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

(D) 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 of the reaction $(\Delta H)$ is given by:
$E_{a,f} - E_{a,r} = \Delta H$.
If the reaction is exothermic $(\Delta H < 0)$,then $E_{a,f} < E_{a,r}$,meaning $E_{a,r} > 50 \, kcal$.
If the reaction is endothermic $(\Delta H > 0)$,then $E_{a,f} > E_{a,r}$,meaning $E_{a,r} < 50 \, kcal$.
Since the nature of the reaction (exothermic or endothermic) is not specified,the activation energy of the reverse reaction can be either greater or less than $50 \, kcal$.

(B) 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: $T_1 = 293 \ K$,$T_2 = 313 \ K$,$k_2 = 4k_1$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$.
Substituting the values: $\log(4) = \frac{E_a}{2.303 \times 8.314} \left[ \frac{313 - 293}{293 \times 313} \right]$
$0.6021 = \frac{E_a}{19.147} \times \frac{20}{91709}$
$E_a = \frac{0.6021 \times 19.147 \times 91709}{20} \approx 52863 \ J \ mol^{-1}$
$E_a = 52.86 \ kJ \ mol^{-1}$.

(C) The Arrhenius equation is given by $K = A e^{-E_a / RT}$.
Taking the natural logarithm on both sides,we get $\ln K = \ln A - \frac{E_a}{RT}$.
Converting to base $10$ logarithm,we get $\log K = \log A - \frac{E_a}{2.303 R} \times \frac{1}{T}$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = \log K$ and $x = 1/T$,the slope $m$ is equal to $-\frac{E_a}{2.303 R}$.
Thus,the plot of $\log K$ versus $1/T$ is a straight line with a slope of $-\frac{E_a}{2.303 R}$,which allows us to calculate the activation energy $E_a$.

(A) Using the Arrhenius equation: $\log \frac{K_2}{K_1} = \frac{E_a}{2.303R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)$
Given $R \approx 8.314 \ J \ K^{-1} \ mol^{-1}$,but using $R = 2 \ cal \ K^{-1} \ mol^{-1}$ as per the provided calculation steps:
$\log K_2 - \log K_1 = \frac{E_a}{2.303 \times 2} [ (1.5 \times 10^{-3}) - (1.3 \times 10^{-3}) ]$
$1.1 - 2.9 = \frac{E_a}{4.606} [ 0.2 \times 10^{-3} ]$
$-1.8 = \frac{E_a \times 0.2 \times 10^{-3}}{4.606}$
$E_a = \frac{1.8 \times 4.606}{0.2 \times 10^{-3}} = 9 \times 4.606 \times 10^3 \approx 4.14 \times 10^4 \ J \ mol^{-1}$.
Rounding to the nearest provided option,the correct value is $4 \times 10^4 \ J \ mol^{-1}$.

(B) According to the Arrhenius equation,the rate constant $k$ is given by $k = A e^{-E_a/RT}$.
As the temperature $T$ increases,the fraction of molecules having energy equal to or greater than the activation energy $(E_a)$ increases significantly.
This leads to a larger number of effective collisions,resulting in a substantial increase in the reaction rate.
Therefore,the correct reason is the increase in the number of activated molecules.

(C) The minimum amount of energy that colliding molecules must possess for a chemical reaction to occur is known as the $Threshold \ energy$.
Activation energy is the additional energy required over the average energy of the reactants to reach the threshold energy level.

(D) It is a general rule in chemical kinetics that for every $10\,^\circ\text{C}$ rise in temperature,the reaction rate increases by a factor of $2$ to $3$.
Assuming a temperature coefficient of $2$ for this calculation:
The temperature increase is $50\,^\circ\text{C} - 10\,^\circ\text{C} = 40\,^\circ\text{C}$.
This corresponds to $4$ intervals of $10\,^\circ\text{C}$ each $(40/10 = 4)$.
The rate increase factor is calculated as $2^n$,where $n$ is the number of $10\,^\circ\text{C}$ intervals.
Rate increase factor $= 2^4 = 16$.

