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
If the half-lives of a first-order reaction at $350 \ K$ and $300 \ K$ are $2 \ s$ and $20 \ s$ respectively,the activation energy of the reaction in $kJ \ mol^{-1}$ is:
Reactant $A$ shows two reactions:
$A \xrightarrow{K_1} B$,activation energy $= Ea_1$
$A \xrightarrow{K_2} C$,activation energy $= Ea_2$
If $Ea_1 = \frac{Ea_2}{3}$,then the relation between $K_1$ and $K_2$ is:
$A \xrightarrow{K_1} B$,activation energy $= Ea_1$
$A \xrightarrow{K_2} C$,activation energy $= Ea_2$
If $Ea_1 = \frac{Ea_2}{3}$,then the relation between $K_1$ and $K_2$ is:
Medium
View SolutionThe graph obtained between $\ln k$ ($k=$ Rate constant) on $y$-axis and $1/T$ on $x$-axis is a straight line. The slope of it is $-4 \times 10^4 \ K$. The activation energy of the reaction (in $kJ \ mol^{-1}$) is $(R=8.3 \ J \ K^{-1} \ mol^{-1})$
Two reactions $R_1$ and $R_2$ have identical pre-exponential factors. Activation energy of $R_1$ exceeds that of $R_2$ by $10 \, kJ \, mol^{-1}.$ If $k_1$ and $k_2$ are rate constants for reactions $R_1$ and $R_2$ respectively at $300 \, K,$ then $\ln (k_2/k_1)$ is equal to :
$(R=8.314 \, J \, mol^{-1} \, K^{-1})$
$(R=8.314 \, J \, mol^{-1} \, K^{-1})$
DifficultJEE MAIN 2017
View SolutionFor a first-order reaction $A \rightarrow P$,the rate constant $k$ depends on temperature $(T)$ according to the equation $\log \, k = - \frac{2000}{T} + 0.6$. The pre-exponential factor $A$ and the activation energy $E_a$ are,respectively:
Medium
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