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

(D) The activation energy $(E_a)$ is the minimum energy required for a chemical reaction to occur.
According to the Arrhenius equation,$k = A e^{-E_a / RT}$,where $k$ is the rate constant,$A$ is the frequency factor,$R$ is the gas constant,and $T$ is the temperature.
For most reactions,the activation energy $(E_a)$ is considered to be a characteristic property of the reaction path and is nearly independent of temperature over a wide range of temperatures.
Therefore,option $D$ is the correct statement.

(C) The Arrhenius equation is given by $k = A e^{-E_{a} / RT}$.
Comparing the given equation $k = (4.5 \times 10^{11} \ s^{-1}) e^{-(28000 \ K) / T}$ with the Arrhenius equation,we identify the exponent term:
$E_{a} / R = 28000 \ K$.
Therefore,the activation energy $E_{a} = 28000 \times R$.
Using the gas constant $R = 8.314 \ J \ K^{-1} \ mol^{-1}$,we get:
$E_{a} = 28000 \times 8.314 \ J/mol = 232792 \ J/mol$.

(D) The Arrhenius equation is given by $k = A e^{-E_a / (RT)}$.
$(1)$ The rate constant $k$ is a function of temperature $T$.
$(2)$ As temperature $T$ increases,the fraction of molecules with energy greater than the threshold energy increases,leading to more effective collisions.
$(3)$ In the standard Arrhenius model,the activation energy $E_a$ and the pre-exponential factor $A$ are considered to be temperature independent constants.
$(4)$ Therefore,all statements $A$,$B$,and $C$ are correct according to the theory,while $D$ is technically an approximation but often treated as a correct premise in basic kinetics; however,if we look for the most accurate theoretical description,the Arrhenius equation assumes $E_a$ and $A$ are independent of $T$. Since the question asks for the incorrect statement,and all provided options are standard interpretations,there might be a misunderstanding. Re-evaluating: all statements are actually consistent with the basic Arrhenius model. Given the standard curriculum,this question is often flawed or implies that $A$ and $E_a$ might vary slightly with $T$ in complex systems. However,based on the provided options,none are strictly 'incorrect' in the context of the basic Arrhenius equation.

(A) The Arrhenius equation is $k = A e^{-E_a / RT}$.
Given $k = (\frac{k_1 k_2}{k_3})^{2/3}$,we substitute $k_i = A_i e^{-E_{a_i} / RT}$:
$A e^{-E_a / RT} = [\frac{A_1 e^{-E_{a_1} / RT} \times A_2 e^{-E_{a_2} / RT}}{A_3 e^{-E_{a_3} / RT}}]^{2/3}$
$A e^{-E_a / RT} = [\frac{A_1 A_2}{A_3}]^{2/3} \times e^{(-E_{a_1} - E_{a_2} + E_{a_3}) \times \frac{2}{3RT}}$
Comparing the exponents of $e$:
$-\frac{E_a}{RT} = \frac{2}{3} \times \frac{(-E_{a_1} - E_{a_2} + E_{a_3})}{RT}$
$E_a = \frac{2}{3} [E_{a_1} + E_{a_2} - E_{a_3}]$
Substituting the values:
$E_a = \frac{2}{3} [180 + 80 - 50]$
$E_a = \frac{2}{3} [210] = 140\ kJ/mol$.

(A) Given the rate constants $k_1 = 10^{15} e^{-25000 / 8.34 T}$ and $k_2 = 10^{14} e^{-15000 / 8.34 T}$.
Setting $k_1 = k_2$:
$10^{15} e^{-25000 / 8.34 T} = 10^{14} e^{-15000 / 8.34 T}$
$\frac{10^{15}}{10^{14}} = \frac{e^{-15000 / 8.34 T}}{e^{-25000 / 8.34 T}}$
$10 = e^{(25000 - 15000) / 8.34 T} = e^{10000 / 8.34 T}$
Taking the natural logarithm on both sides:
$\ln(10) = \frac{10000}{8.34 T}$
$2.303 = \frac{10000}{8.34 T}$
$T = \frac{10000}{8.34 \times 2.303} \approx \frac{10000}{19.207} \approx 520.6 \ K$
Rounding to the nearest integer,$T = 520 \ K$.
Hence,the correct answer is $(A)$.

(D) According to the Arrhenius equation,$K = Ae^{-E_a/RT}$.
$(1)$ The term $e^{-E_a/RT}$ represents the fraction of molecules having energy greater than or equal to activation energy $(E_a)$. This is correct.
$(2)$ As $E_a$ decreases,the value of $e^{-E_a/RT}$ increases,making the rate constant $K$ larger,thus the reaction is faster. This is correct.
$(3)$ The Arrhenius factor $A$ (frequency factor) is indeed the product of the collision frequency $(Z)$ and the steric/probability factor $(P)$,i.e.,$A = PZ$. This is correct.
$(4)$ The temperature dependence of a reaction is determined by the activation energy. $A$ reaction with a high $E_a$ is more sensitive to temperature changes because the exponential term $e^{-E_a/RT}$ changes significantly with temperature. Therefore,the statement that it is 'less temperature dependent' is incorrect.

