Tag: order of reactions

Questions Related to order of reactions

The inversion of cane sugar to produce glucose and fructose is represented by the reaction
     ${ C } _{ 12 }{ H } _{ 22 }{ O } _{ 11 }+{ H } _{ 2 }O\quad \xrightarrow { { H }^{ + } } { C } _{ 6 }{ H } _{ 12 }{ O } _{ 6 }+{ C } _{ 6 }{ H } _{ 12 }{ O } _{ 6 }$
The reaction is:

chemistry chemical kinetics dependence of reaction rate on concentration of reactants order of reactions factors influencing rate of a reaction

  1. second order

  2. unimolecular

  3. pseudo uni-molecular

  4. bi molecular

Show answer
Correct Option: C
Explanation:

Inversion of Cane Sugar to produce glucose and fructose is a Pseudo uni-molecular Reaction.

Here the rate of reaction is made first order by changing the concentration of sucrose ($C _{12}H _{22}O _{11}$).

Units of the rate constant of first and zero order reactions in terms of molarity M unit are respectively:

chemistry chemical kinetics dependence of reaction rate on concentration of reactants order of reactions factors influencing rate of a reaction

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

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

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

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

Show answer
Correct Option: A
Explanation:

For first order, $M/sec=k[M].$
$\therefore k=sec^{-1}$
For zero order, $M/sec = k[M]^0.$
$\therefore k=M sec^{-1}.$

The unit of rate constant obeying the rate expression $r=K[A]^{1}[B]^{2/3}$ is:

chemistry chemical kinetics dependence of reaction rate on concentration of reactants order of reactions factors influencing rate of a reaction

  1. $mole^{-2/3}lit^{2/3}time^{-1}$

  2. $mole^{2/3}lit^{-2/3}time^{-1}$

  3. $mole^{-5/3}lit^{2/3}time^{-1}$

  4. none of the above

Show answer
Correct Option: A
Explanation:

$r=K[A][B]^{2/3}$


$K=\dfrac{r}{[A][B]^{2/3}}$

$K=\dfrac{mol}{liter.sec}$$\left ( \dfrac{liter}{mole} \right )\left ( \dfrac{liter}{mole} \right )^{2/3}$

$K=mole^{-2/3} liter^{2/3} sec^{-1}$

The reaction, $2A+ B \rightarrow$ Products, follows the mechanism:


$2A \leftrightharpoons A _2$ (at equilibrium)
$A _2 + B \rightarrow$ Products (slow) 

The order of the reaction is:

chemistry chemical kinetics dependence of reaction rate on concentration of reactants order of reactions factors influencing rate of a reaction

  1. $2$

  2. $1$

  3. $3$

  4. $1\dfrac{1}{2}$

Show answer
Correct Option: C
Explanation:

From the slow step, $rate = k [A _2][B]$ ...... (1)
From the equilibrium reaction, equilibrium  constant  $K = \dfrac {[A _2]} {[A]^2}$
$[A _2]=K [A]^2$......(2)


Substitute equation (2) in equation (1).
$rate = k K [A]^2[B]$

Thus the overall order of the reaction is $2+1=3$.

Hence, option C is correct.

For the elementary reaction 2A $\rightarrow $ C ,the concentration of A after 30 minutes was found to be 0.01 mole/lit. If the rate constant of the reaction is $2.5 \times 10^{-2}$ lit mole$^{-1}$ sec$^{-1}$. The rate of the reaction at 30 minutes is:

chemistry chemical kinetics dependence of reaction rate on concentration of reactants order of reactions factors influencing rate of a reaction

  1. $2.5 \times 10^{-4}$ lit mole$^{-1}$ sec$^{-1}$

  2. $2.5 \times 10^{-6}$ lit mole$^{-1}$ sec$^{-1}$

  3. $2.5 \times 10^{-2}$ lit mole$^{-1}$ sec$^{-1}$

  4. $2.5 \times 10^{-8}$ lit mole$^{-1}$ sec$^{-1}$

Show answer
Correct Option: B
Explanation:
Rate of Reaction = Rate Constant  x ${ [A] }^{ 2 }$

 $= 2.5 \times { 10 }^{ -2 } \times   { [0.01] }^{ 2 }$

$ =  2.5  \times   { 10 }^{ -2 }  \times   { 10 }^{ -4 }$

$=  2.5  \times   { 10 }^{ -6 }$  lit   mol $^{ -1 }$ sec$^{ -1 }$

Hence, the correct option is $\text{B}$

The specific rate of a reaction is $1.51 \times10^{-4}$ lit mole$^{-1}$ sec$^{-1}$. If the reaction is commenced with 0.2 mole lit$^{-1}$ of the reactant, the initial rate of the reaction in mole lit$^{-1}$ sec$^{-1}$ is:

chemistry chemical kinetics dependence of reaction rate on concentration of reactants order of reactions factors influencing rate of a reaction

  1. $1.5 \times 10^{-4}$

  2. $3 \times 10^{-5}$

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

  4. $6 \times 10^{-5}$

Show answer
Correct Option: C
Explanation:

For a general reaction of order $n$, the units of rate constant is given as $(mol/litre)^{1-n}sec^{-1}$.
So for our question, the unit of rate constant is $(mol/litre)^{-1}sec^{-1}$ so $ 1-n = -1$.


