The reaction mechanism for the reaction $P \to R$ is as follows:
$P k_{1} \Leftarrow \Rightarrow k_{2} 2 Q (fast), 2 Q + p k_{3} - - \to R (slow)$ the rate law for the net reaction $(p \to R)$ is:

$k_{1} [P] [Q]$

$k_{1} k_{2} [P]$

$\frac{k_{1} k_{3} {[p]}^{2}}{k_{2}}$

$K_{1} k_{2} [P] [R]$

Video Solution

Text Solution

Verified by Experts

The correct Answer is:C

Answer

Step by step video, text & image solution for The reaction mechanism for the reaction PtoR is as follows: Punderset(k_(2))overset(k_(1))hArr2Q("fast"),2Q+poverset(k_(3))toR("slow") the rate law for the net reaction (ptoR)is: by Chemistry experts to help you in doubts & scoring excellent marks in Class 12 exams.

Updated on:21/07/2023

Class 12 CHEMISTRY CHEMICAL KINETICS

Knowledge Check

Question 1 - Select One
A reaction proceeds by a two step mechanism
$A_{2} k_{1} \Leftarrow \Rightarrow k_{2} 2 A$ (fast reaction)
$A + B \to$ Products (slow reaction)
What is the rate law for the overall reaction ?
Arate $= k [A_{2}] [B]$
Brate $= k {[A_{2}]}_{2}^{2} [B]$
Crate $= k {[A_{2}]}_{2}^{\frac{1}{2}}$
Drate $= k {[A_{2}]}_{2}^{\frac{1}{2}} [B]$
Question 1 - Select One
Conisder the reaction mechanism:
$A_{2} k_{e q} \Leftarrow \Rightarrow 2 A (fast)$ (where $A$ is the intermediate.)
$A + B - - \to k_{1} P (s l o w)$
The rate law for the reaction is
A $k_{1} [A] [B]$
B $k_{1} k^{1 / 2} {[A_{2}]}_{2}^{1 / 2} [B]$
C $k_{1} k^{1 / 2} [A] [B]$
D $k_{1} k^{1 / 2} {[A]}^{2} [B]$
Question 1 - Select One
The rate law expresison is given for a typical reaction, $n_{1} A + n_{2} B \to P$ as $r = k {[A]}^{n} {[B]}^{n 2}$ . The reaction completes only in one step and $A$ and $B$ are present in the solution. If the reaction occurs in more than one step, then the rate law is expressed by consdering the slowest step, i.e., for $S_{N} l$ reaction $r = k [R X]$ . If the eraction occurs in more than one step and the rates of the steps involved are comparable, then steady state approximation is conisdered, i.e., the rate of formation of intermediate is always equal to the rate of decompoistion of the intermediate. Conisder the reaction:
$⎡ ⎢ ⎢ ⎣ \begin{matrix} I_{2} k_{2} \Leftarrow \Rightarrow k_{1} 2 I (rapid equilibrium) H_{2} + 2 I k_{3} - - \to 2 H I (slow) \end{matrix} ⎤ ⎥ ⎥ ⎦$
If we increase the concentration of $I_{2}$ two times, then the rate of formation of $H I$ will
AIncrease four times
BIncrease two times
CRemain same
DCannot predict
Question 1 - Select One
For hypothetical reaction $A \to B$ takes placed according to
$A k_{1} \Leftarrow \Rightarrow C (fast), A + C k_{2} - - \to D (slow)$
Rate law will be :
A $k_{2} [A] [C]$
B $k_{1} k_{2} [A]$
C $k_{1} k_{2} [A]$
D $k_{1} k_{2} [A] [C]$
Question 1 - Select One
The mechanism of the reaction
$2 N O + O_{2} \to 2 N O_{2}$ is
$N O + N O k_{1} \Leftarrow \Rightarrow k_{- 1} N_{2} O_{2} (fast)$
$N_{2} O_{2} + O_{2} k_{2} - - \to 2 N O_{2} (s l o w)$
The rate constant of the reaction is
A $k_{2}$
B $k_{2} k_{1} (k_{- 1})$
C $k_{2} k_{1}$
D $k_{2} (\frac{k_{1}}{k_{- 1}})$
Question 1 - Select One
Consider the reaction mechanism:
$A_{2} k_{e q} \Leftarrow \Rightarrow 2 A (fast)$ (where $A$ is the intermediate.)
$A + B - - \to k_{1} P (s l o w)$
The rate law for the reaction is
A $k_{1} [A] [B]$
B $k_{1} K^{1 / 2} {[A_{2}]}_{2}^{1 / 2} [B]$
C $k_{1} K^{1 / 2} [A] [B]$
D $k_{1} K^{1 / 2} {[A]}^{2} [B]$
Question 1 - Select One
The mechanism of the reaction $A + 2 B \to C + D$ is:
Step I: $A + B k_{1} \Leftarrow \Rightarrow k_{2} I$
Step II: $B + I k_{3} - - \to C + D$
In the first step is a fast equilibruim , then the incorrect statement is :
AOrder of reacton with respect to A is 1.
BOrder of reacton with respect to B is 2.
COverall rate of reaction is , $r = K_{3} . [A] {[B]}^{2}$
DRates of forward and backward reaction of step I is much greater than the rate of reaction of step II.

