Tag: rate of a chemical reaction

Questions Related to rate of a chemical reaction

When a chemical reaction takes place, during the course of the reaction the rate of reaction?

chemistry reaction kinetics determining rates graphically calculating rate of reaction graphically rate of a chemical reaction

  1. Keeps on increasing with time

  2. Remains constant with time

  3. Keeps on decreasing with time

  4. Shows irregular trend with time

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Correct Option: C
Explanation:

Rate of reaction $\propto$ conc. of reactants
As the reaction proceeds, concentration of the reactants decreases hence the rate also keeps on decreasing with time.

In a reaction $2HI \rightarrow H _{2} + I _{2}$, the concentration of $HI$ decreases from $0.5\ mol\ L^{-1}$ to $0.4\ mol\ L^{-1}$ in $10$ minutes. What is the rate of reaction during this interval?

chemistry reaction kinetics determining rates graphically calculating rate of reaction graphically rate of a chemical reaction

  1. $5\times 10^{-3} M\ min^{-1}$

  2. $2.5\times 10^{-3} M\ min^{-1}$

  3. $5\times 10^{-2} M\ min^{-1}$

  4. $2.5\times 10^{-2} M\ min^{-1}$

Show answer
Correct Option: A
Explanation:

Average rate $= -\dfrac {1}{2} \dfrac {\triangle [R]}{\triangle t} = -\dfrac {1}{2}\times \dfrac {0.4 - 0.5}{10}$


$= \dfrac {1}{2}\times \dfrac {0.1}{10} = 5\times 10^{-3}M\ min^{-1}$.

Which of the following statements is correct?

chemistry reaction kinetics determining rates graphically calculating rate of reaction graphically rate of a chemical reaction

  1. The rate of a reaction decreases with passage of time as the concentration of reactants decreases.

  2. The rate of a reaction is same at any time during the reaction.

  3. The rate of a reaction is independent of temperature change.

  4. The rate of a reaction decreases with increase in concentration of reaction(s).

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Correct Option: A
Explanation:

According to rate law expression,
$Rate \propto [R]$
Thus, rate of a reaction decreases with passage of time as the concentration of reactants decreases.

The rate constant of a zero order reaction is 0.2 mol $d{ m }^{ -3 }{ h }^{ -3 }$. If the concentration minutes is 0.05 mol $d{ m }^{ -3 }$. Then its initial concentration would be:

chemistry reaction kinetics determining rates graphically calculating rate of reaction graphically rate of a chemical reaction

  1. $6.05 mol d{ m }^{ -3 }$

  2. $0.15 mol d{ m }^{ -3 }$

  3. $0.25 mol d{ m }^{ -3 }$

  4. none of the above

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Correct Option: B

Consider the chemical reaction:
$N _2(g)+3H _2(g)\rightarrow 2NH _3(g)$


The rate of this reaction can be expressed; in terms of time and of concentration of $N _2(g), H _2(g)$ $NH _3(g)$. Identify the correct relationship amongst the rate expressions.

chemistry reaction kinetics determining rates graphically calculating rate of reaction graphically rate of a chemical reaction

  1. Rate $=-\dfrac{d[N _2]}{dt}=-\dfrac{1}{3}\dfrac{d[H _2]}{dt}=+\dfrac{1}{2}\dfrac{d[NH _3]}{dt}$

  2. Rate $=-\dfrac{d[N _2]}{dt}=-\dfrac{3d[H _2]}{dt}=\dfrac{2d[NH _3]}{dt}$

  3. Rate $=-\dfrac{d[N _2]}{dt}=-\dfrac{1}{3}\dfrac{d[H _2]}{dt}=\dfrac{d[NH _3]}{dt}$

  4. Rate $=-\dfrac{d[N _2]}{dt}=\dfrac{d[H _2]}{dt}=\dfrac{d[NH _3]}{dt}$

Show answer
Correct Option: A
Explanation:

As the reaction proceed, the concentration of the reactant decrease and that of the product increases.
Thus, the rate of the reaction of the reactant and product can be given as follow:
$Rate =-\cfrac{d}{dt}[N _2]=-\cfrac{1}{3}\cfrac{d}{dt}[H _2]=+\cfrac{1}{2}\cfrac{d}{dt}[NH _3]$

Rate of formation of $SO _3$ in the following reaction $2SO _2+O _2\rightarrow 2SO _3$ is $100g$ $min^{-1}$. 