So, the value of $n =2$, so the reaction is of second order.

The rate is given as $r = k[A]^{2}$. Specific rate of equation is the rate for concentration of 1 $mol/litre$. 

From this, the value of k is $1.51 \times1 0^{-4}$.
$r = 1.51 \times 10^{-4}[A]^{2}$.
[A] =  $0.2 mol/litre$. 

Initial rate is $6 \times1 0^{-6}$.

Hence, option C is correct.

Read the following table and chose the appropriate option


Rate equation Units of K
I) rate $=$ k[A] a) mol lit$^{-1}$ sec $^{-1}$
II) rate $=$ k[A][B] b) mol$^{-2}$ lit$^{2}$ sec $^{-1}$
III) rate $=$ k[A][B]$^2$ c) sec $^{-1}$
IV) rate $=$ k d) lit mol$^{-1}$ sec $^{-1}$

chemistry chemical kinetics dependence of reaction rate on concentration of reactants order of reactions factors influencing rate of a reaction

  1. I - d, II - c, III - a, IV - b

  2. I - c, II - d, III - b, IV - a

  3. I - a, II - b, III - c, IV - d

  4. I - b, II - a, III - d, IV - c

Show answer
Correct Option: B
Explanation:

Units of k $= mole^{1-n} lit^{n-1} sec^{-1}$ , where n is the order of reaction.


For option A, its first order reaction. Rate constant $=sec^{-1}$

For option B, its second order reaction. Rate constant $=mole^{-1} lit sec^{-1}$


For option C, its third order reaction. Rate constant $=mole^{-2} lit^{2} sec^{-1}$

For option D, its zero order reaction. rate constant $=mole/lit sec^{-1}$

Hence, the correct option is $\text{B}$

Identify the reaction order from each of the following rate constants.
(i) $k=2.3 \times 10^{-5} L \quad mol^{-1} \quad s^{-1}$
(ii) $k=3 \times 10^{-4} \quad s^{-1}$

chemistry chemical kinetics dependence of reaction rate on concentration of reactants order of reactions factors influencing rate of a reaction

  1. (i) First order (ii) Second order

  2. (i)Second order (ii) First order

  3. (i) Zero order (ii) First order

  4. None

Show answer
Correct Option: B
Explanation:

(i) Rate of the reaction is given by,

$Rate=k\times [A]^x$, where x is the order of reaction
$(\dfrac{mol}{L})=2.3\times 10^{-5}L(mol^{-1})(s)^{-1}[A]^x$
$({\dfrac{mol}{L}})^2=constant(s^{-1})[\dfrac{mol}{L}]^x$
Therefore, $x=2$ and the reaction is second order
(ii)$Rate=k\times [A]^x$, where x is the order of reaction
$(\dfrac{mol}{L})=3\times 10^{-4}(s)^{-1}[A]^x$
$({\dfrac{mol}{L}})=constant(s^{-1})[\dfrac{mol}{L}]^x$
Therefore, $x=1$ and reaction is first order

Units of rate constant for the first and zero order reactions in terms of molarity M units are respectively:

chemistry chemical kinetics dependence of reaction rate on concentration of reactants order of reactions factors influencing rate of a reaction

  1. $sec^{-1},\;Msec^{-1}$

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

  3. $Msec^{-1},\;sec^{-1}$

  4. None of the above

Show answer
Correct Option: A
Explanation:

The unit of the rate constant K is $\left (\text {Molarity} \right )^{1-n}time^{-1}.$


Here, n is the order of the reaction.

For first order reaction, $n=1.$

The unit of the rate constant K is $\left ( \text{Molarity} \right )^{1-1}sec^{-1}=sec^{-1}.$

For zero order reaction, $n=0.$

The unit of the rate constant K is $\left ( \text{Molarity} \right )^{1-0}sec^{-1}=\left ( \text{Molarity} \right )sec^{-1}=\left (M \right )sec^{-1}.$

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

chemistry chemical kinetics dependence of reaction rate on concentration of reactants order of reactions factors influencing rate of a reaction

  1. $s^{-1}$

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

  3. Unitless

  4.  $mol\ L^{-1}s^{-1}$

Show answer
Correct Option: A
Explanation:
When the concentration of B alone was doubled, the rate did not change. Hence the reaction is zero order in B.

When the concentration of A alone was doubled, the rate increased by two times. Hence, the reaction is of first-order in A.

$r=k\left [ A \right ]^{1}\left [ B \right ]^{0}$

The overall order of the reaction is 1.

For the first-order reaction, the unit of k is $sec^{-1}$.

Hence, the correct option is $\text{A}$