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

(B) The Arrhenius equation is given by: $\log \frac{K_2}{K_1} = \frac{E_a}{2.303R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]$
Given: $\frac{K_2}{K_1} = 2$,$T_1 = 300 \, K$,$T_2 = 310 \, K$,$R = 8.314 \, J \, K^{-1} \, mol^{-1}$.
Substituting the values: $\log 2 = \frac{E_a}{2.303 \times 8.314} \left[ \frac{310 - 300}{300 \times 310} \right]$
$0.3010 = \frac{E_a}{19.147} \times \frac{10}{93000}$
$E_a = \frac{0.3010 \times 19.147 \times 93000}{10} \approx 53598 \, J \, mol^{-1} \approx 53.6 \, kJ \, mol^{-1}$.
Rounding to the nearest integer,the activation energy is $54 \, kJ \, mol^{-1}$.

(D) According to the Arrhenius equation,$k = A e^{-E_a / RT}$.
When the activation energy $E_a = 0$,the equation becomes $k = A e^0 = A$.
Since $A$ (the frequency factor) is a constant,the rate constant $k$ becomes independent of temperature.

(C) The rate constant $k$ is given by the Arrhenius equation: $k = p Z e^{-E_a/RT}$.
For the reaction to proceed more rapidly,the rate constant $k$ must increase.
In the expression $k = p Z e^{-E_a/RT}$,$p$ is the steric factor,$Z$ is the collision frequency,$E_a$ is the activation energy,$R$ is the gas constant,and $T$ is the temperature.
To increase $k$,the value of the exponential term $e^{-E_a/RT}$ must increase.
This occurs when the exponent $-E_a/RT$ becomes less negative,which happens if the activation energy $E_a$ decreases or the temperature $T$ increases.
Therefore,a decrease in the activation energy $E_a$ will cause the reaction to proceed more rapidly.

(C) In an endothermic reaction,the potential energy of the products $(P)$ is higher than that of the reactants $(R)$.
Activation energy $(E_a)$ is the energy difference between the transition state (peak of the curve) and the reactants $(R)$.
Diagram $C$ shows that the energy of the products is higher than the reactants (endothermic) and the energy barrier (peak) is significantly higher than the reactant energy,indicating high activation energy.

(A) For an endothermic reaction,the enthalpy change $\Delta H$ is positive $(+ve)$.
The relationship between enthalpy change,forward activation energy $(E_f)$,and backward activation energy $(E_b)$ is given by the equation: $\Delta H = E_f - E_b$.
Since $\Delta H > 0$,it follows that $E_f - E_b > 0$,which implies $E_f > E_b$ or $E_b < E_f$.

(A) The Arrhenius equation describes the dependence of the rate constant $(k)$ on temperature $(T)$ as:
$k = A e^{-E_a/RT}$
Taking the natural logarithm on both sides:
$\ln k = \ln(A e^{-E_a/RT})$
Using the properties of logarithms:
$\ln k = \ln A + \ln(e^{-E_a/RT})$
$\ln k = \ln A - \frac{E_a}{RT}$
Therefore,the correct expression is given by option $(A)$.

(C) The Arrhenius equation is given by $K = A \, e^{-E_a / RT}$.
Taking the natural logarithm on both sides,we get $\ln \, K = \ln \, A - \frac{E_a}{RT}$.
To convert this to base $10$,we divide by $2.303$,resulting in $\log \, K = \log \, A - \frac{E_a}{2.303 \, RT}$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = \log \, K$ and $x = \frac{1}{T}$,the slope $m$ is equal to $-\frac{E_a}{2.303 \, R}$.

(C) For an endothermic reaction,the enthalpy change $\Delta H$ is positive.
The relationship between the activation energy of the forward reaction $(E_{a(f)})$,the activation energy of the backward reaction $(E_{a(b)})$,and the enthalpy of the reaction is given by the equation: $E_{a(f)} = E_{a(b)} + \Delta H$.
Since $E_{a(b)}$ must always be a positive value for any reaction to occur,it follows that $E_{a(f)} > \Delta H$.
Therefore,the minimum value for the energy of activation is more than $\Delta H$.