(D) According to the Arrhenius equation,$k = A e^{-E_a / RT}$.
Here,$k$ 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.
As the activation energy $(E_a)$ decreases,the value of the exponent $-E_a / RT$ becomes less negative (closer to zero).
Consequently,the value of $e^{-E_a / RT}$ increases,leading to a larger rate constant $(k)$.
$A$ larger rate constant indicates a faster reaction rate.
Therefore,reactions with low activation energy are generally fast.

(D) The activation energy for the reverse reaction depends on the enthalpy change $(\Delta H)$ of the reaction.
For an endothermic reaction,$\Delta H > 0$,so $E_{a(\text{reverse})} = E_{a(\text{forward})} - \Delta H$,which is less than $E_a$.
For an exothermic reaction,$\Delta H < 0$,so $E_{a(\text{reverse})} = E_{a(\text{forward})} - \Delta H$,which is greater than $E_a$.
Therefore,the activation energy in the reverse direction can be less than or more than $E_a$.

(B) The relationship between enthalpy change $(\Delta H)$,activation energy of the forward reaction $(E_f)$,and activation energy of the reverse reaction $(E_b)$ is given by: $\Delta H = E_f - E_b$.
Given: $\Delta H = -20 \ kJ \ mol^{-1}$ and $E_f = 85 \ kJ \ mol^{-1}$.
Substituting the values: $-20 = 85 - E_b$.
Solving for $E_b$: $E_b = 85 + 20 = 105 \ kJ \ mol^{-1}$.

(C) The activation energy $(E_a)$ is defined as the minimum amount of extra energy absorbed by the reactant molecules to form the activated complex,which allows them to overcome the potential energy barrier of the reaction.
Therefore,it is the minimum external energy required for the reactants to reach the transition state.

(C) The rate of a chemical reaction is inversely related to the activation energy $(E_a)$ of the reaction.
According to the Arrhenius equation,$k = Ae^{-E_a/RT}$,a lower activation energy leads to a higher rate constant $(k)$ and thus a faster reaction rate.
The reaction $2NO + O_2 \rightarrow 2NO_2$ has a significantly lower activation energy compared to the reaction $2CO + O_2 \rightarrow 2CO_2$ at room temperature.
Therefore,the reaction between $NO$ and $O_2$ is fast,while the reaction between $CO$ and $O_2$ is slow.

(C) For an exothermic reaction,the enthalpy change $\Delta H$ is given by $\Delta H = E_{a(f)} - E_{a(b)}$,where $E_{a(f)}$ is the activation energy of the forward reaction and $E_{a(b)}$ is the activation energy of the backward reaction.
Given: $\Delta H = -5 \, K \, cal/mol$ (since it is exothermic),
$E_{a(f)} = 15 \, K \, cal/mol$.
Substituting the values: $-5 = 15 - E_{a(b)}$.
Therefore,$E_{a(b)} = 15 + 5 = 20 \, K \, cal/mol$.

(B) According to the Arrhenius equation,the rate constant is given by $K = A e^{-E_a/RT}$.
For two reactions at the same temperature $T$,assuming the frequency factor $A$ is the same for both:
$\frac{K_1}{K_2} = \frac{e^{-E_1/RT}}{e^{-E_2/RT}} = e^{(E_2 - E_1)/RT}$.
Given $K_1 = 2K_2$,we have $\frac{K_1}{K_2} = 2$.
Taking the natural logarithm on both sides: $\ln(2) = \frac{E_2 - E_1}{RT}$.
Since $\ln(2) > 0$,it follows that $E_2 - E_1 > 0$,which means $E_2 > E_1$ or $E_1 < E_2$.

