Tag: dependence of reaction rate on concentration of reactants

Questions Related to dependence of reaction rate on concentration of reactants

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

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

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

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

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


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:

  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


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:

  1. $2$

  2. $1$

  3. $3$

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


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:

  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$^{-1}$


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:

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

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

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

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


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 $^{-1}$

  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


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) $k=3 \times 10^{-4} \quad s^{-1}$

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

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

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

  4. None


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:

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

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

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

  4. None of the above


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:

  1. $s^{-1}$

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

  3. Unitless

  4.  $mol\ L^{-1}s^{-1}$


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 $\text{A}$

The following mechanism has been proposed for the reaction of $NO$ with $\displaystyle Br _{2}$ to form $NOBr$


$NO(g)+Br _{2}(g)\rightleftharpoons NOBr _{2}(g)$
$NOBr _{2}(g)+NO(g)\rightarrow 2NOBr(g)$


If the second step is the rate determining step,the order of the reaction with respect to $NO(g)$ is:
  1. 2

  2. 1

  3. 0

  4. 3


Correct Option: A