Tag: rate of chemical reaction

Questions Related to rate of chemical reaction

Which of the following expressions is correct for the rate of reaction given below?


$5Br^{-} _{(aq)} + BrO _{3(aq)}^{-} + 6H^{+} _{(aq)} \rightarrow 3Br _{2(aq)} + 3H _{2}O _{(l)}$

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

  1. $\dfrac {\triangle [Br^{-}]}{\triangle t} = 5\dfrac {\triangle [H^{+}]}{\triangle t}$

  2. $\dfrac {\triangle [Br^{-}]}{\triangle t} = \dfrac {6}{5}\dfrac {\triangle [H^{+}]}{\triangle t}$

  3. $\dfrac {\triangle [Br^{-}]}{\triangle t} = \dfrac {5}{6}\dfrac {\triangle [H^{+}]}{\triangle t}$

  4. $\dfrac {\triangle [Br^{-}]}{\triangle t} = 6\dfrac {\triangle [H^{+}]}{\triangle t}$

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Correct Option: C
Explanation:

For the reaction,


$5Br^{-} _{(aq)} + BrO _{3(aq)}^{-} + 6 _{(aq)}^{+} \rightarrow 3Br _{2(aq)} + 3H _{2}O _{(l)}$


Rate of disappearance $= \dfrac {-1}{5}\dfrac {\triangle [Br^{-}]}{\triangle t} = -\dfrac {\triangle [BrO _{3}^{-}]}{\triangle t} = \dfrac {-1}{6} \dfrac {\triangle [H^{+}]}{\triangle t}$

$ \dfrac {\triangle [Br^{-}]}{\triangle t} = \dfrac {5}{6} \dfrac {\triangle [H^{+}]}{\triangle t}$.

Hence, the correct answer is option $\text{C}$.

The rate of reaction usually decreases with time.

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

  1. True

  2. False

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Correct Option: A
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. Hence rate of reaction is inversely related as the time increases rate of reaction decreases. Hence given statement is true.

The rate of a gaseous reaction is given by the expression $k[A]^{2}[B]^{3}$. The volume of the reaction vessel is reduced to one half of the initial volume. What will be the reaction rate as compared to the original rate $a$?

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

  1. $\dfrac {1}{8}a$

  2. $\dfrac {1}{2}a$

  3. $2a$

  4. $32a$

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Correct Option: D
Explanation:

$Rate = k[A]^{2} [B]^{3} = a$


When volume is reduced to one half then conc. of reactants will be doubled.

$Rate = k[2A]^{2}[2B]^{2}$

         $=32 k[A]^{2} [B]^{3} = 32a$.

So, the correct option is $D$

In a reaction, $2X \rightarrow Y$, the concentration of $X$ decreases from $0.50\ M$ to $0.38\ M$ in $10\ min$. What is the rate of reaction in $M\ s^{-1}$ during this interval?

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

  1. $2\times 10^{-4}$

  2. $4\times 10^{-2}$

  3. $2\times 10^{-2}$

  4. $1\times 10^{-2}$

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Correct Option: A
Explanation:

Rate of reaction $= \dfrac {\triangle [X]}{\triangle t}$


$\triangle [X] = X _{i} - X _{f} = 0.50 - 0.38 = 0.12\ M$

$Rate = \dfrac {0.12}{10\times 60} = 2\times 10^{-4} M\ s^{-1}$.

The rate equation for a reaction is r =  $K[A]^{\circ}[B]^3$. Which of the following statements are true?

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

  1. Doubling the concentration of B quadruples the rate of reaction

  2. The units of rate constant are mole$^{-2} L^2 S^{-1}$

  3. The plot of concentration of A Vs time is parallel to the time axis

  4. If the volume of the reaction vessel is decreased to $\frac{1}{3}$, the rate of reaction is $ \frac{1^th}{27}$ of the original rate

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Correct Option: B,C
Explanation:

$r= K[A]^{o} $ $[B]^3$

The reaction is zero order with respect $A$  and order with respect to $B$ is 3. The overall order being 3, the units of the rate constant are mole $^{-2}$l$^2$ s$^{-1}$. The rate of reaction does not change with a change in concentration of $A$. Therefore, the plot of concentration of $A$ vs time is parallel to a time axis. If the volume of the reaction vessel is decreased to $\frac{1}{3}$, the rate of reaction is 27 times of the original rate.

The reaction $A(g)+2B(g)\rightarrow C(g)+D(g)$ is an elementary process. In an experiment, the initial partial pressure of $A$ and $B$ are $P _A=0.6$ and $P _B=0.8$ atm when $P _C=0.2$ atm the rate of reaction relative to the initial rate is:

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

  1. $\dfrac{1}{48}$

  2. $\dfrac{1}{24}$

  3. $\dfrac{9}{16}$

  4. $\dfrac{1}{16}$

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Correct Option: D
Explanation:

$ \implies \space \space \space A(g) \space \space \space \space \space \space+ \space \space \space \space \space \space 2B(g) \rightarrow \space \space \space \space \space \space \space \space C(g) + D(g) $
$ At \space t=0 \space0.6 \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space 0.8 \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space 0 \space \space \space \space \space \space \space \space \space \space \space \space \space 0 $
$ At \space t=t \space (0.6-0.2) \space \space \space \space (0.8 - 2 \times 0.2) \space \space 0.2 \space \space \space \space \space \space \space \space \space \space0.2$

$ Rate _i = [0.6][0.8]^2$
$ Rate _t = [0.4][0.4]^2$
$ \dfrac{Rate _t}{Rate _i} = \dfrac{1}{6}$

Which does not affect the rate of a reaction?

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

  1. Nature of the reactants

  2. Time

  3. Concentrations

  4. Surface area exposed

  5. Temperature

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Correct Option: B
Explanation:

Time does not affect the rate of a reaction. Hence, in the rate law expression $ \displaystyle rate = k[A]^n$, there is no term for time.
Nature of the reactants, Concentrations, surface area exposed and temperature affects  the rate of a 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. $2mm\,{\text{ - }}{\min ^{ - 1}}$

Show answer
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, $Rate=K[A][B]^2$