(B) Let the rate constant without a catalyst be $k_1$ and with a catalyst be $k_2$.
Activation energy without catalyst $(E_1) = 30\,kJ\,mol^{-1}$.
Activation energy with catalyst $(E_2) = 24\,kJ\,mol^{-1}$.
Temperature $(T) = 25 + 273 = 298\,K$.
According to the Arrhenius equation:
$\ln(\frac{k_2}{k_1}) = \frac{E_1 - E_2}{RT}$
$\log(\frac{k_2}{k_1}) = \frac{E_1 - E_2}{2.303RT}$
$\log(\frac{k_2}{k_1}) = \frac{(30 - 24) \times 10^3\,J\,mol^{-1}}{2.303 \times 8.314\,J\,K^{-1}\,mol^{-1} \times 298\,K}$
$\log(\frac{k_2}{k_1}) = \frac{6000}{5705.84} \approx 1.0515$
$\frac{k_2}{k_1} = \text{antilog}(1.0515) \approx 11.26$
Thus,the rate of reaction increases by approximately $11$ times.

(C) For any reaction,the enthalpy change $(\Delta H)$ is related to the activation energy of the forward reaction $(E_{a(f)})$ and the reverse reaction $(E_{a(b)})$ by the equation: $\Delta H = E_{a(f)} - E_{a(b)}$.
Given: $E_{a(f)} = 30 \ kJ \ mol^{-1}$ and $\Delta H = -20 \ kJ \ mol^{-1}$.
Substituting the values: $-20 = 30 - E_{a(b)}$.
Rearranging the equation: $E_{a(b)} = 30 + 20 = 50 \ kJ \ mol^{-1}$.

(A) Using the Arrhenius equation: $\log \frac{K_2}{K_1} = \frac{E_a}{2.303R} \left[ \frac{1}{T_1} - \frac{1}{T_2} \right]$
Given: $K_1 = 0.325$,$K_2 = 6.735$,$T_1 = 30 + 273 = 303 \ K$,$T_2 = 60 + 273 = 333 \ K$,$R = 1.987 \ cal \ K^{-1} \ mol^{-1}$.
Substituting the values:
$\log \left( \frac{6.735}{0.325} \right) = \frac{E_a}{2.303 \times 1.987} \left[ \frac{1}{303} - \frac{1}{333} \right]$
$\log(20.723) = \frac{E_a}{4.575} \left[ \frac{333 - 303}{303 \times 333} \right]$
$1.3164 = \frac{E_a}{4.575} \times \frac{30}{100899}$
$E_a = \frac{1.3164 \times 4.575 \times 100899}{30} \approx 20260 \ cal \ mol^{-1}$.

(A) catalyst is a substance that increases the rate of a chemical reaction without itself undergoing any permanent chemical change.
It functions by providing an alternative pathway for the reaction with a lower activation energy $(E_a)$.
It does not change the Gibbs free energy $(\Delta G)$ of the reaction,nor does it alter the equilibrium constant $(K_{eq})$ of the reaction.
Therefore,statement $A$ is correct.

(C) The influence of temperature on the rate of reaction is described by the Arrhenius equation.
The equation is given by: $K = Ae^{-E_a/RT}$
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.

(A) From the given energy profile diagram:
$1$. Energy of reactants $(E_R)$ = $20 \, kJ$
$2$. Energy of products $(E_P)$ = $40 \, kJ$
$3$. Threshold energy $(E_T)$ = $80 \, kJ$
Calculations:
- Activation energy for forward reaction $(E_{a(f)})$ = $E_T - E_R = 80 - 20 = 60 \, kJ$
- Activation energy for backward reaction $(E_{a(b)})$ = $E_T - E_P = 80 - 40 = 40 \, kJ$
- Enthalpy change $(\Delta H)$ = $E_P - E_R = 40 - 20 = 20 \, kJ$
Since $\Delta H > 0$,the reaction is endothermic.
Evaluating the options:
- Option $(A)$: $E_{a(b)} = 40 \, kJ$. The statement says $80 \, kJ$,which is incorrect.
- Option $(B)$: Correct,as $\Delta H > 0$.
- Option $(C)$: Correct,$\Delta H = 20 \, kJ$.
- Option $(D)$: Correct,$E_{a(f)} = 60 \, kJ$.
Therefore,the incorrect statement is $(A)$.

