Arrhenius Collision Theory

The activation energy for the reaction $2\mathrm{HI}\left(\mathrm{g}\right)\to {\mathrm{H}}_2\left(\mathrm{g}\right)+{\mathrm{I}}_2\left(\mathrm{g}\right)$  is $209.5{\mathrm{kJmol}}^{-1}$ at $581\mathrm{K}$. The fraction of molecules having energy equal to or greater than activation energy is: $[\mathrm{R}=8.31{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}]$

The factors which influence the rate of reaction are :

Concentration: Greater the concentrations of the reactants, faster is the rate of reaction.

Temperature: The rate of reaction increases with increase in temperature. For most of the reactions, the rate of reaction becomes almost double with 10o rise in temperature.

Presence of catalyst: A catalyst generally increases the speed of a reaction.

Answer the following question :

A small increase in temperature of the reacting system, the rate of reaction exceed to large extend. The most appropriate reason for this is-

In a reversible reaction, the enthalpy change and the activation energy in the forward direction are respectively $-\mathrm{x} \mathrm{kJ} {\mathrm{mol}}^{-1}$ and $\mathrm{y} \mathrm{kJ} {\mathrm{mol}}^{-1}$. Therefore, the energy of activation in the backward direction is

Among the following graphs showing variation of rate constant with temperature ($\mathrm{T}$) for a reaction, the one that exhibits Arrhenius behaviour over the entire temperature range is

Addition of a catalyst at $300 \mathrm{K}$ increases the rate of a chemical reaction by a factor of $36$. By how many $\mathrm{kJ}$, the activation energy of the catalysed pathway is less than the activation energy of the original pathway approximately $\left(\log \left(6\right)=0.78 \mathrm{and} \mathrm{R}=8.3 \mathrm{J} {\mathrm{k}}^{-1}{\mathrm{mol}}^{-1}\right)$.

Two reactions ${\mathrm{A}}_1$ and ${\mathrm{A}}_2$ have identical pre-exponential factors. The activation energy of ${\mathrm{A}}_1$ is more than ${\mathrm{A}}_2$ by $10 \mathrm{kJ} {\mathrm{mol}}^{-1}$ . If ${\mathrm{k}}_1$ and ${\mathrm{k}}_2$ are the rate constants for reactions ${\mathrm{A}}_1$ and ${\mathrm{A}}_2$, respectively at $300 \mathrm{K}$, then $\ln \left(\frac{{\mathrm{k}}_2}{{\mathrm{k}}_1}\right)$ is equal to

$\left(\mathrm{R}=8.314\mathrm{ }\mathrm{J}\mathrm{ }{\mathrm{mol}}^{-1}{\mathrm{K}}^{-1}\right)$

For the given chemical reactions,

$\mathrm{A}\to \mathrm{B}; {\mathrm{k}}_1={10}^{10}{\mathrm{e}}^{-20\mathrm{,}000/\mathrm{T}}$

$\mathrm{C}\to \mathrm{D}; {\mathrm{k}}_2={10}^{12}{\mathrm{e}}^{-24\mathrm{,}606/\mathrm{T}}$

Calculate the temperature $\mathrm{T}$ at which ${\mathrm{k}}_1$ becomes equal to ${\mathrm{k}}_2$.

In Arrhenius equation, $\mathrm{k}={\mathrm{Ae}}^{-{\mathrm{E}}_{\mathrm{a}}/\mathrm{RT}};\mathrm{A}$  may be called the rate constant at:

Activation energy of a chemical reaction can be determined by:

The rate constant of the first-order reaction, i.e., decomposition of ethylene oxide into ${\mathrm{CH}}_4$ and $\mathrm{CO}$ may be described by the following equation: $\log \mathrm{k} \left({\mathrm{s}}^{-1}\right)=14.34-\frac{1.25\times {10}^4}{\mathrm{T}}\mathrm{K}.$

Find the energy of activation (in kJ/mole). Report answer till the nearest integer.

A certain physiologically important first-order reaction has an activation energy equal to $45.0 \mathrm{kJ}/\mathrm{mol}$ at normal body temperature $\left(37°\mathrm{C}\right).$ Without a catalyst, the rate constant for the reaction is $5.0\times {10}^{-4} {\mathrm{s}}^{-1}$. To be effective in the human body, where the reaction is catalysed by an enzyme, the rate constant must be at least  $2.0\times {10}^{-2} {\mathrm{s}}^{-1}$. If the activation energy is the only factor affected by the presence of the enzyme, by how much $\mathrm{kJ}$ must the enzyme lower the activation energy of the reaction to achieve the desired rate?

Report the answer by multiplying the value with 10 and rounding off to two significant figures.

The temperature coefficient of the rate of a reaction is $2.3$. How many times will the rate of the reaction increase if the temperature is raised by $25\mathrm{K}?$ Give answer to the nearest integer.

The rate of a reaction decreased by $3.555$ times when the temperature was changed from $40°\mathrm{C}$ to $30°\mathrm{C}$. The activation energy (in ${\mathrm{KJmol}}^{-1}$) of the reaction is.........(Report the answer in the nearest integer value)

[Take; $\mathrm{R}=8.314 \mathrm{J} {\mathrm{mol}}^{-1} {\mathrm{K}}^{-1}$ and  $\ln$ $3.555=1.268$]

For the reaction, $\mathrm{aA}+\mathrm{bB}\to \mathrm{cC}+\mathrm{dD}$, the plot of $\log \mathrm{k}$ vs $\frac{1}{ \mathrm{T}}$ is given below:

In the $\ln \mathrm{K }v\mathrm{s }\frac{1}{T}$ plot of a chemical process having $∆S^0>0$ and $∆H^o<0$ the slope is proportional to (where $K$ is equilibrium constant)

Collision theory is applicable to 

For a first order chemical reaction,