, option $(D)$ is correct answer.

Which of the following is not a valid way to describe the rate of the following reaction?
$A + B + C \rightarrow D + E$

  1. $\dfrac {-\triangle [A]}{\triangle t}$

  2. $\dfrac {-\triangle [B]}{\triangle t}$

  3. $\dfrac {-\triangle [C]}{\triangle t}$

  4. $\dfrac {-\triangle [D]}{\triangle t}$

  5. $\dfrac {-\triangle [E]}{\triangle t}$


Correct Option: D,E
Explanation:

$A+B+C\longrightarrow D+E$

 Rate of reaction is defined as the change in concentration of reactant or product to time.
 Rate $ (R)=\cfrac { -d[A] }{ dt } =\cfrac { -d[B] }{ dt } =\cfrac { -d[C] }{ dt } =\cfrac { d[D] }{ dt } =\cfrac { d[E] }{ dt } $
Therefore, (D)  &  (E) i.e. $ \cfrac { -d[D] }{ dt } & \quad \cfrac { -d[E] }{ dt } $ respectively are not valid ways of describing the rate of the following reaction.

What is the rate-determining step?

  1. The slowest step in the reaction.

  2. The fastest step in the reaction.

  3. The overall rate of the reaction.

  4. A law relating the steps of a reaction.

  5. The step which keeps on changing


Correct Option: A
Explanation:

The slowest step in the mechanism of any reaction is the rate-determining step of the reaction.


Hence, the correct option is A.

Rate of reaction depends upon:

  1. Temperature

  2. Concentration

  3. Catalyst

  4. All of these


Correct Option: D
Explanation:

The rate of reaction depends upon the nature of reactants, the concentration of the reactants, temperature, and catalyst.

So, the correct option is $D$

${H} _{2}(g)+{I} _{2}(g)+51.9\ kilojoules\rightarrow 2HI(g)$
Which of the following can be expected to increase the rate of the reaction given by the equation above?
$I$. Adding some helium gas
$II$. Adding a catalyst
$III$. Increasing the temperature

  1. $I$ only

  2. $III$ only

  3. $II$ only

  4. $II$ and $III$ only

  5. $I,II$ and $III$


Correct Option: D
Explanation:

(i)   Addition of Helium gas will not increase the rate of reaction. It will either remain same or decrease depending upon 

      whether addition is done in constant volume or constant pressure conditions.
(ii)  Adding a catalyst will lower the activation energy and thus increase the value of the rate constant which will increase the rate.
(iii)  Increasing the temperature will also increase the value of rate constant and thus the rate of the reaction will increase.
                       $K=A{ e }^{ -Ea/RT }$
$\therefore $ only (ii) & (iii).

The rate of reaction for a concentrated strong acid with a concentrated strong base is least affected by which of the following?

  1. The use of a catalyst.

  2. A change in temperature.

  3. A change in reactant concentration.

  4. A change in pressure.


Correct Option: A
Explanation:

A change in pressure effects only if there is a difference between the number of moles of gaseous products & the number of moles of gaseous reactant. In short, a change in pressure affects only if there is at least one gas either in products or reactants. 

Reaction could be such as:


$ HCl(aq) + NaOH (aq) \rightarrow  NaCl(aq) + H _2O (l)$

As the reaction between concentrated acid and a base reactant or product takes place, there is least effect of change in pressure on the rate of reaction.

Hence, the correct option is D.


Which factor has no influence on the rate of reaction?

  1. Molecularity

  2. Temperature

  3. Concentration of reactant

  4. Nature of reactant


Correct Option: A

The rate constant of the relation $ A \rightarrow B $ is $ 0.6 \times 10^{-3} $ mole per second. If the concentration of $B$ after $20$ minutes is :

  1. $0.36$ M

  2. $0.72$ M

  3. $1.08$ M

  4. $3.60$ M


Correct Option: B
Explanation:

By unit of rate constant it is clear,that the $reaction$ is zero order. 