(C) According to the Arrhenius equation,$K = A e^{-E_a / RT}$.
Taking the natural logarithm on both sides,$\ln K = \ln A - \frac{E_a}{RT}$.
This shows that the rate constant $K$ is inversely related to the activation energy $E_a$ for a given temperature and frequency factor $A$.
Since $K' = 2K''$,it implies that $K' > K''$.
Therefore,the reaction with the higher rate constant must have a lower activation energy.
Thus,${E_a}' < {E_a}''$.

(B) The Arrhenius equation is given by $k = A e^{-E_a / RT}$.
As $T \to \infty$,the term $\frac{E_a}{RT} \to 0$.
Therefore,the rate constant $k$ approaches the pre-exponential factor $A$ (Arrhenius parameter).
Given that the Arrhenius parameter $A = 6.0 \times 10^{14}\,s^{-1}$,the value of the rate constant as $T \to \infty$ is $6.0 \times 10^{14}\,s^{-1}$.

(A) For the given reaction,$\Delta H = -44.12 \ kcal$. Since $\Delta H < 0$,the reaction is exothermic.
We know the relationship between enthalpy change and activation energies is given by $\Delta H = E_f - E_b$,where $E_f$ is the activation energy of the forward reaction $(E_2)$ and $E_b$ is the activation energy of the backward reaction $(E_1)$.
Thus,$\Delta H = E_2 - E_1$.
Since $\Delta H$ is negative,$E_2 - E_1 < 0$,which implies $E_1 > E_2$.

(A) The Arrhenius equation is given by $k = A e^{-E^*/RT}$.
Taking the logarithm on both sides,we get $\log \, k = \log \, A - \frac{E^*}{2.303 R} \cdot \frac{1}{T}$.
This equation follows the linear form $y = mx + c$,where $y = \log \, k$ and $x = \frac{1}{T}$.
Plotting $\log \, k$ versus $\frac{1}{T}$ yields a straight line with a slope equal to $-\frac{E^*}{2.303 R}$.
Thus,the activation energy $E^*$ can be determined from the slope of this plot.

(C) The activation energy $(E_a)$ is determined using the Arrhenius equation at two different temperatures ($T_1$ and $T_2$).
According to the Arrhenius equation: $\log \frac{K_2}{K_1} = \frac{E_a}{2.303R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]$.
By measuring the rate constants ($K_1$ and $K_2$) at two different temperatures,the activation energy can be calculated.

(A) Using the Arrhenius equation: $\log \frac{K_2}{K_1} = \frac{E_a}{2.303 \times R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]$
Given: $E_a = 9.0 \, kcal/mol = 9000 \, cal/mol$,$R = 1.987 \, cal \, K^{-1} \, mol^{-1} \approx 2 \, cal \, K^{-1} \, mol^{-1}$,$T_1 = 298 \, K$,$T_2 = 308 \, K$.
$\log \frac{K_2}{K_1} = \frac{9000}{2.303 \times 2} \left[ \frac{308 - 298}{308 \times 298} \right]$
$\log \frac{K_2}{K_1} = \frac{9000}{4.606} \times \frac{10}{91784} \approx 1954.19 \times 0.0001089 \approx 0.2129$
$\frac{K_2}{K_1} = \text{antilog}(0.2129) \approx 1.632$
Percentage increase = $\frac{K_2 - K_1}{K_1} \times 100 = (1.632 - 1) \times 100 = 63.2 \% \approx 63 \%$.
The closest option is $65 \%$.

