Collision theory, Energy of activation and Arrhenius equation Questions in English

(A) Using the Arrhenius equation: $\ln(\frac{K_1}{K_2}) = \frac{E_a}{R} (\frac{1}{T_2} - \frac{1}{T_1})$
Given: $T_1 = 300 \, K$,$T_2 = 280 \, K$,$E_a = 1.157 \times 10^4 \, cal \, mol^{-1}$,$R = 1.987 \, cal \, K^{-1} \, mol^{-1}$.
$\ln(\frac{K_1}{K_2}) = \frac{1.157 \times 10^4}{1.987} (\frac{1}{280} - \frac{1}{300})$
$\ln(\frac{K_1}{K_2}) = 5822.85 \times (\frac{20}{84000}) = 5822.85 \times 0.000238 \approx 1.386$
Since $\ln(4) \approx 1.386$,we have $\frac{K_1}{K_2} = 4$,which implies $K_1 = 4 K_2$ or $K_2 = 0.25 K_1$.

(C) catalyst provides an alternative reaction pathway with a lower activation energy.
In a reversible reaction,it lowers the activation energy for both the forward and backward reactions equally,thereby increasing the rate of both reactions without affecting the equilibrium constant.

(A) The correct answer is $(A)$.
$A$ catalyst increases the rate of both forward and backward reactions equally by providing an alternative reaction pathway with a lower activation energy.
As a result,the system reaches the state of equilibrium more quickly.
It does not change the position of the equilibrium or the equilibrium constant.

(C) The heat of reaction $(\Delta H)$ is defined as the difference between the activation energy of the forward reaction $(E_{a,f})$ and the activation energy of the backward reaction $(E_{a,b})$.
$\Delta H = E_{a,f} - E_{a,b}$
Given $E_{a,f} = 19 \ kJ/mol$ and $E_{a,b} = 9 \ kJ/mol$.
$\Delta H = 19 \ kJ/mol - 9 \ kJ/mol = 10 \ kJ/mol$.
Therefore,the correct option is $C$.

(A) The temperature coefficient is given as $2$ for a $10^\circ C$ rise in temperature.
If the temperature is increased by $\Delta T = 50^\circ C$,the number of $10^\circ C$ intervals is $n = \frac{50}{10} = 5$.
The rate of reaction increases by a factor of $2^n$.
Therefore,the increase in velocity is $2^5 = 32$ times.

(A) For many reactions,the rate constant approximately doubles or triples when the temperature is increased by $10\,^{\circ}C$. This factor is known as the temperature coefficient. Therefore,the most appropriate answer is $2$.

(B) The temperature coefficient is given as $2$ for every $10^\circ C$ rise.
The number of $10^\circ C$ intervals is calculated as: $\Delta n = \frac{100^\circ C - 10^\circ C}{10^\circ C} = \frac{90}{10} = 9$.
The increase in reaction rate is given by the formula: $\text{Rate increase} = (2)^{\Delta n}$.
Substituting the value: $\text{Rate increase} = 2^9 = 512$.

(B) Catalyst increases the rate of a chemical reaction by providing an alternative pathway with a lower activation energy $(E_a)$.
By decreasing the activation energy,a larger fraction of reactant molecules can cross the energy barrier at a given temperature,thereby increasing the reaction rate.

(A) For every $10 \ K$ rise in temperature,the rate constant of a reaction approximately doubles.
Given that the temperature increases from $290 \ K$ to $310 \ K$,which is a rise of $20 \ K$.
This corresponds to two intervals of $10 \ K$ each.
Therefore,the rate constant will increase by a factor of $2^2 = 4$.
New rate constant $= 3.2 \times 10^{-3} \times 4 = 12.8 \times 10^{-3} = 1.28 \times 10^{-2}$.

