Tag: chemistry

Questions Related to chemistry

The reaction, $CH _3COOC _2H _5+NaOH\rightarrow CH _3COONa+C _2H _5OH$ is:

  1. bimolecular reaction

  2. second order reaction

  3. third order reaction

  4. none of these


Correct Option: A,B
Explanation:

The reaction, $CH _3COOC _2H _5+NaOH\rightarrow CH _3COONa+C _2H _5OH$ is bimolecular and second order reaction,as it contains two diiferent types of molecules.
Rate = $ k$ [$CH _3COOC _2H _5$] [$NaOH$]

For the reaction: $2NO+Cl _2\rightarrow 2NOCl$, the following mechanism was proposed on the basis of experimental observation.
$NO+Cl _2\overset {Fast}{\rightleftharpoons}NOCl _2$
$NOCl _2+NO\xrightarrow {Slow}2NOBr$
The order of reaction is:

  1. 2nd order

  2. pseudo 2nd order

  3. 3rd order

  4. first order


Correct Option: C
Explanation:

From the slow step, the rate law for the reaction is rate $\displaystyle = k' [NOCl _2][NO]$.....(1)


 $\displaystyle NOCl _2$ is an intermediate and its concentration is obtained from fast step (equilibrium step)

 $\displaystyle [NOCl _2] = K[NO][Cl _2]$......(2)

Substitute (2) in (1)
 $\displaystyle rate = k'K [NO][Cl _2][NO]$

Hence, rate $\displaystyle = k [NO]^2[Cl]$

The reaction is of second order in NO and first order in $\displaystyle Cl _2$. The overall reaction order is  $\displaystyle 2 + 1 = 3$

The unit of rate constant for a given reaction is $M^{1-n}L^{n-1}t^{-1}$ where n is order of reaction.

  1. True

  2. False


Correct Option: A
Explanation:

The statement is true. The unit of rate constant for a given reaction is  $M^{1-n}L^{n-1}t^{-1}$ where $n$ is order of reaction.It is generally form for expressing rate constant. Rates are usually given in concentrations units over time units. The units of the rate constants depend upon the order of the reaction. The concentration is molarity and time is in seconds.

A 22.4 litre flask contains 0.76 mm of ozone at $25^oC$. Calculate:
(i) the concentration of oxygen atom needed so that the reaction, $O+O _3\rightarrow 2O _2$ having rate constant equal to $1.5\times 10^7$ litre $mol^{-1} sec^{-1}$ can proceed with a rate of 0.15 mol $litre^{-1} sec^{-1}$
(ii) the rate of formation of oxygen under this condition.

  1. (i) $2.45\times 10^{-4}$, (ii) $0.30 mol L^{-1} t^{-1}$

  2. (i) $4.9\times 10^{-4}$, (ii) $0.60 mol L^{-1} t^{-1}$

  3. (i) $2.45\times 10^{-5}$, (ii) $0.30 mol L^{-1} t^{-1}$

  4. None of these


Correct Option: A
Explanation:

$\displaystyle [O _3] = \frac {P}{RT} = \frac {0.76}{760 \times 0.0821 \times 298} = 4.08 \times 10^{-5} $
$\displaystyle [O] = \frac {rate}{k \times [O _3]} = \frac {0.15}{1.5 \times 10^{7} \times 4.08 \times 10^{-5} = 2.45 \times 10^{-4}  }$
The rate of formation of oxygen under this condition is equal to twice the rate of the reaction.
It is equal to $\displaystyle 2 \times 0.15 : mol litre^{-1} sec^{-1} = 0.30 : mol litre^{-1} sec^{-1} $

The unit and value of rate constant and that of rate of reaction are same for:

  1. first order

  2. zero order

  3. second order

  4. all are wrong


Correct Option: B
Explanation:

As we know,
For zero order reaction, $\frac {dx}{dt}=K$.
so, unit and value of rate constant and that of rate of reaction are same for zero order.

The rate constant of $n^{th}$ order has units:

  1. $Litre^{-1} mol^{1-n} sec^{-1}$

  2. $Mol^{1-n} litre^{1-n} sec^{-1}$

  3. $Mol^{1-n} litre^{n-1} sec^{-1}$

  4. $Mol^{1-n^{2}} litre^{n^{2}} sec^{-1}$


Correct Option: C
Explanation:

As we know,
For nth order; unit of rate constant may be derived by
$K=\frac {rate}{[reactant]^n}$
$=Mol^{1-n} litre^{n-1} sec^{-1}$

A reaction proceeds in three stages. The first stage is slow and involves two molecules of reactants. The second and third stage are fast. The overall order of the reaction is:

  1. first order

  2. second order

  3. third order

  4. zero order


Correct Option: B
Explanation:

The overall order of the reaction is second order.
Reactions are proceeded in many steps, it does not means that no. of step will be equal to order of reaction. But r
ate is determined only by slow step of mechanism.

The rate of the reaction, $A+B+C\rightarrow P$; is given by; $r=-\frac {d[A]}{dt}=K[A]^{1/2}[B]^{1/2}[C]^{1/4}$. The order of the reaction is:

  1. 1

  2. 2

  3. 1/2

  4. 5/4


Correct Option: D
Explanation:

As we know,

For a reaction, $A+B\rightarrow Product$, if rate $=K[A]^m[B]^n$, the order of reaction $=m+n$.

So here for the given reaction, order of reaction $= 1/2+1/2+1/4 = 5/4$

The rate constant of nth order has units:

  1. $Litre^{-1} mol^{1-n} sec^{-1}$

  2. $Mol^{1-n} litre^{1-n} sec^{-1}$

  3. $Mol^{1-n} litre^{n-1} sec^{-1}$

  4. none of the above


Correct Option: C
Explanation:

For nth order; unit of rate constant may be derived by
$rate  = k[a]^n$
$K=\frac {rate}{[reactant]^n}$.

If the concentration is measured in mol $L^{-1}$ and time in minutes, the unit for the rate constant of a third order reaction is:

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

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

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

  4. $min^{-1}$


Correct Option: B
Explanation:

As we know,
$k\, =\, [conc]^{1-n}\, min^{-1}$
For third order reaction = $[mol\, L^{-1}]^{1-3}\, min^{-1}$
$L^2\, mol^{-2}\, min^{-2}$