(D) The enthalpy change $(\Delta H)$ of a reaction is given by the difference between the activation energy of the forward reaction $(E_{af})$ and the activation energy of the backward reaction $(E_{ab})$.
$\Delta H = E_{af} - E_{ab}$
Initially,$\Delta H = 180 \ kJ \ mol^{-1} - 200 \ kJ \ mol^{-1} = -20 \ kJ \ mol^{-1}$.
$A$ catalyst lowers the activation energy of both the forward and backward reactions by the same amount $(100 \ kJ \ mol^{-1})$.
New $E_{af}' = 180 - 100 = 80 \ kJ \ mol^{-1}$.
New $E_{ab}' = 200 - 100 = 100 \ kJ \ mol^{-1}$.
New $\Delta H' = E_{af}' - E_{ab}' = 80 - 100 = -20 \ kJ \ mol^{-1}$.
Since a catalyst does not change the enthalpy of the reaction,the value remains $-20 \ kJ \ mol^{-1}$.

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

(B) Given that $K_1 = K_2$,we have:
$10^{16} \cdot e^{-2000/T} = 10^{15} \cdot e^{-1000/T}$
Divide both sides by $10^{15}$ and $e^{-1000/T}$:
$\frac{10^{16}}{10^{15}} = \frac{e^{-1000/T}}{e^{-2000/T}}$
$10 = e^{(-1000/T) - (-2000/T)}$
$10 = e^{1000/T}$
Taking the natural logarithm $(\ln)$ on both sides:
$\ln(10) = \frac{1000}{T}$
$2.303 \cdot \log_{10}(10) = \frac{1000}{T}$
$2.303 = \frac{1000}{T}$
Therefore,$T = \frac{1000}{2.303} \, K$.

(B) According to the collision theory of chemical reactions,for a reaction to occur,reactant molecules must collide with each other.
However,not all collisions lead to the formation of products.
Only those collisions where molecules possess energy equal to or greater than the activation energy $(E_a)$ and have the proper orientation are considered 'effective' or 'successful' collisions.
Therefore,the statement that 'all collisions between reactant molecules result in the formation of products' is incorrect.

(D) For the forward reaction $A \rightarrow B$,the activation energy $E_a(f) = 17 \, kJ/mol$.
Since the reaction is exothermic,the enthalpy change $\Delta H = -40 \, kJ/mol$.
The relationship between activation energy of forward $(E_a(f))$ and reverse $(E_a(r))$ reactions is given by: $\Delta H = E_a(f) - E_a(r)$.
Substituting the values: $-40 = 17 - E_a(r)$.
Therefore,$E_a(r) = 17 + 40 = 57 \, kJ/mol$.

(C) Given: $T_1 = 27\,^oC = 300\,K$,$k_1 = k$
$T_2 = 37\,^oC = 310\,K$,$k_2 = 2k$
Using the Arrhenius equation: $\log \frac{k_2}{k_1} = \frac{E_a}{2.303R} \left( \frac{T_2 - T_1}{T_1 T_2} \right)$
Substituting the values:
$\log \frac{2k}{k} = \frac{E_a}{2.303 \times 8.314} \times \frac{310 - 300}{300 \times 310}$
$\log 2 = \frac{E_a}{2.303 \times 8.314} \times \frac{10}{93000}$
$0.3010 = \frac{E_a}{19.147} \times 0.0001075$
$E_a = \frac{0.3010 \times 19.147}{0.0001075} \approx 53598.6\,J\,mol^{-1}$
$E_a \approx 53.6\,kJ\,mol^{-1}$

(A) According to the Arrhenius equation: $k = Ae^{-E_a/RT}$
Taking the logarithm on both sides: $\log \, k = \log \, A - \frac{E_a}{2.303R} \times \frac{1}{T}$
Comparing this with the equation of a straight line $y = mx + c$,the slope $(m)$ is given by: $m = -\frac{E_a}{2.303R}$
Given that the slope is $-8000$,we have: $\frac{E_a}{2.303R} = 8000$
Using $R = 1.987 \, cal \, K^{-1} \, mol^{-1}$:
$E_a = 8000 \times 2.303 \times 1.987$
$E_a = 36608 \, cal \, mol^{-1}$

