Tag: rate of chemical reaction

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Questions Related to rate of chemical reaction

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
D Correct answer
Multiple choice chemistry rates of reaction introduction to rate of reaction rate of chemical reaction rate of reaction

Negative sign denotes that the concentration of reactant is                     with time.

  1. decreasing

  2. increasing

  3. heating up

  4. cooling up
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The speed of a chemical reaction may be defined as the change in concentration of a substance divided by the time interval during which this change is observed. Definition of a rate of reaction. We, in general, use a negative sign for reactants because its concentration is decreasing as the chemical reaction is taking place.

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

In a reaction $2X \rightarrow Y$, the concentration of $X$ decreases from $3.0$ moles/ litre to $2.0\ moles/ litre$ in $5$ minutes. The rate of reaction is :

  1. $0.1\ mol\ L^{-1} min^{-1}$
  2. $5\ mol\ L^{-1} min^{-1}$
  3. $1\ mol\ L^{-1} min^{-1}$
  4. $0.5\ mol\ L^{-1} min^{-1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$Rate = -\dfrac {1}{2} \dfrac {\triangle [X]}{\triangle t}$
$= -\dfrac {1}{2} \dfrac {(3 - 2)}{5} = -0.1\ mol\ L^{-1} min^{-1}$
Negative sign signifies the decrease in concentration.

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

The rate law for a reaction, $A + B \rightarrow C + D$ is given by the expression $k[A]$. The rate of reaction will be:

  1. doubled on doubling the concentration of $B$
  2. halved on reducing the concentration of $A$ to half

  3. decreased on increasing the temperature of the reaction

  4. unaffected by any change in concentration of temperature
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Rate of reaction- The speed at which a chemical reaction proceeds, It is often expressed in terms of either the concentration (amount per unit volume) of a product that is formed in a unit of time or the concentration of a reactant that is consumed in a unit of time.

other terms of expression are produced or consumed $\dfrac{mol}{time}$ and in case of gas we can use pressure term also.
So the correct option is $[B]$