Tag: rates of reaction

Questions Related to rates 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:

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

  1. $8mm\,{\text{ - }}{\min ^{ - 1}}$

  2. $4mm\,{\text{ - }}{\min ^{ - 1}}$

  3. $16mm\,{\text{ - }}{\min ^{ - 1}}$

  4. $2mm\,{\text{ - }}{\min ^{ - 1}}$

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Correct Option: B
Explanation:

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

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

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?

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

  1. The rate is doubled

  2. The rate becomes four times

  3. The rate becomes six times

  4. The rate becomes eight times

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Correct Option: D
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.

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.

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

  1. rate = k[A][B]

  2. rate = $k[A]^2$[B]

  3. rate = k[A]$[B]^2$

  4. rate = k$[A]^2[B]^2$

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Correct Option: C
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, $Rate=K[A][B]^2$

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) 

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

  1. $16\ { R } _{ 0 }$

  2. $64\ { R } _{ 0 }$

  3. $\dfrac { { R } _{ 0 } }{ 32 }$

  4. $\dfrac { { R } _{ 0 } }{ 16 }$

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Correct Option: A
Explanation:
${ k } _{ { r } _{ 1 } }={ R } _{ 0 }$ at ${ T } _{ 1 }=273K,{ k } _{ { r } _{ 2 } }=?$ at ${ T } _{ 2 }=313K$ given $\mu =2$
$\cfrac { { k } _{ { r } _{ 2 } } }{ { k } _{ { r } _{ 1 } } } ={ \mu  }^{ { { T } _{ 2 }-{ T } _{ 1 } }/{ 10 } }\quad \quad \cfrac { { T } _{ 2 }-{ T } _{ 1 } }{ 10 } =\cfrac { 313-273 }{ 10 } =\cfrac { 40 }{ 10 } =4$
$\cfrac { { k } _{ { r } _{ 2 } } }{ { R } _{ 0 } } ={ 2 }^{ 4 }$
${ k } _{ { r } _{ 2 } }=16{ R } _{ 0 }$