(C) The activation energy $(E_a)$ is the minimum energy required for a reaction to occur.
It is a characteristic property of the reaction mechanism and the transition state,which depends on the nature of the reactants and the pathway,not on the stoichiometric coefficients used to balance the chemical equation.
When we multiply or divide the stoichiometric coefficients of a balanced chemical equation by a constant factor,the reaction pathway and the transition state remain the same.
Therefore,the activation energy $E_a$ remains unchanged regardless of the stoichiometric coefficients.
Thus,$E_a = E_a'$.

(B) The rate constant $(K)$ is independent of the concentration of the reactants.
It depends only on the temperature of the reaction.
According to the Arrhenius equation,$K = Ae^{-E_a/RT}$,increasing the temperature $(T)$ increases the value of the rate constant $(K)$.

(D) The Arrhenius equation is given by $\log \, k = \log \, A - \frac{E_a}{2.303 \, RT}$.
Comparing this with the given equation $\log \, k = - \frac{2000}{T} + 0.6$:
$\log \, A = 0.6 \implies A = 10^{0.6} \approx 3.98 \, s^{-1}$ (Note: The provided option $10^6$ suggests the equation was $\log \, k = - \frac{2000}{T} + 6$).
Assuming the equation is $\log \, k = - \frac{2000}{T} + 6$:
$\log \, A = 6 \implies A = 10^6 \, s^{-1}$.
$\frac{E_a}{2.303 \, R} = 2000$.
$E_a = 2000 \times 2.303 \times 8.314 \, J \, mol^{-1} \approx 38297 \, J \, mol^{-1} = 38.3 \, kJ \, mol^{-1}$.

(C) For a reversible reaction $A \rightleftharpoons B$,the relationship between the activation energy of the forward reaction $(E_{a,f})$ and the backward reaction $(E_{a,b})$ is given by $\Delta H = E_{a,f} - E_{a,b}$.
If the reaction is exothermic $(\Delta H < 0)$,then $E_{a,b} > E_{a,f}$.
If the reaction is endothermic $(\Delta H > 0)$,then $E_{a,b} < E_{a,f}$.
Therefore,the activation energy of the backward reaction can be more or less than $E_a$ depending on the enthalpy change of the reaction.

(B) The Arrhenius equation is given by $k = A \cdot e^{-E_a / (RT)}$.
As $T \rightarrow \infty$,the exponent term $-E_a / (RT) \rightarrow 0$.
Therefore,$e^{-E_a / (RT)} \rightarrow e^0 = 1$.
Thus,the rate constant $k$ approaches the Arrhenius factor $A$.
Given $A = 6.0 \times 10^{14} \, s^{-1}$,the value of the rate constant as $T \rightarrow \infty$ is $6.0 \times 10^{14} \, s^{-1}$.

(A) The rate of reaction increases by a factor of $2^{\Delta T/10}$,where $\Delta T$ is the change in temperature.
Given,initial temperature $T_1 = 0\,^{\circ}C$ and final temperature $T_2 = 50\,^{\circ}C$.
Change in temperature $\Delta T = T_2 - T_1 = 50\,^{\circ}C - 0\,^{\circ}C = 50\,^{\circ}C$.
The number of $10\,^{\circ}C$ intervals is $n = \frac{50}{10} = 5$.
Therefore,the rate of reaction increases by $2^n = 2^5 = 32$ times.

(B) The rate of reaction doubles with every $10^\circ C$ rise in temperature.
The number of $10^\circ C$ intervals is $n = \frac{100^\circ C - 10^\circ C}{10^\circ C} = 9$.
The increase in rate is given by the formula $2^n$.
Therefore,the rate increase factor is $2^9 = 512$ times.