(D) . The minimum amount of extra energy required by the reacting molecules to get converted into products is known as Activation energy.

(B) The minimum energy that the colliding molecules must possess for a chemical reaction to occur is known as the threshold energy. Molecules must cross this energy barrier to convert into products.

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

(C) catalyst provides an alternative pathway for the reaction with lower activation energy. By lowering the activation energy,a larger fraction of molecules can cross the energy barrier,which effectively increases the rate of the reaction. While the frequency of collisions is determined by temperature and concentration,the catalyst's primary role is to alter the reaction mechanism to a lower energy path. However,in the context of standard multiple-choice questions regarding the effect of catalysts on reaction rates,option $C$ is the most scientifically accurate description of how a catalyst functions.

(A) The enthalpy change of a reaction is given by the difference between the activation energy of the forward reaction $(E_a)_f$ and the activation energy of the reverse reaction $(E_a)_b$.
$\Delta H = (E_a)_f - (E_a)_b$
Given that the energies of activation for forward and reverse reactions are equal,i.e.,$(E_a)_f = (E_a)_b$.
Therefore,$\Delta H = 0$.

(C) In an endothermic reaction,the energy of the reactants is less than that of the products.
From the potential energy diagram for an endothermic reaction,the relationship between the activation energy of the forward reaction $(E_a)$,the activation energy of the backward reaction $(E_a')$,and the enthalpy of reaction $(\Delta H)$ is given by:
$E_a = E_a' + \Delta H$
Since the activation energy of the backward reaction $(E_a')$ must always be a positive value $(E_a' > 0)$,it follows that:
$E_a > \Delta H$
Therefore,the minimum value of $E_a$ is greater than $\Delta H$.

(B) catalyst is a substance that increases the rate of a chemical reaction without itself undergoing any permanent chemical change.
It functions by providing an alternative reaction pathway with a lower activation energy $(E_a)$.
As shown in the potential energy diagram,the catalyst reduces the energy barrier between reactants and products,but it does not change the energy of the reactants or products,meaning the enthalpy change $(\Delta H)$ remains unaffected.

(A) 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 equation follows the linear form $y = mx + c$,where $y = \ln k$,$x = \frac{1}{T}$,and the slope $m = -\frac{E_{a}}{R}$.
Therefore,the activation energy $E_{a}$ can be determined from the slope of the graph of $\ln k$ vs. $\frac{1}{T}$.

(A) The Arrhenius equation is given by: $\log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 \,R} \left( \frac{1}{T_{1}} - \frac{1}{T_{2}} \right)$
Given: $k_{2} = 2 \,k_{1}$,$T_{1} = 20 + 273 = 293 \,K$,$T_{2} = 35 + 273 = 308 \,K$,and $R = 8.314 \,J \,mol^{-1} \,K^{-1}$.
Substituting the values:
$\log(2) = \frac{E_{a}}{2.303 \times 8.314} \left( \frac{1}{293} - \frac{1}{308} \right)$
$0.3010 = \frac{E_{a}}{19.147} \times \left( \frac{308 - 293}{293 \times 308} \right)$
$0.3010 = \frac{E_{a}}{19.147} \times \frac{15}{90244}$
$E_{a} = \frac{0.3010 \times 19.147 \times 90244}{15} \approx 34673 \,J \,mol^{-1}$
Converting to $kJ \,mol^{-1}$: $E_{a} = 34.673 \,kJ \,mol^{-1} \approx 34.7 \,kJ \,mol^{-1}$.