(C) The temperature coefficient of a reaction is defined as the ratio of the rate constants of the reaction at two temperatures differing by $10\,^{\circ}C$.
Usually,it is taken as the ratio of rate constants at $35\,^{\circ}C$ and $25\,^{\circ}C$.
Mathematically,$\text{Temperature coefficient} = \frac{K_{T+10}}{K_T} = \frac{K_{35\,^{\circ}C}}{K_{25\,^{\circ}C}} \approx 2$ to $3$ for most reactions.

(A) The rate of a reaction is given by the expression $r = k[reactant]^n$.
According to the Arrhenius equation,the rate constant $k$ is given by $k = Ae^{-E_a/RT}$.
As the temperature $T$ increases,the term $e^{-E_a/RT}$ increases,which leads to an increase in the rate constant $k$.
Since the rate $r$ is directly proportional to the rate constant $k$,the rate of the reaction increases with an increase in temperature.

(D) The correct answer is $D$.
Catalysts provide an alternative reaction pathway with a lower activation energy $(E_a)$.
By lowering the activation energy,the rate of reaction increases.
Catalysts do not change the enthalpy of reaction $(\Delta H)$ or the equilibrium constant $(K_{eq})$ of the reaction.

(B) The rate constant of a reaction generally doubles for every $10 \ K$ rise in temperature.
Given that at $T_1 = 290 \ K$,the rate constant $k_1 = 3.2 \times 10^{-3}$.
For a temperature increase of $10 \ K$ to $T_2 = 300 \ K$,the rate constant $k_2$ becomes approximately double the initial value.
$k_2 = 2 \times k_1$
$k_2 = 2 \times 3.2 \times 10^{-3} = 6.4 \times 10^{-3}$.

(B) The rate of reaction generally increases with an increase in temperature for most reactions due to an increase in the number of effective collisions. However,the effect can be complex depending on the reaction mechanism. For a simple elementary reaction,the rate constant $k$ increases with temperature according to the Arrhenius equation,$k = Ae^{-E_a/RT}$. While the rate of reaction typically increases,the statement that it 'may increase or decrease' is sometimes used in specific contexts involving complex mechanisms or reversible reactions where equilibrium shifts significantly. However,in standard chemical kinetics,the rate of reaction is primarily expected to increase with temperature.

(A) The Arrhenius equation is given by $K = A e^{-E_a / RT}$.
From this equation,it is clear that the rate constant $K$ is a function of temperature $T$,where $A$ is the frequency factor and $E_a$ is the activation energy.

(A) According to the Arrhenius equation: $k = A e^{-E_a / (RT)}$.
The rate constant $k$ is independent of the concentration of reactants.
It depends only on the frequency factor $A$,activation energy $E_a$,and temperature $T$.
Increasing the temperature $T$ increases the value of the rate constant $k$.

(C) The rate constant $(k)$ of a reaction is a characteristic property that depends only on the temperature of the system,as described by the Arrhenius equation,$k = Ae^{-E_a/RT}$. It is independent of the concentration of reactants or the pressure of the system.

(C) According to the Arrhenius equation,
$K = A e^{-E_a/RT}$
where $K$ is the rate constant,$A$ is the frequency factor,$E_a$ is the activation energy,$R$ is the gas constant,and $T$ is the temperature.
Since $A$ and $E_a$ are constants for a given reaction,the rate constant $K$ depends only upon the temperature of the system.

(D) The specific rate constant $(k)$ of a reaction is independent of the concentration of reactants or products and the time of the reaction.
It is a characteristic constant for a given reaction at a specific temperature.
According to the Arrhenius equation,$k = A e^{-E_a / RT}$,the rate constant depends significantly on the temperature of the reaction.

(B) According to the Arrhenius theory and Maxwell-Boltzmann distribution,as temperature increases,the fraction of molecules possessing energy greater than or equal to the activation energy $(E_a)$ increases significantly.
This leads to a larger number of effective collisions,which increases the rate of the reaction.
Therefore,the correct option is $(b)$.

