Tag: rate of chemical reaction

Questions Related to rate of chemical reaction

The fraction of collisions that posses the energy $E _a$ is given by:

chemistry how far? how fast? collision theory collision theory of chemical reactions rate of chemical reaction

  1. $f = e^{\frac{-Ea}{RT}}$

  2. $f = e^{\frac{Ea}{RT}}$

  3. $f = e^{- Ea.RT}$

  4. $f = e^{Ea.RT}$

Show answer
Correct Option: A
Explanation:
$k=Pze^{\cfrac {-Ea}{RT}}$
$\cfrac kA$ or $\cfrac k{Pz}= $ Fraction of collision possessing $E _a$
$\therefore$ Fraction of collision possessing energy $E _a= \cfrac kA$
$=\cfrac {Ae^{-E _a/RT}}{A}$
$f=e^{-E _a/RT}$

For a chemical reaction to occur, all of the following must happen except.

chemistry how far? how fast? collision theory collision theory of chemical reactions rate of chemical reaction

  1. A large enough number of collisions must occur

  2. Chemical bonds in the reactants must break

  3. Reactant particles must collide which enough energy for change to occur

  4. Reactant particles must collide with correct orientation

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Correct Option: A
Explanation:
For a chemical reaction to occur, bonds must break in reactant so that new bonds are formed. Moreover, reactants must collide with energy greater than activation energy for change to ocur.
Also, reactant must have proper orientation for reaction.
However, it is not necessary that enough number of collisions must occur.

According to the collision theory, most molecular collisions do not lead to a reaction. Which of the following is(are) necessary for collisions to successfully lead to the reaction?

chemistry how far? how fast? collision theory collision theory of chemical reactions rate of chemical reaction

  1. The total kinetic energy of the collision must be greater than some minimum value.

  2. A catalyst must be present at the collision.

  3. The colliding particles must be properly oriented in space when they collide.

  4. None of the above.

Show answer
Correct Option: A,C
Explanation:

The conditions (A) and (C) are necessary for the collisions to successfully lead to reaction .
For an effective collision, molecules must possess sufficient energy (called activation orientation) and proper orientation. The presence of catalyst is not essential for collision as reaction can occur even in absence of a catalyst.

What does it mean when a collision is elastic?

chemistry how far? how fast? collision theory collision theory of chemical reactions rate of chemical reaction

  1. No energy is gained or lost.

  2. Energy is gained.

  3. Energy is lost.

  4. The particles can stretch out.

  5. The particles slow down.

Show answer
Correct Option: A
Explanation:

(A) , No energy is gained or lost.
An elastic collision is an encounter between two bodies in which the total kinetic energy of the two bodies after the encounter is equal to their total kinetic energy before the encounter. Elastic collisions occur only if there is no net conversion of kinetic energy into other forms

Among the following which will decrease the rate of the reaction?
i. Using highly concentrated reactants
ii. Decreasing the temperature by $25\ K$
iii. Stirring the reactants

chemistry how far? how fast? collision theory collision theory of chemical reactions rate of chemical reaction

  1. i only

  2. ii only

  3. i and iii only

  4. ii and iii only

  5. i, ii, and iii

Show answer
Correct Option: B
Explanation:

$\bullet $  Using highly concentrated reactants will increase the rate of the reaction as rate is directly proportional to the 

    concentration of reactants.
$\bullet $  Decreasing the temperature by 25 K will decrease the rate constant and hence the rate of reaction will decrease.
$\bullet $  Stirring the reactants increases the rate of interaction between the reactants and hence the rate of reaction increases.
$\therefore $ The correct answer is (ii) only.

_________ increases effective collisions without increasing average energy.

chemistry how far? how fast? collision theory collision theory of chemical reactions rate of chemical reaction

  1. An increase in the reactant concentration

  2. An increase in the temperature

  3. A decrease in pressure

  4. Catalysts

  5. $\displaystyle pH$

Show answer
Correct Option: A
Explanation:

An increase in the reactant concentration increase effective collisions without increasing average energy.

This is due to the fact that molecules comes more and more closer, hence, they tends to collide more easily.