(B) The number of $10\,^{\circ}C$ intervals $(n)$ is calculated as:
$n = \frac{T_2 - T_1}{10} = \frac{100 - 10}{10} = \frac{90}{10} = 9$
Since the rate doubles for every $10\,^{\circ}C$ rise,the rate increases by a factor of $2^n$.
Rate factor $= 2^9 = 512$
Therefore,the rate of the reaction will become $512$ times.

(D) The Arrhenius equation is given by: $k = A e^{-E_a / RT}$.
For two different temperatures $T_1$ and $T_2$ with corresponding rate constants $k_1$ and $k_2$:
$\ln k_1 = \ln A - \frac{E_a}{RT_1}$
$\ln k_2 = \ln A - \frac{E_a}{RT_2}$
Subtracting the first equation from the second:
$\ln k_2 - \ln k_1 = \left( \ln A - \frac{E_a}{RT_2} \right) - \left( \ln A - \frac{E_a}{RT_1} \right)$
$\ln \frac{k_2}{k_1} = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)$
Alternatively,this can be written as:
$\ln \frac{k_2}{k_1} = \frac{-E_a}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right)$
Thus,both expressions $(b)$ and $(c)$ are correct.

(A) The rate constant $k$ is a characteristic property of a reaction at a given temperature.
It is independent of the concentrations of the reactants.
According to the Arrhenius equation,$k = Ae^{-E_a/RT}$,the rate constant $k$ increases with an increase in temperature.

(B) Given the rate constants: $k_1 = 10^{16} e^{-2000/T}$ and $k_2 = 10^{15} e^{-1000/T}$.
At the temperature where $k_1 = k_2$,we have:
$10^{16} e^{-2000/T} = 10^{15} e^{-1000/T}$
Dividing both sides by $10^{15} e^{-2000/T}$:
$\frac{10^{16}}{10^{15}} = \frac{e^{-1000/T}}{e^{-2000/T}}$
$10 = e^{1000/T}$
Taking the natural logarithm $(\ln)$ on both sides:
$\ln(10) = \frac{1000}{T}$
Since $\ln(10) = 2.303 \log_{10}(10) = 2.303$,we get:
$2.303 = \frac{1000}{T}$
$T = \frac{1000}{2.303} \ K$

(C) In the Arrhenius equation $k = A e^{-E_a / RT}$,$E_a$ represents the activation energy.
Activation energy is defined as the minimum extra amount of energy absorbed by the reactant molecules so that their energy becomes equal to the threshold energy,allowing them to react and form products.
Therefore,it is the energy barrier that colliding molecules must overcome to react.

$(A)$ The enthalpy change of a reaction $(\Delta H)$ is defined as the difference between the activation energy of the forward reaction $(E_{af})$ and the activation energy of the reverse reaction $(E_{ab})$.
$\Delta H = E_{af} - E_{ab} = 180 \, kJ \, mol^{-1} - 200 \, kJ \, mol^{-1} = -20 \, kJ \, mol^{-1}$.
$A$ catalyst provides an alternative reaction pathway with lower activation energy, but it does not change the energy of the reactants or the products.
Since $\Delta H$ is a state function, it depends only on the initial and final states of the system.
Therefore, the enthalpy change of the reaction remains unchanged in the presence of a catalyst, which is $-20 \, kJ \, mol^{-1}$.