(B) According to the collision theory,for a reaction to occur,molecules must collide with sufficient energy (activation energy) and proper orientation.
Therefore,not all collisions are effective.
Only those collisions where molecules possess energy equal to or greater than the activation energy $(E_a)$ and have the correct orientation lead to product formation.
Thus,statement $(B)$ is incorrect.

(B) According to the collision theory,the rate of a reaction is directly proportional to the number of collisions per unit time (collision frequency).
An increase in collision frequency leads to an increase in the number of effective collisions,thereby increasing the rate of the reaction.

(C) . According to the collision theory,the rate of reaction increases with temperature primarily because a larger fraction of molecules possess energy equal to or greater than the activation energy $(E_a)$.
While the collision frequency also increases,the exponential increase in the number of molecules with energy $\ge E_a$ is the dominant factor.

(A) According to the Arrhenius equation,$k = A e^{-E_a / RT}$.
As the reaction rate decreases,the rate constant $k$ decreases.
For a given temperature $T$,a decrease in $k$ implies an increase in the activation energy $E_a$.
Therefore,the free energy of activation is higher.

(A) The rate of a reaction increases with a rise in temperature primarily because a larger fraction of molecules possess kinetic energy equal to or greater than the activation energy $(E_a)$.
According to the Arrhenius equation,$k = A e^{-E_a/RT}$,the rate constant $k$ increases exponentially with temperature $T$.
While the number of collisions increases slightly,the most significant factor is the increase in the number of molecules having energy $\ge E_a$,which effectively lowers the energy barrier requirement for the reaction to proceed at a higher rate.

(D) The number of collisions in a chemical reaction depends on the frequency of collisions,which is influenced by the concentration of reactants,the pressure of the system (for gases),and the temperature of the system. Therefore,all the given factors affect the number of collisions. Hence,the correct option is $D$.

(B) For an exothermic reaction,the energy of the products is lower than the energy of the reactants.
The enthalpy change of the reaction is given by $\Delta H = E_f - E_r$.
Since the reaction is exothermic,$\Delta H < 0$.
Therefore,$E_f - E_r < 0$,which implies $E_f < E_r$.

(C) According to the Arrhenius theory,activation energy $(E_a)$ is defined as the additional energy that the reactant molecules must acquire over and above their average kinetic energy to reach the threshold energy required for an effective collision.
Therefore,the correct statement is that it is the energy the molecules have to acquire further so that they can enter into an effective collision.

(B) $Activation \ energy$ is the additional energy required by reactant molecules to reach the $Threshold \ energy$ level.
Mathematically,$E_a = E_{threshold} - E_{average}$.

(D) The Arrhenius equation is given by $k = Ae^{-E_a/RT}$.
Taking the natural logarithm on both sides: $\ln k = \ln A - \frac{E_a}{RT}$.
Converting to base $10$: $\log_{10} k = \log_{10} A - \frac{E_a}{2.303 RT}$.
Option $D$ $(k = AE^{-RT})$ is incorrect as it does not represent the exponential decay relationship with activation energy $E_a$ and temperature $T$.

(D) The rate of a chemical reaction increases with an increase in temperature primarily because the fraction of molecules possessing energy greater than the activation energy $(E_a)$ increases.
This leads to an increase in the number of effective collisions per unit time,thereby increasing the reaction rate.

(A) The activation energy $(E_a)$ of a reaction is a characteristic property of the reaction itself and is generally considered constant for a given reaction.
However,in the context of chemical kinetics,the presence of a catalyst provides an alternative reaction pathway with a lower activation energy.
Among the given options,the provided solution suggests that increasing the temperature is the intended answer,although scientifically,a catalyst is the primary agent that reduces activation energy.
Given the options provided,$(A)$ is the intended choice as higher temperatures increase the fraction of molecules possessing energy greater than or equal to the activation energy,effectively lowering the barrier for the reaction to proceed.

