Difficult
Explain why proper orientation is required for any chemical reaction to occur.
(N/A) According to collision theory,for a chemical reaction to occur,the reactant molecules must collide with sufficient kinetic energy (activation energy) and in the proper orientation.
The proper orientation of colliding molecules facilitates the breaking of bonds between reacting species and the formation of new bonds to yield products.
For example,the reaction between bromomethane $(CH_3Br)$ and the hydroxide ion $(OH^-)$ to form methanol $(CH_3OH)$ depends on the orientation of the reactant molecules. If the $OH^-$ ion approaches the carbon atom from the side opposite to the bromine atom,it results in the formation of products. If it approaches from the same side,no reaction occurs due to steric hindrance and electrostatic repulsion.
The proper orientation of colliding molecules facilitates the breaking of bonds between reacting species and the formation of new bonds to yield products.
For example,the reaction between bromomethane $(CH_3Br)$ and the hydroxide ion $(OH^-)$ to form methanol $(CH_3OH)$ depends on the orientation of the reactant molecules. If the $OH^-$ ion approaches the carbon atom from the side opposite to the bromine atom,it results in the formation of products. If it approaches from the same side,no reaction occurs due to steric hindrance and electrostatic repulsion.
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Similar Questions
Consider the following transformation involving first order elementary reactions in each step at constant temperature as shown below.
$A + B \underset{\text{Step } 3}{\overset{\text{Step } 1}{\rightleftharpoons}} C \xrightarrow{\text{Step } 2} P$
Some details of the above reaction are listed below.
Step Rate constant $(s^{-1})$ Activation energy $(kJ \ mol^{-1})$ $1$ $k_1$ $300$ $2$ $k_2$ $200$ $3$ $k_3$ $Ea_3$
If the overall rate constant of the above transformation $(k)$ is given as $k = \frac{k_1 k_2}{k_3}$ and the overall activation energy $(E_a)$ is $400 \ kJ \ mol^{-1}$,then the value of $Ea_3$ is $\qquad$ $kJ \ mol^{-1}$ (nearest integer).
$A + B \underset{\text{Step } 3}{\overset{\text{Step } 1}{\rightleftharpoons}} C \xrightarrow{\text{Step } 2} P$
Some details of the above reaction are listed below.
| Step | Rate constant $(s^{-1})$ | Activation energy $(kJ \ mol^{-1})$ |
|---|---|---|
| $1$ | $k_1$ | $300$ |
| $2$ | $k_2$ | $200$ |
| $3$ | $k_3$ | $Ea_3$ |
If the overall rate constant of the above transformation $(k)$ is given as $k = \frac{k_1 k_2}{k_3}$ and the overall activation energy $(E_a)$ is $400 \ kJ \ mol^{-1}$,then the value of $Ea_3$ is $\qquad$ $kJ \ mol^{-1}$ (nearest integer).
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Easy
View SolutionFor a chemical reaction at $27^{\circ} C$, the activation energy is $600 R$. The ratio of the rate constants at $327^{\circ} C$ to that of at $27^{\circ} C$ will be
MediumWBJEE 2013
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Difficult
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