Then the rate of disappearance of $O _2$ is:

chemistry reaction kinetics determining rates graphically calculating rate of reaction graphically rate of a chemical reaction

  1. $50g$ $min^{-1}$

  2. $40g$ $min^{-1}$

  3. $200g$ $min^{-1}$

  4. $20g$ $min^{-1}$

Show answer
Correct Option: A
Explanation:

$2{ SO } _{ 2 }+{ O } _{ 2 }\rightarrow 2{ SO } _{ 3 }$


rate $=\dfrac { 1 }{ 2 } \dfrac { \left[ { \triangle SO } _{ 2 } \right]  }{ \triangle t } =\dfrac { \triangle \left[ { O } _{ 2 } \right]  }{ \triangle t } =\dfrac { 1 }{ 2 } \dfrac { \triangle \left[ { SO } _{ 3 } \right]  }{ \triangle t } $

rate of formation of ${ SO } _{ 3 }=$ rate of disappearance of ${ O } _{ 2 }$

$\dfrac { 1 }{ 2 } \dfrac { \triangle \left[ { SO } _{ 3 } \right]  }{ \triangle t } =-\dfrac { \triangle \left[ { O } _{ 2 } \right]  }{ \triangle t } $

$-\dfrac { \triangle w _{{ O } _{ 2 }}   }{ \triangle t }=\dfrac { 1 }{ 2 } \times \dfrac{100}{80}\times 32g{ min }^{ -1 }=20g{ min }^{ -1 }$

So, the correct option is $D$

If concentration of reactants is increased by a factor x then the rate constant k becomes:

chemistry reaction kinetics determining rates graphically calculating rate of reaction graphically rate of a chemical reaction

  1. $\ln{\frac{k}{x}}$

  2. $\frac{k}{x}$

  3. $k+x$

  4. $k$

Show answer
Correct Option: D
Explanation:

Rate Constant doesn't depend on the concentration of the reactants.

So, If the conc of reactants is increased by a factor x, then the rate constant k becomes k.

An aqueous solution of $CH _3COOH$ has a pH = $3$ and acid dissociation constant of $CH _3COOH$ is $10^{-5}$. What will be the concentration of acid taken initially?

chemistry reaction kinetics determining rates graphically calculating rate of reaction graphically rate of a chemical reaction

  1. $0.1$M

  2. $0.11$M

  3. $0.09$M

  4. $0.101$M

Show answer
Correct Option: A
Explanation:

${H}^{+} = \sqrt{{K} _{a} \times {C}}$


${10}^{-6} = {10}^{-5} \times {C}$

${C} = 0.1\space M$

For the non-equilibrium process, $A + B \rightarrow Products$, the rate is first order with respect to $A$ and second-order with respect to $B$. If $1.0$ mole each of $A$ and $B$ are introduced into a 1-litre vessel and the initial rate was $1.0 \times 10^{-2}$ mol/litre-sec. The rate (in mol $litre^{-1} sec^{-1}$) when half of the reactants have been used:

chemistry reaction kinetics determining rates graphically calculating rate of reaction graphically rate of a chemical reaction

  1. $1.2 \times 10^{-3}$

  2. $1.2 \times 10^{-2}$

  3. $2.5 \times 10^{-4}$

  4. none of these

Show answer
Correct Option: A
Explanation:

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


Initially 1 mol each of A and B are present and the rate was $1 \times {10}^{-2}\space mol/lit-sec$

When Half of the Reactants are used, the rate becomes $\dfrac{1}{2} \times {[\dfrac{1}{2}]}^{2} \times {10}^{-2}$ = $1.2 \times {10}^{-3}$

 Time  0  5min  10min  15min
 [A]  20mol  18mol  16mol  16 mol

For the reaction $A\longrightarrow Products$; $\frac { -d[A] }{ dt } =k$ and at different time interval, IAI values are given. At $20$ minute, rate will be :

chemistry reaction kinetics determining rates graphically calculating rate of reaction graphically rate of a chemical reaction

  1. $12 mol /min$

  2. $10 mol/min$

  3. $8 mol/min$

  4. $0.4 mol/min$

Show answer
Correct Option: D
Explanation:

The concentration decreases linearly with time

Hence moles of A at $20 min$
$=12$
Therefore, $k=\dfrac{-d[A]}{dt}$
$=-\dfrac{12-14}{20-15}$
$=0.4\ mol/min$