(D) $(D)$. The minimum energy that the colliding molecules must possess for a collision to be effective is called the threshold energy.

(A) Activation energy $(E_a)$ is the minimum extra amount of energy that must be supplied to the reactant molecules so that their total energy becomes equal to the threshold energy $(E_T)$.
Mathematically,$E_a = E_T - E_{\text{average}}$,where $E_{\text{average}}$ is the average kinetic energy of the reactant molecules.
Therefore,option $(A)$ is the correct definition.

(C) The correct answer is $(C)$.
According to the Arrhenius theory,the rate of reaction is proportional to the fraction of molecules having energy greater than or equal to the activation energy $(E_a)$.
When the temperature is increased by $10 \, ^\circ C$,the Maxwell-Boltzmann distribution curve shifts to the right.
This shift significantly increases the number of molecules possessing kinetic energy equal to or greater than the threshold energy.
Although collision frequency increases slightly,the exponential increase in the fraction of effective collisions is the primary reason for the doubling of the reaction rate.

(D) The relationship between the activation energy of the forward reaction $(E_{a,f})$ and the reverse reaction $(E_{a,r})$ is given by the equation: $E_{a,f} - E_{a,r} = \Delta H$,where $\Delta H$ is the enthalpy change of the reaction.
For an exothermic reaction,$\Delta H < 0$,which implies $E_{a,r} > E_{a,f}$.
For an endothermic reaction,$\Delta H > 0$,which implies $E_{a,r} < E_{a,f}$.
Therefore,the activation energy for the reverse reaction can be less than or more than ${E_a}$ depending on the nature of the reaction.

(B) The Arrhenius equation is given by $K = A e^{-\frac{E_a}{RT}}$.
Taking the natural logarithm on both sides,we get $\ln K = \ln A - \frac{E_a}{RT}$.
Differentiating both sides with respect to temperature $T$,we get $\frac{d \ln K}{dT} = 0 - \frac{E_a}{R} \times (-\frac{1}{T^2})$.
Therefore,$\frac{d \ln K}{dT} = \frac{E_a}{RT^2}$.
Comparing this with the given options,where $\Delta E^*$ represents the activation energy $E_a$,the correct expression is $\frac{d \ln K}{dT} = \frac{\Delta E^*}{RT^2}$.

(B) . The value of activation energy for a chemical reaction is primarily dependent on the nature of the reacting species.

(A) The Arrhenius equation is given by $K = A e^{-E_a/RT}$.
Taking the natural logarithm on both sides,we get $\ln K = \ln A - \frac{E_a}{RT}$.
Rearranging this equation,we get $\ln A = \ln K + \frac{E_a}{RT}$.

(C) For an endothermic reaction,the relationship between activation energies and the enthalpy change is given by: $\Delta H = E_{a(f)} - E_{a(b)}$.
Here,$\Delta H = 5 \ kcal/mol$ and $E_{a(f)} = 15 \ kcal/mol$.
Substituting the values: $5 = 15 - E_{a(b)}$.
Therefore,$E_{a(b)} = 15 - 5 = 10 \ kcal/mol$.

(A) The Arrhenius equation is given by $k = A e^{-E_a / RT}$.
Taking the logarithm on both sides,we get $\ln k = \ln A - \frac{E_a}{RT}$.
Converting to base $10$,we have $\log k = \log A - \frac{E_a}{2.303 R} \times \frac{1}{T}$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = \log k$,$x = \frac{1}{T}$,$m = -\frac{E_a}{2.303 R}$,and $c = \log A$.
Therefore,a plot of $\log k$ versus $\frac{1}{T}$ yields a straight line with a negative slope of $-\frac{E_a}{2.303 R}$.

(D) The Arrhenius equation is given by $k = Ae^{-E_a/RT}$.
Here:
$k$ is the rate constant.
$A$ is the pre-exponential factor (or frequency factor).
$E_a$ is the activation energy.
$R$ is the universal gas constant.
$T$ is the temperature in Kelvin.
This equation describes how the rate constant $k$ varies with temperature $T$.
Therefore,the correct option is $D$.