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

(A) The Arrhenius equation is given by $\log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 R} \left( \frac{T_{2} - T_{1}}{T_{1} T_{2}} \right).$
Given: $\frac{k_{2}}{k_{1}} = 2, T_{1} = 300 \, K, T_{2} = 310 \, K, R = 8.314 \, J \, K^{-1} \, mol^{-1}, \log 2 = 0.301.$
Substituting the values:
$0.301 = \frac{E_{a}}{2.303 \times 8.314} \left( \frac{310 - 300}{310 \times 300} \right)$
$0.301 = \frac{E_{a}}{19.147} \times \frac{10}{93000}$
$E_{a} = \frac{0.301 \times 19.147 \times 93000}{10}$
$E_{a} = 53598.6 \, J \, mol^{-1} = 53.6 \, kJ \, mol^{-1}.$

(D) According to the Arrhenius equation,$k = A e^{-E_a / RT}$.
For reactions $R_1$ and $R_2$ with identical pre-exponential factors $A$:
$k_1 = A e^{-E_{a1} / RT}$
$k_2 = A e^{-E_{a2} / RT}$
Dividing $k_2$ by $k_1$:
$\frac{k_2}{k_1} = \frac{A e^{-E_{a2} / RT}}{A e^{-E_{a1} / RT}} = e^{(E_{a1} - E_{a2}) / RT}$
Taking the natural logarithm on both sides:
$\ln(k_2/k_1) = \frac{E_{a1} - E_{a2}}{RT}$
Given $E_{a1} - E_{a2} = 10 \, kJ \, mol^{-1} = 10,000 \, J \, mol^{-1}$,$R = 8.314 \, J \, mol^{-1} \, K^{-1}$,and $T = 300 \, K$:
$\ln(k_2/k_1) = \frac{10,000}{8.314 \times 300} \approx \frac{10,000}{2494.2} \approx 4.009 \approx 4$.

(A) The Arrhenius equation is given by: $\ln \frac{K_{2}}{K_{1}} = \frac{E_{a}}{R} \left( \frac{T_{2} - T_{1}}{T_{1} T_{2}} \right)$.
Given: $K_{1} = 0.04 \, \text{min}^{-1}$,$K_{2} = 0.08 \, \text{min}^{-1}$,$T_{1} = 27 + 273 = 300 \, \text{K}$,$T_{2} = 37 + 273 = 310 \, \text{K}$,and $R = 2 \, \text{cal} \, \text{K}^{-1} \, \text{mol}^{-1}$.
Substituting the values: $\ln \left( \frac{0.08}{0.04} \right) = \frac{E_{a}}{2} \left( \frac{310 - 300}{300 \times 310} \right)$.
$\ln(2) = \frac{E_{a}}{2} \left( \frac{10}{93000} \right)$.
$0.7 = \frac{E_{a}}{2} \times \frac{1}{9300}$.
$E_{a} = 0.7 \times 2 \times 9300 = 13020 \, \text{cal} / \text{mol}$.
Converting to $\text{kcal} / \text{mol}$: $E_{a} = 13.02 \, \text{kcal} / \text{mol}$.

(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 in Kelvin.
As the temperature $T$ increases,the term $e^{-E_{a} / RT}$ increases,which leads to an increase in the rate constant $k$ for all elementary reactions.
Option $A$ and $B$ depend on the sign of $\Delta C_{p}$,which varies for different reactions.
Option $D$ is incorrect because activation energy is generally considered independent of temperature for elementary reactions.

(A) According to the Arrhenius equation,$k = A e^{-E_a/RT}$,the activation energy $(E_a)$ is a characteristic property of a reaction and is generally considered independent of temperature. Therefore,the statement that activation energy decreases on decreasing temperature is incorrect.
$(b)$ The order of reaction can change with temperature if the mechanism of the reaction changes.
$(c)$ If the first step is the rate-determining step (slowest step),the rate law of the overall reaction is determined by the rate law of that step.
$(d)$ The rate of a photochemical reaction is directly proportional to the intensity of the incident light (photons).

(A) The enthalpy change of a reaction is given by 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 $\Delta H = -22 \ kcal$ and $(E_a)_f = 70 \ kcal$:
$-22 = 70 - (E_a)_b$
Rearranging the equation to solve for $(E_a)_b$:
$(E_a)_b = 70 + 22 = 92 \ kcal$.