$\therefore \left [ B \right ]=k\times t$
         $=0.72M$

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

Negative sign denotes that the concentration of reactant is                     with time.

  1. decreasing

  2. increasing

  3. heating up

  4. cooling up


Correct Option: A
Explanation:

The speed of a chemical reaction may be defined as the change in concentration of a substance divided by the time interval during which this change is observed. Definition of a rate of reaction. We, in general, use a negative sign for reactants because its concentration is decreasing as the chemical reaction is taking place.

In a reaction $2X \rightarrow Y$, the concentration of $X$ decreases from $3.0$ moles/ litre to $2.0\ moles/ litre$ in $5$ minutes. The rate of reaction is :

  1. $0.1\ mol\ L^{-1} min^{-1}$

  2. $5\ mol\ L^{-1} min^{-1}$

  3. $1\ mol\ L^{-1} min^{-1}$

  4. $0.5\ mol\ L^{-1} min^{-1}$


Correct Option: A
Explanation:

$Rate = -\dfrac {1}{2} \dfrac {\triangle [X]}{\triangle t}$
$= -\dfrac {1}{2} \dfrac {(3 - 2)}{5} = -0.1\ mol\ L^{-1} min^{-1}$
Negative sign signifies the decrease in concentration.

The rate law for a reaction, $A + B \rightarrow C + D$ is given by the expression $k[A]$. The rate of reaction will be:

  1. doubled on doubling the concentration of $B$

  2. halved on reducing the concentration of $A$ to half

  3. decreased on increasing the temperature of the reaction

  4. unaffected by any change in concentration of temperature


Correct Option: B
Explanation:

Rate of reaction- The speed at which a chemical reaction proceeds, It is often expressed in terms of either the concentration (amount per unit volume) of a product that is formed in a unit of time or the concentration of a reactant that is consumed in a unit of time.

other terms of expression are produced or consumed $\dfrac{mol}{time}$ and in case of gas we can use pressure term also.
So the correct option is $[B]$

Which of the following expressions is correct for the rate of reaction given below?


$5Br^{-} _{(aq)} + BrO _{3(aq)}^{-} + 6H^{+} _{(aq)} \rightarrow 3Br _{2(aq)} + 3H _{2}O _{(l)}$

  1. $\dfrac {\triangle [Br^{-}]}{\triangle t} = 5\dfrac {\triangle [H^{+}]}{\triangle t}$

  2. $\dfrac {\triangle [Br^{-}]}{\triangle t} = \dfrac {6}{5}\dfrac {\triangle [H^{+}]}{\triangle t}$

  3. $\dfrac {\triangle [Br^{-}]}{\triangle t} = \dfrac {5}{6}\dfrac {\triangle [H^{+}]}{\triangle t}$

  4. $\dfrac {\triangle [Br^{-}]}{\triangle t} = 6\dfrac {\triangle [H^{+}]}{\triangle t}$


Correct Option: C
Explanation:

For the reaction,


$5Br^{-} _{(aq)} + BrO _{3(aq)}^{-} + 6 _{(aq)}^{+} \rightarrow 3Br _{2(aq)} + 3H _{2}O _{(l)}$


Rate of disappearance $= \dfrac {-1}{5}\dfrac {\triangle [Br^{-}]}{\triangle t} = -\dfrac {\triangle [BrO _{3}^{-}]}{\triangle t} = \dfrac {-1}{6} \dfrac {\triangle [H^{+}]}{\triangle t}$

$ \dfrac {\triangle [Br^{-}]}{\triangle t} = \dfrac {5}{6} \dfrac {\triangle [H^{+}]}{\triangle t}$.

Hence, the correct answer is option $\text{C}$.

The rate of reaction usually decreases with time.

  1. True

  2. False


Correct Option: A
Explanation:

The speed of a chemical reaction may be defined as the change in concentration of a substance divided by the time interval during which this change is observed. Hence rate of reaction is inversely related as the time increases rate of reaction decreases. Hence given statement is true.