(B) For an endothermic reaction,the energy of the products is higher than the energy of the reactants.
Mathematically,the enthalpy change $\Delta H = E_f - E_b > 0$.
This implies $E_f > E_b$,which can be rearranged as $E_b < E_f$.

(B) Given: Temperature coefficient $= \frac{K_2}{K_1} = 2$.
$T_1 = 25^{\circ}C = 298 \ K$,$T_2 = 35^{\circ}C = 308 \ K$.
Using the Arrhenius equation: $\log \frac{K_2}{K_1} = \frac{E_a}{2.303R} \left( \frac{T_2 - T_1}{T_1 T_2} \right)$.
Substituting the values: $\log 2 = \frac{E_a}{2.303 \times 8.314} \left( \frac{308 - 298}{298 \times 308} \right)$.
$0.3010 = \frac{E_a}{19.147} \left( \frac{10}{91784} \right)$.
$E_a = \frac{0.3010 \times 19.147 \times 91784}{10} \approx 52897 \ J/mol \approx 52.9 \ kJ/mol$.
Rounding to the nearest provided option,$E_a = 52.31 \ kJ/mol$.

(B) According to the Arrhenius equation,the rate constant $k$ at temperature $T$ is given by $k = A e^{-E_a/RT}$.
For two different temperatures $T_1$ and $T_2$,we have:
$\ln k_1 = \ln A - \frac{E_a}{RT_1}$ and $\ln k_2 = \ln A - \frac{E_a}{RT_2}$.
Subtracting the two equations:
$\ln k_2 - \ln k_1 = \frac{E_a}{RT_1} - \frac{E_a}{RT_2} = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) = \frac{E_a}{R} \left( \frac{T_2 - T_1}{T_1 T_2} \right)$.
Converting to base $10$ logarithm:
$\log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left( \frac{T_2 - T_1}{T_1 T_2} \right)$.
Multiplying by $-1$ on both sides:
$\log \frac{k_1}{k_2} = \frac{E_a}{2.303 R} \left( \frac{T_1 - T_2}{T_1 T_2} \right)$.

(A) The Arrhenius equation is given by $k = Ae^{-E_a/RT}$.
As temperature $(T)$ increases,the term $e^{-E_a/RT}$ increases exponentially.
Therefore,the rate constant $(k)$ increases exponentially with an increase in temperature $(T)$.
This relationship is represented by an exponential growth curve,which corresponds to the graph in option $A$.

(A) Using the Arrhenius equation:
$\log_{10} \frac{K_2}{K_1} = \frac{E_a}{2.303 \times R} \left[ \frac{T_2 - T_1}{T_1 \times T_2} \right]$
Given:
$\frac{K_2}{K_1} = 3$,
$R = 8.314 \, J \, K^{-1} \, mol^{-1}$,
$T_1 = 20 + 273 = 293 \, K$,
$T_2 = 50 + 273 = 323 \, K$,
$\log_{10} 3 \approx 0.4771$.
Substituting the values:
$0.4771 = \frac{E_a}{2.303 \times 8.314} \left[ \frac{323 - 293}{323 \times 293} \right]$
$E_a = \frac{0.4771 \times 2.303 \times 8.314 \times 323 \times 293}{30}$
$E_a \approx 28811.8 \, J \, mol^{-1} = 28.81 \, kJ \, mol^{-1}$.

(C) The temperature coefficient $\mu = 2$.
The relationship between the rate constant and temperature change is given by:
$\frac{K_2}{K_1} = \mu^{\frac{\Delta T}{10}}$
Given $\Delta T = 50^\circ \text{C}$ and $\mu = 2$:
$\frac{K_2}{K_1} = 2^{\frac{50}{10}}$
$\frac{K_2}{K_1} = 2^5 = 32$
Thus,the rate of reaction increases by a factor of $32$.