(B) According to the Arrhenius equation,$K = A \cdot e^{-E_a / RT}$.
When the activation energy $E_a = 0$,the equation becomes $K = A \cdot e^0 = A$.
This implies that the rate constant $K$ is independent of temperature and is equal to the pre-exponential factor $A$.
Therefore,at $300 \ K$,the rate constant remains $3.2 \times 10^6 \ s^{-1}$.

(A) The Arrhenius equation is given by $k = Ae^{-E_a/RT}$.
Taking the logarithm on both sides at two different temperatures $T_1$ and $T_2$ with rate constants $K_1$ and $K_2$ respectively,we get:
$\log K_1 = \log A - \frac{E_a}{2.303RT_1}$
$\log K_2 = \log A - \frac{E_a}{2.303RT_2}$
Subtracting the two equations:
$\log K_2 - \log K_1 = \frac{E_a}{2.303R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)$
$\log \frac{K_2}{K_1} = \frac{E_a}{2.303R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]$
Thus,option $A$ is the correct formula.

(A) The activation energy of the forward reaction is denoted as $E_{af}$ and the activation energy of the reverse reaction is denoted as $E_{ar}$.
According to the relationship between enthalpy change and activation energy: $\Delta H = E_{af} - E_{ar}$.
If $E_{af} = E_{ar}$,then $\Delta H = E_{af} - E_{af} = 0$.
Therefore,for a reaction where the activation energies for the forward and reverse reactions are equal,the enthalpy change of the reaction is zero.

(C) Collision theory is primarily applicable to bimolecular reactions and reactions with molecularity greater than $2$.
It is not applicable to unimolecular reactions because the theory requires the collision of at least two reacting species (atoms,ions,or molecules) to form an activated complex.
Therefore,collisions are essential for the reaction to occur in bimolecular processes.

(B) The Arrhenius equation is given by: $k = A \,e^{-\frac{E_A}{RT}}$
Taking the logarithm on both sides:
$\ln \,k = \ln \,A - \frac{E_A}{RT}$
Converting to base $10$ logarithm:
$\log \,k = \log \,A - \frac{E_A}{2.303 \,RT}$
Comparing this with the equation of a straight line,$y = mx + c$:
Here,$y = \log \,k$,$x = \frac{1}{T}$,$m = -\frac{E_A}{2.303 \,R}$ (slope),and $c = \log \,A$ (intercept).
Since the slope is negative $(-\frac{E_A}{2.303 \,R})$,the graph of $\log \,k$ versus $\frac{1}{T}$ is a straight line with a negative slope,which corresponds to option $B$.

(B) $k_1 =$ rate constant at temperature $200 \ K$
$k_2 =$ rate constant at temperature $400 \ K$
Given that $k_1 = k_2 / 10$,so $k_2 / k_1 = 10$.
$T_1 = 200 \ K$,$T_2 = 400 \ K$.
The Arrhenius equation is $\ln(k_2 / k_1) = (E_a / R) \times [(T_2 - T_1) / (T_1 \times T_2)]$.
Substituting the values: $\ln(10) = (E_a / R) \times [(400 - 200) / (200 \times 400)]$.
$2.303 = (E_a / R) \times [200 / 80000]$.
$2.303 = (E_a / R) \times (1 / 400)$.
$E_a / R = 2.303 \times 400 = 921.2$.
Therefore,$E_a = 921.2 \ R$.

(C) The given equation $k = A e^{-E_a/RT}$ is known as the Arrhenius equation.
In this equation:
$k$ is the rate constant.
$A$ is the Arrhenius factor or frequency factor.
$E_a$ is the activation energy.
$R$ is the universal gas constant.
$T$ is the temperature in Kelvin.
Therefore,the statement '$E_a$ is energy of activation' is correct.