The reaction mechanism for the reaction $P \to R$ is as follows:
$P k_{1} \Leftarrow \Rightarrow k_{2} 2 Q (fast), 2 Q + p k_{3} - - \to R (slow)$ the rate law for the net reaction $(p \to R)$ is:

The correct Answer is:C

Knowledge Check

A reaction proceeds by a two step mechanism
$A_{2} k_{1} \Leftarrow \Rightarrow k_{2} 2 A$ (fast reaction)
$A + B \to$ Products (slow reaction)
What is the rate law for the overall reaction ?

Conisder the reaction mechanism:
$A_{2} k_{e q} \Leftarrow \Rightarrow 2 A (fast)$ (where $A$ is the intermediate.)
$A + B - - \to k_{1} P (s l o w)$
The rate law for the reaction is

For hypothetical reaction $A \to B$ takes placed according to
$A k_{1} \Leftarrow \Rightarrow C (fast), A + C k_{2} - - \to D (slow)$
Rate law will be :

The mechanism of the reaction
$2 N O + O_{2} \to 2 N O_{2}$ is
$N O + N O k_{1} \Leftarrow \Rightarrow k_{- 1} N_{2} O_{2} (fast)$
$N_{2} O_{2} + O_{2} k_{2} - - \to 2 N O_{2} (s l o w)$
The rate constant of the reaction is

Consider the reaction mechanism:
$A_{2} k_{e q} \Leftarrow \Rightarrow 2 A (fast)$ (where $A$ is the intermediate.)
$A + B - - \to k_{1} P (s l o w)$
The rate law for the reaction is

The mechanism of the reaction $A + 2 B \to C + D$ is:
Step I: $A + B k_{1} \Leftarrow \Rightarrow k_{2} I$
Step II: $B + I k_{3} - - \to C + D$
In the first step is a fast equilibruim , then the incorrect statement is :

Similar Questions

Recommended Questions

The reaction mechanism for the reaction P→R is as follows: Pk1⇐⇒k22Q(fast),2Q+pk3−−→R(slow) the rate law for the net reaction (p→R)is:

The correct Answer is:C

Knowledge Check

A reaction proceeds by a two step mechanism A2k1⇐⇒k22A (fast reaction) A+B→ Products (slow reaction) What is the rate law for the overall reaction ?

Conisder the reaction mechanism: A2keq⇐⇒2A(fast) (where A is the intermediate.) A+B−−→k1P(slow) The rate law for the reaction is

For hypothetical reaction A→B takes placed according to Ak1⇐⇒C(fast),A+Ck2−−→D(slow) Rate law will be :

The mechanism of the reaction 2NO+O2→2NO2 is NO+NOk1⇐⇒k−1N2O2(fast) N2O2+O2k2−−→2NO2(slow) The rate constant of the reaction is

Consider the reaction mechanism: A2keq⇐⇒2A(fast) (where A is the intermediate.) A+B−−→k1P(slow) The rate law for the reaction is

The mechanism of the reaction A+2B→C+D is: Step I: A+Bk1⇐⇒k2I Step II: B+Ik3−−→C+D In the first step is a fast equilibruim , then the incorrect statement is :

Similar Questions

Recommended Questions

The reaction mechanism for the reaction $P \to R$ is as follows:
$P k_{1} \Leftarrow \Rightarrow k_{2} 2 Q (fast), 2 Q + p k_{3} - - \to R (slow)$ the rate law for the net reaction $(p \to R)$ is:

A reaction proceeds by a two step mechanism
$A_{2} k_{1} \Leftarrow \Rightarrow k_{2} 2 A$ (fast reaction)
$A + B \to$ Products (slow reaction)
What is the rate law for the overall reaction ?

Conisder the reaction mechanism:
$A_{2} k_{e q} \Leftarrow \Rightarrow 2 A (fast)$ (where $A$ is the intermediate.)
$A + B - - \to k_{1} P (s l o w)$
The rate law for the reaction is

For hypothetical reaction $A \to B$ takes placed according to
$A k_{1} \Leftarrow \Rightarrow C (fast), A + C k_{2} - - \to D (slow)$
Rate law will be :

The mechanism of the reaction
$2 N O + O_{2} \to 2 N O_{2}$ is
$N O + N O k_{1} \Leftarrow \Rightarrow k_{- 1} N_{2} O_{2} (fast)$
$N_{2} O_{2} + O_{2} k_{2} - - \to 2 N O_{2} (s l o w)$
The rate constant of the reaction is

Consider the reaction mechanism:
$A_{2} k_{e q} \Leftarrow \Rightarrow 2 A (fast)$ (where $A$ is the intermediate.)
$A + B - - \to k_{1} P (s l o w)$
The rate law for the reaction is

The mechanism of the reaction $A + 2 B \to C + D$ is:
Step I: $A + B k_{1} \Leftarrow \Rightarrow k_{2} I$
Step II: $B + I k_{3} - - \to C + D$
In the first step is a fast equilibruim , then the incorrect statement is :