(B) The Arrhenius equation is given by $K = A e^{-E_a/RT}$.
Taking the natural logarithm on both sides,we get $\ln K = \ln A - \frac{E_a}{RT}$.
Differentiating both sides with respect to temperature $T$,we get $\frac{d}{dT}(\ln K) = \frac{d}{dT}(\ln A) - \frac{d}{dT}(\frac{E_a}{RT})$.
Since $A$ and $E_a$ are constants,$\frac{d}{dT}(\ln A) = 0$.
Thus,$\frac{d}{dT}(\ln K) = -E_a \cdot \frac{d}{dT}(T^{-1}) = -E_a \cdot (-T^{-2}) = \frac{E_a}{RT^2}$.

(A) Enzymes act as biological catalysts.
Catalysts provide an alternative reaction pathway with a lower activation energy $(E_a)$.
By lowering the activation energy,the number of effective collisions increases,which significantly enhances the rate of the reaction.
Therefore,in the presence of enzymes,the activation energy $(E_a)$ of the reaction decreases.

(C) The Arrhenius equation is given by $k = A e^{-E_a/RT}$.
For $k$ to be equal to $A$,the exponential term $e^{-E_a/RT}$ must be equal to $1$.
This happens when the exponent $-E_a/RT = 0$.
As $T \to \infty$,the term $E_a/RT$ approaches $0$,and $e^0 = 1$.
Therefore,$k = A$ when $T \to \infty$.

(D) Enzymes act as biological catalysts that increase the rate of reaction by lowering the activation energy $(E_a)$.
Since the rate of reaction increases significantly in the presence of an enzyme,the activation energy of the catalyzed reaction is lower than that of the uncatalyzed reaction.
However,the exact value of the activation energy cannot be determined solely from the rate enhancement factor without knowing the specific Arrhenius parameters or the temperature.

(B) The Arrhenius equation is given by: $\log \frac{K_2}{K_1} = \frac{E_a}{2.303R} \left( \frac{T_2 - T_1}{T_1 T_2} \right)$.
Given: $T_1 = 25 + 273 = 298 \, K$,$T_2 = 35 + 273 = 308 \, K$,and $K_2 = 2K_1$.
Substituting these values: $\log \frac{2K_1}{K_1} = \frac{E_a}{2.303R} \left( \frac{308 - 298}{298 \times 308} \right)$.
$\log 2 = \frac{E_a}{2.303R} \left( \frac{10}{298 \times 308} \right)$.
Rearranging for $E_a$: $E_a = \frac{2.303 \times R \times 298 \times 308 \times \log 2}{10}$.

(A) The relationship between the activation energy of the forward reaction $(E_{a,f})$,the activation energy of the reverse reaction $(E_{a,r})$,and the enthalpy change $(\Delta H)$ is given by the equation: $\Delta H = E_{a,f} - E_{a,r}$.
Given: $E_{a,f} = 60 \ kJ \ mol^{-1}$ and $\Delta H = -20 \ kJ \ mol^{-1}$.
Substituting these values into the equation: $-20 = 60 - E_{a,r}$.
Rearranging to solve for $E_{a,r}$: $E_{a,r} = 60 + 20 = 80 \ kJ \ mol^{-1}$.

(B) The Arrhenius equation is given by: $\log K = \log A - \frac{E_a}{2.303 \, RT}$.
Rearranging the equation: $\log K - \log A = -\frac{E_a}{2.303 \, RT}$.
Using the property $\log K - \log A = \log(K/A)$,we get: $\log(K/A) = -\frac{E_a}{2.303 \, RT}$.
Given $E_a = 2.303 \, RT$,substituting this value:
$\log(K/A) = -\frac{2.303 \, RT}{2.303 \, RT} = -1$.
Therefore,the ratio $\frac{K}{A} = \text{antilog}(-1) = 10^{-1} = 0.1$.

