Tag: rates of reaction

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Questions Related to rates of reaction

Multiple choice chemistry rates of reaction introduction to rate of reaction rate of chemical reaction rate of reaction

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. ${2}^{(n-m)}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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}$

Multiple choice chemistry rates of reaction introduction to rate of reaction rate of chemical reaction rate of reaction

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $K$=$\cfrac {mol L^{-1} s^{-1}}{mol L^{-1} mol L^{-1}}$
$=mol^{-1} L$ $s^{-1}$.

Multiple choice chemistry rates of reaction introduction to rate of reaction rate of chemical reaction rate of reaction

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. $\displaystyle\frac{1}{18}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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
$r _f = K(0.4) (0.4)^2 = 0.064K$
$r _1/r _2 = 0.064/0.384 = 1/6$
Multiple choice chemistry rates of reaction introduction to rate of reaction rate of chemical reaction rate of reaction

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
Reveal answer Fill a bubble to check yourself
C Correct answer
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

Multiple choice chemistry rates of reaction introduction to rate of reaction rate of chemical reaction rate of reaction

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. $- \frac{1}{V} \frac{n _A}{t} = \frac{1}{V} \frac{n _B}{t}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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} $
Multiple choice chemistry rates of reaction introduction to rate of reaction rate of chemical reaction rate of reaction

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}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Decomposition of N2O5 is a known first-order reaction. The rate constant can be determined from the time-volume data using the first-order integrated rate equation.

Multiple choice chemistry rates of reaction introduction to rate of reaction rate of chemical reaction rate of reaction

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

Multiple choice chemistry rates of reaction introduction to rate of reaction rate of chemical reaction rate of reaction

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
Reveal answer Fill a bubble to check yourself
C Correct answer
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
Multiple choice chemistry rates of reaction introduction to rate of reaction rate of chemical reaction rate of reaction

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
Reveal answer Fill a bubble to check yourself
C Correct answer
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
$t = 30\space min$

Multiple choice chemistry rates of reaction introduction to rate of reaction rate of chemical reaction rate of reaction

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 $min^{-1}$
Reveal answer Fill a bubble to check yourself
B Correct answer