In a reaction carried out at 400 k, $0.0001\%$ of the total number of collisions are effective. The energy of activation of the reaction is:

chemistry how far? how fast? collision theory collision theory of chemical reactions rate of chemical reaction

  1. zero

  2. 7.37 k cal/mol

  3. 9.212 k cal/mol

  4. 11.05 k cal/mol

Show answer
Correct Option: D
Explanation:

We know that, Arrhenius equation for calculation of energy of activation of reaction with rate constant $K$ and temeperature $T$ is 

$K=A$ $e^{-Ea/RT}$
where, $Ea$= Arrhenius activation energy
$A$= pre exponential factor (frequency factor)

Now, $e^{-Ea/K _BT}$= Fraction of collision having more than activation energy
where, $K _B=$ Boltzmann constant
Given, $T=400$ $K$       and effective collision= $0.0001$%

$\Rightarrow$ Effective Collision= $e^{-Ea/K _BT}$
$\Rightarrow$ $0.0001$%= $e^{-Ea/1.3\times 10^{-23}\times 400}$
$\Rightarrow$ $10^{-6}$= $e^{-Ea/1.3\times 10^{-23}\times 400}$

$\Rightarrow$ $2.303\times \log 10^{-6}$= $\cfrac {-Ea}{1.3\times 10^{-23}\times 400}$
$\Rightarrow$ $2.303\times (-6)$= $\cfrac {-Ea}{1.3\times 10^{-23}\times 400}$

$\Rightarrow$ $Ea$= $1.3\times 10^{-23}\times 400\times6\times 2.303=7.19\times 10^{-20}$ $J/mol$

For a chemical reaction, $A \rightarrow products$, the rate of reaction doubles when the concentration of A is increased by a factor of 4, the order of reaction is :

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

  1. 2

  2. 0.5

  3. 4

  4. 1

Show answer
Correct Option: B
Explanation:
let x be order of reaction and K be rate of constant.
* A$\rightarrow $ products.
$ \cfrac { K\cfrac { { \left( A \right)  }^{ x } }{ time }  }{ K\cfrac { { \left( 4A \right)  }^{ x } }{ time }  } or,\quad \cfrac { 1 }{ 2 } = { \left( \cfrac { A }{ 4A }  \right)  }^{ x }\\ or,\quad \cfrac { 1 }{ 2 } ={ \left( \cfrac { 1 }{ 4 }  \right)  }^{ x }$      (reciprocating)
$\\ or,\quad 2={ \left( 4 \right)  }^{ x }\\ or,\quad 2={ { 2 }^{ 2 } }^{ x }\\ or,\quad 2x=1\\ or,\quad x=\cfrac { 1 }{ 2 } \\ x=0.5$

The term $-\dfrac{dc}{dt}$ in a rate equation refers to:

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

  1. the concentration of a reactant

  2. the decrease in concentration of the reactant with time

  3. the velocity constant of reaction

  4. none of the above

Show answer
Correct Option: B
Explanation:

The term $-\dfrac{dc}{dt}$ in a rate expression indicates the decrease in concentration of the reactant with time. It is the minus sign which is used to show the decrease in concentration of the reactant.

For a first order reaction, A$\rightarrow$ products, the concentration of A changes from $0.1$M to $0.025$ M $80$ minutes. The rate of reaction when the concentration of A is $0.01$M, is:

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

  1. $1.73\times 10^{-5}$M/min

  2. $3.47\times 10^{-4}$M/min

  3. $3.47\times 10^{-5}$M/min

  4. $1.73\times 10^{-4}$M/min

Show answer
Correct Option: D
Explanation:

K$= \frac { 2.303 }{ t } \log { \left[ \dfrac { { A } _{ 0 } }{ A }  \right]  } $


$\quad =\dfrac { 2.303 }{ 80 } \log [{ \dfrac { 0.1 }{ 0.025 }  }]$

$k= 0.0173\quad { min }^{ -1 }$

$\therefore $ Rate$= K\left[ A \right]$

$\quad \quad \quad =0.0173\left[ 0.01 \right]$

$\quad \quad \quad = 1 .73 \times { 10 }^{ -4 } $

$\quad \quad \quad =1.73\times{ 10 }^{ -4 }M/min$

Hence, the correct option is $(D)$