(C) The Arrhenius equation is given by: $\log \frac{k_2}{k_1} = \frac{E_a}{2.303R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]$
Given: $k_2 = 2k_1$,$T_1 = 298 \, K$,$T_2 = 308 \, K$,$R = 8.314 \, J \, mol^{-1} \, K^{-1}$.
Substituting the values: $\log(2) = \frac{E_a}{2.303 \times 8.314} \left[ \frac{308 - 298}{308 \times 298} \right]$
$0.3010 = \frac{E_a}{19.147} \left[ \frac{10}{91784} \right]$
$E_a = \frac{0.3010 \times 19.147 \times 91784}{10} \approx 52897 \, J \, mol^{-1} \approx 52.9 \, kJ \, mol^{-1}$.

(C) For any chemical reaction,the activation energy $(E_a)$ is the minimum energy required for the reactants to form the activated complex.
In an exothermic reaction,the energy of the products is lower than the energy of the reactants.
The relationship between activation energy $(E_a)$,enthalpy of reaction $(\Delta H)$,and the activation energy of the reverse reaction $(E_{a,reverse})$ is given by:
$E_a - E_{a,reverse} = \Delta H$.
Since $E_{a,reverse}$ must always be a positive value,$E_a$ must be greater than $\Delta H$ (considering the magnitude of $\Delta H$ for an exothermic reaction).
Therefore,the activation energy is always a positive value and is greater than the enthalpy change of the reaction.

(A) Activation energy $(E_a)$ is defined as the minimum extra amount of energy absorbed by the reactant molecules so that their total energy becomes equal to the threshold energy $(E_T)$.
Mathematically,$E_a = E_T - E_{\text{average}}$.

(A) Using the Arrhenius equation: $\log \frac{k_2}{k_1} = \frac{E_a}{2.303R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]$
Given: $k_2 = 2k_1$,$T_1 = 300 \ K$,$T_2 = 310 \ K$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$.
$\log 2 = \frac{E_a}{2.303 \times 8.314} \left[ \frac{310 - 300}{300 \times 310} \right]$
$0.301 = \frac{E_a}{19.147} \left[ \frac{10}{93000} \right]$
$E_a = \frac{0.301 \times 19.147 \times 9300}{10} \approx 53598 \ J \ mol^{-1} \approx 53.6 \ kJ \ mol^{-1}$.

(D) According to the Arrhenius equation,$K_1 = A_1 e^{-Ea_1/RT}$ and $K_2 = A_2 e^{-Ea_2/RT}$.
Dividing the two equations: $\frac{K_1}{K_2} = \frac{A_1}{A_2} e^{(Ea_2 - Ea_1)/RT}$.
Given $Ea_2 = 2 Ea_1$,substituting this into the exponent gives: $\frac{K_1}{K_2} = \frac{A_1}{A_2} e^{(2 Ea_1 - Ea_1)/RT} = \frac{A_1}{A_2} e^{Ea_1/RT}$.
Therefore,$K_1 = A' K_2 e^{Ea_1/RT}$,where $A' = \frac{A_1}{A_2}$.

(D) The rate constant $(k)$ of a reaction is a characteristic property that depends primarily on the temperature of the reaction,as described by the Arrhenius equation: $k = A e^{-E_a / RT}$.
It is independent of the concentration of reactants,concentration of products,or time.

(B) According to the Arrhenius equation: $k = A e^{-E_a / RT}$.
Taking the logarithm on both sides:
$\ln \, k = \ln \, A - \frac{E_a}{RT}$
Converting to base $10$:
$\log \, k = \log \, A - \frac{E_a}{2.303 \, R} \left( \frac{1}{T} \right)$
This equation is in the form of a straight line $y = mx + c$,where $y = \log \, k$,$x = 1/T$,and the slope $m = -\frac{E_a}{2.303 \, R}$.
Since the slope is negative,the graph of $\log \, k$ versus $1/T$ is a straight line with a negative slope,as shown in option $B$.