The rate of a gaseous reaction is given by the expression $k[A]^{2}[B]^{3}$. The volume of the reaction vessel is reduced to one half of the initial volume. What will be the reaction rate as compared to the original rate $a$?

  1. $\dfrac {1}{8}a$

  2. $\dfrac {1}{2}a$

  3. $2a$

  4. $32a$


Correct Option: D
Explanation:

$Rate = k[A]^{2} [B]^{3} = a$


When volume is reduced to one half then conc. of reactants will be doubled.

$Rate = k[2A]^{2}[2B]^{2}$

         $=32 k[A]^{2} [B]^{3} = 32a$.

So, the correct option is $D$

In a reaction, $2X \rightarrow Y$, the concentration of $X$ decreases from $0.50\ M$ to $0.38\ M$ in $10\ min$. What is the rate of reaction in $M\ s^{-1}$ during this interval?

  1. $2\times 10^{-4}$

  2. $4\times 10^{-2}$

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

  4. $1\times 10^{-2}$


Correct Option: A
Explanation:

Rate of reaction $= \dfrac {\triangle [X]}{\triangle t}$


$\triangle [X] = X _{i} - X _{f} = 0.50 - 0.38 = 0.12\ M$

$Rate = \dfrac {0.12}{10\times 60} = 2\times 10^{-4} M\ s^{-1}$.

The rate equation for a reaction is r =  $K[A]^{\circ}[B]^3$. Which of the following statements are true?

  1. Doubling the concentration of B quadruples the rate of reaction

  2. The units of rate constant are mole$^{-2} L^2 S^{-1}$

  3. The plot of concentration of A Vs time is parallel to the time axis

  4. If the volume of the reaction vessel is decreased to $\frac{1}{3}$, the rate of reaction is $ \frac{1^th}{27}$ of the original rate


Correct Option: B,C
Explanation:

$r= K[A]^{o} $ $[B]^3$

The reaction is zero order with respect $A$  and order with respect to $B$ is 3. The overall order being 3, the units of the rate constant are mole $^{-2}$l$^2$ s$^{-1}$. The rate of reaction does not change with a change in concentration of $A$. Therefore, the plot of concentration of $A$ vs time is parallel to a time axis. If the volume of the reaction vessel is decreased to $\frac{1}{3}$, the rate of reaction is 27 times of the original rate.

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

  1. $\dfrac{1}{48}$

  2. $\dfrac{1}{24}$

  3. $\dfrac{9}{16}$

  4. $\dfrac{1}{16}$


Correct Option: D
Explanation:

$ \implies \space \space \space A(g) \space \space \space \space \space \space+ \space \space \space \space \space \space 2B(g) \rightarrow \space \space \space \space \space \space \space \space C(g) + D(g) $
$ At \space t=0 \space0.6 \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space 0.8 \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space 0 \space \space \space \space \space \space \space \space \space \space \space \space \space 0 $
$ At \space t=t \space (0.6-0.2) \space \space \space \space (0.8 - 2 \times 0.2) \space \space 0.2 \space \space \space \space \space \space \space \space \space \space0.2$

$ Rate _i = [0.6][0.8]^2$
$ Rate _t = [0.4][0.4]^2$
$ \dfrac{Rate _t}{Rate _i} = \dfrac{1}{6}$

Which does not affect the rate of a reaction?

  1. Nature of the reactants

  2. Time

  3. Concentrations

  4. Surface area exposed

  5. Temperature


Correct Option: B
Explanation:

Time does not affect the rate of a reaction. Hence, in the rate law expression $ \displaystyle rate = k[A]^n$, there is no term for time.
Nature of the reactants, Concentrations, surface area exposed and temperature affects  the rate of a reaction.

