Tag: rates of reaction

Quizzes (1) Questions (44)

Questions Related to rates of reaction

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

A gaseous phase reaction ${A _2} \to B + \frac{1}{2}C$ shows an increase in pressure from 100 mm to 120 mm in 5 min. Now, $ - \dfrac{{\Delta \left[ {{A _2}} \right]}}{{\Delta t}}$ should be:

  1. $8mm\,{\text{ - }}{\min ^{ - 1}}$
  2. $4mm\,{\text{ - }}{\min ^{ - 1}}$
  3. $16mm\,{\text{ - }}{\min ^{ - 1}}$
  4. $2mm\,{\text{ - }}{\min ^{ - 1}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Change in pressure $=120-100=20 mmHg$

Change in time $=5 min$
$\therefore$ Rate$=\cfrac{20}{5}mmHgmin^{-1}$
$=4mm\ Hg\ min^{-1}$

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

Two gases A and B are filled in a container. The experimental rate law for the reaction for the reaction between them has been found to be $Rate = k [A]^2 [B]$. Predict the effect on the rate of the reaction when pressure is doubled?

  1. The rate is doubled

  2. The rate becomes four times

  3. The rate becomes six times

  4. The rate becomes eight times
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

If, $Rate=kx[A]^2[B]^1$

$order=3$.

If pressure is increased by factor of $2$, then rate will be increased by factor of $2^3=8$.

$\therefore $ Rate becomes eight times.

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

For the reaction A + B $\rightarrow$ products, it is observed that :-
(a) on doubling the initial concentration of A only, the rate of reaction is also doubled and 
(b) on doubling the initial concentrations of both A and B, there is a change by a factor of 8 in the rate of the reaction.

  1. rate = k[A][B]
  2. rate = $k[A]^2$[B]
  3. rate = k[A]$[B]^2$
  4. rate = k$[A]^2[B]^2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The given reaction is $A+B\longrightarrow$ Product

Let us suppose the rate law of this reaction is:-
$Rate=K[A]^a[B]^b$
where $K$ is a rate constant.
$a$ and $b$ are order of the reaction with respect to the reactants $A$ and $B$ respectively.
Given that,
When [A] is doubled, the rate of the reaction is also doubled, so the reaction is first order $w.r.t. A$ and hence $a=1$
When $[A],[B]$ is doubled, the rate of reaction becomes $8$ times. Now,
$(Rate) _{new}=K [2A]^1[2B]^b$          $- (ii)$
$Rate=K[A]^1[B]^b$          $-(iii)$
Now, $\because$ New rate of reaction is $8$ times, so dividing $(ii)$ by $(iii)$ :-
$\Rightarrow 8=2.2^b$
$2^3=2^{1+b}$
Equating the exponents:-
$\Rightarrow 3=1+b\Rightarrow b=2$
So, order of reaction $w.r.t$ to $B$ is $2$
So, $Rate=K[A][B]^2$

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

The rate of reaction at 273 K is ${ R } _{ 0 }$. The rate of reaction at 313 K will be : (Assuming temperature coefficient equal to 2) 

  1. $16\ { R } _{ 0 }$
  2. $64\ { R } _{ 0 }$
  3. $\dfrac { { R } _{ 0 } }{ 32 }$
  4. $\dfrac { { R } _{ 0 } }{ 16 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
${ k } _{ { r } _{ 1 } }={ R } _{ 0 }$ at ${ T } _{ 1 }=273K,{ k } _{ { r } _{ 2 } }=?$ at ${ T } _{ 2 }=313K$ given $\mu =2$
$\cfrac { { k } _{ { r } _{ 2 } } }{ { k } _{ { r } _{ 1 } } } ={ \mu  }^{ { { T } _{ 2 }-{ T } _{ 1 } }/{ 10 } }\quad \quad \cfrac { { T } _{ 2 }-{ T } _{ 1 } }{ 10 } =\cfrac { 313-273 }{ 10 } =\cfrac { 40 }{ 10 } =4$
$\cfrac { { k } _{ { r } _{ 2 } } }{ { R } _{ 0 } } ={ 2 }^{ 4 }$
${ k } _{ { r } _{ 2 } }=16{ R } _{ 0 }$