(D) Given that $K_1 = K_2$,we have:
$10^{15} \exp \left( \frac{-2000}{T} \right) = 10^{14} \exp \left( \frac{-1000}{T} \right)$
Dividing both sides by $10^{14}$:
$10 \exp \left( \frac{-2000}{T} \right) = \exp \left( \frac{-1000}{T} \right)$
Rearranging the exponential terms:
$10 = \frac{\exp \left( \frac{-1000}{T} \right)}{\exp \left( \frac{-2000}{T} \right)} = \exp \left( \frac{-1000}{T} - \left( \frac{-2000}{T} \right) \right) = \exp \left( \frac{1000}{T} \right)$
Taking the natural logarithm $(\ln)$ on both sides:
$\ln(10) = \frac{1000}{T}$
Using $\ln(10) \approx 2.303$:
$2.303 = \frac{1000}{T}$
$T = \frac{1000}{2.303} \approx 434.22 \ K$

(B) For an exothermic reaction,the enthalpy change $\Delta_{r} H$ is negative. 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 $\Delta_{r} H$ is given by:
$\Delta_{r} H = (E_{a})_{f} - (E_{a})_{r}$
Given that the reaction is exothermic,the heat evolved is $280 \ kJ \ mol^{-1}$,so $\Delta_{r} H = -280 \ kJ \ mol^{-1}$.
The activation energy of the forward reaction $(E_{a})_{f} = 200 \ kJ \ mol^{-1}$.
Substituting these values into the equation:
$-280 \ kJ \ mol^{-1} = 200 \ kJ \ mol^{-1} - (E_{a})_{r}$
$(E_{a})_{r} = 200 \ kJ \ mol^{-1} + 280 \ kJ \ mol^{-1} = 480 \ kJ \ mol^{-1}$.

(D) The Arrhenius equation is given by $\ln k = \ln A - \frac{E_a}{RT}$.
Converting to base $10$,we get $\log_{10} k = \log_{10} A - \frac{E_a}{2.303 RT}$.
Comparing this with the given equation $\log_{10} k = -2000 \left(\frac{1}{T}\right) + 6$,we identify the slope as $-\frac{E_a}{2.303 R} = -2000$.
Thus,$E_a = 2000 \times 2.303 \times R$.
Using $R = 8 \, J \, mol^{-1} K^{-1}$ and the approximation $\ln x \approx 2.3 \log_{10} x$ provided in the question:
$E_a = 2000 \times 2.3 \times 8 = 36800 \, J \, mol^{-1}$.
Converting to $kJ \, mol^{-1}$,we get $E_a = \frac{36800}{1000} = 36.8 \, kJ \, mol^{-1}$.

(B) The Arrhenius equation is given by $k = A e^{-E_a / RT}$.
Taking the natural logarithm,$\ln k = \ln A - \frac{E_a}{RT}$.
Differentiating with respect to $T$,$\frac{d(\ln k)}{dT} = \frac{E_a}{RT^2}$.
For a small temperature change $\Delta T$,the change in $\ln k$ is $\Delta(\ln k) \approx \frac{E_a}{RT^2} \Delta T$.
The percentage increase in rate constant is proportional to $\frac{\Delta k}{k} \approx \Delta(\ln k) = \frac{E_a \Delta T}{RT^2}$.
To maximize the percentage increase,we need to maximize the value of $\frac{E_a}{T^2}$ (since $\Delta T$ is constant at $10 \ K$ for all cases).
Comparing the cases:
$I$: $\frac{40}{200^2} = \frac{40}{40000} = 1 \times 10^{-3}$
$II$: $\frac{80}{200^2} = \frac{80}{40000} = 2 \times 10^{-3}$
$III$: $\frac{40}{300^2} = \frac{40}{90000} \approx 0.44 \times 10^{-3}$
$IV$: $\frac{80}{300^2} = \frac{80}{90000} \approx 0.89 \times 10^{-3}$
The value is maximum for case $II$.