A gaseous phase reaction ${A _2} \to B + \frac{1}{2}C$ shows an increase in pressure from 100 mm to 120 mm in 5 min. Now, $ - \dfrac{{\Delta \left[ {{A _2}} \right]}}{{\Delta t}}$ should be:

  1. $8mm\,{\text{ - }}{\min ^{ - 1}}$

  2. $4mm\,{\text{ - }}{\min ^{ - 1}}$

  3. $16mm\,{\text{ - }}{\min ^{ - 1}}$

  4. $2mm\,{\text{ - }}{\min ^{ - 1}}$


Correct Option: B
Explanation:

Change in pressure $=120-100=20 mmHg$

Change in time $=5 min$
$\therefore$ Rate$=\cfrac{20}{5}mmHgmin^{-1}$
$=4mm\ Hg\ min^{-1}$

Two gases A and B are filled in a container. The experimental rate law for the reaction for the reaction between them has been found to be $Rate = k [A]^2 [B]$. Predict the effect on the rate of the reaction when pressure is doubled?

  1. The rate is doubled

  2. The rate becomes four times

  3. The rate becomes six times

  4. The rate becomes eight times


Correct Option: D
Explanation:

If, $Rate=kx[A]^2[B]^1$

$order=3$.

If pressure is increased by factor of $2$, then rate will be increased by factor of $2^3=8$.

$\therefore $ Rate becomes eight times.

For the reaction A + B $\rightarrow$ products, it is observed that :-
(a) on doubling the initial concentration of A only, the rate of reaction is also doubled and 
(b) on doubling the initial concentrations of both A and B, there is a change by a factor of 8 in the rate of the reaction.

  1. rate = k[A][B]

  2. rate = $k[A]^2$[B]

  3. rate = k[A]$[B]^2$

  4. rate = k$[A]^2[B]^2$


Correct Option: C
Explanation:

The given reaction is $A+B\longrightarrow$ Product

Let us suppose the rate law of this reaction is:-
$Rate=K[A]^a[B]^b$
where $K$ is a rate constant.
$a$ and $b$ are order of the reaction with respect to the reactants $A$ and $B$ respectively.
Given that,
When [A] is doubled, the rate of the reaction is also doubled, so the reaction is first order $w.r.t. A$ and hence $a=1$
When $[A],[B]$ is doubled, the rate of reaction becomes $8$ times. Now,
$(Rate) _{new}=K [2A]^1[2B]^b$          $- (ii)$
$Rate=K[A]^1[B]^b$          $-(iii)$
Now, $\because$ New rate of reaction is $8$ times, so dividing $(ii)$ by $(iii)$ :-
$\Rightarrow 8=2.2^b$
$2^3=2^{1+b}$
Equating the exponents:-
$\Rightarrow 3=1+b\Rightarrow b=2$
So, order of reaction $w.r.t$ to $B$ is $2$
So, $Rate=K[A][B]^2$

The rate of reaction at 273 K is ${ R } _{ 0 }$. The rate of reaction at 313 K will be : (Assuming temperature coefficient equal to 2) 

  1. $16\ { R } _{ 0 }$

  2. $64\ { R } _{ 0 }$

  3. $\dfrac { { R } _{ 0 } }{ 32 }$

  4. $\dfrac { { R } _{ 0 } }{ 16 }$


Correct Option: A
Explanation:
${ k } _{ { r } _{ 1 } }={ R } _{ 0 }$ at ${ T } _{ 1 }=273K,{ k } _{ { r } _{ 2 } }=?$ at ${ T } _{ 2 }=313K$ given $\mu =2$
$\cfrac { { k } _{ { r } _{ 2 } } }{ { k } _{ { r } _{ 1 } } } ={ \mu  }^{ { { T } _{ 2 }-{ T } _{ 1 } }/{ 10 } }\quad \quad \cfrac { { T } _{ 2 }-{ T } _{ 1 } }{ 10 } =\cfrac { 313-273 }{ 10 } =\cfrac { 40 }{ 10 } =4$
$\cfrac { { k } _{ { r } _{ 2 } } }{ { R } _{ 0 } } ={ 2 }^{ 4 }$
${ k } _{ { r } _{ 2 } }=16{ R } _{ 0 }$
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