(B) The rate of a chemical reaction is proportional to the number of effective collisions,which depends on the fraction of molecules having energy greater than or equal to the activation energy $(E_a)$.
According to the Arrhenius equation,the fraction of molecules with energy $\geq E_a$ is given by $f = e^{-E_a/RT}$.
For a $10^{\circ}C$ rise in temperature,this fraction approximately doubles,leading to a doubling of the reaction rate.
Option $(A)$ is incorrect because the average speed $u$ is proportional to $\sqrt{T}$,so it does not double.
Option $(C)$ is incorrect as concentration is independent of temperature.
Option $(D)$ is incorrect because collision frequency increases only by a small factor (proportional to $\sqrt{T}$),not by a factor of $2$.

(A) Given: $T_{1} = 27\,^{\circ}C = 300\,K$,$(K_{eq})_{1} = 30$.
$T_{2} = 127\,^{\circ}C = 400\,K$,$(K_{eq})_{2} = 40$.
Since the equilibrium constant increases with an increase in temperature,the reaction is endothermic.
For an endothermic reaction,the enthalpy change $\Delta H = (E_{a})_{f} - (E_{a})_{b} > 0$.
This implies $(E_{a})_{f} > (E_{a})_{b}$.
Therefore,the ratio $\frac{(E_{a})_{f}}{(E_{a})_{b}} > 1$.

(D) According to the Arrhenius equation:
$k = A \cdot e^{-E_a/RT}$ (Catalyst absent)
$k' = A \cdot e^{-E'_a/RT}$ (Catalyst present)
Given $E'_a = 4.15 \, kJ \, mol^{-1}$ and $\frac{k'}{k} = 2.718 = e^1$.
Taking the ratio:
$\frac{k'}{k} = e^{(E_a - E'_a)/RT}$
$e^1 = e^{(E_a - E'_a)/RT}$
Therefore,$1 = \frac{E_a - E'_a}{RT}$
$E_a = E'_a + RT$
Substituting the values ($R = 8.314 \times 10^{-3} \, kJ \, K^{-1} \, mol^{-1}$,$T = 500 \, K$):
$E_a = 4.15 + (8.314 \times 10^{-3} \times 500)$
$E_a = 4.15 + 4.157 = 8.307 \, kJ \, mol^{-1} \approx 8.3 \, kJ \, mol^{-1}$.

(D) $1$. In the given energy profile diagram,curve $N$ represents the reaction without a catalyst,and curve $M$ represents the reaction with a catalyst.
$2$. The activation energy for the backward reaction is the energy difference between the transition state (peak of the curve) and the energy level of the products $(C+D)$.
$3$. In the presence of a catalyst,the transition state is at the energy level corresponding to curve $M$.
$4$. The energy level of the products $(C+D)$ is at a distance $P$ below the energy level of the reactants $(A+B)$.
$5$. The energy level of the transition state for the catalyzed reaction $(M)$ is at a distance $Q$ above the reactants $(A+B)$.
$6$. Therefore,the activation energy for the backward reaction in the presence of a catalyst is the sum of the energy difference between the products and reactants $(P)$ and the energy difference between the reactants and the catalyzed transition state $(Q)$,which is $P+Q$.

(B) For an exothermic reaction,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 change $(\Delta H)$ is given by:
$\Delta H = E_{a(f)} - E_{a(b)}$
Given:
$E_{a(f)} = 65 \ kJ \ mol^{-1}$
$\Delta H = -42 \ kJ \ mol^{-1}$
Substituting the values:
$-42 = 65 - E_{a(b)}$
$E_{a(b)} = 65 + 42 = 107 \ kJ \ mol^{-1}$

(B) The Arrhenius equation is given by $k = A \times e^{-E_a/RT}$.
Statement $(i)$ is incorrect because the rate of a zero-order reaction is independent of the concentration of the reactant.
Statement $(ii)$ is correct because if $E_a = 0$,then $k = A \times e^0 = A$.
Statement $(iii)$ is incorrect because if $E_a = \infty$,then $k = A \times e^{-\infty} = 0$.
Statement $(iv)$ is incorrect because $\ln k = \ln A - \frac{E_a}{RT}$,which shows that $\ln k$ is a function of $1/T$,not $T$.
Statement $(v)$ is correct because the equation $\ln k = \ln A - \frac{E_a}{R} \times (1/T)$ represents a straight line with a slope of $-E_a/R$ when plotting $\ln k$ versus $1/T$.
Therefore,statements $(ii)$ and $(v)$ are correct.

(D) Given that $k_1 = k_2$,we have $A_1 e^{-E_{a_1} / RT} = A_2 e^{-E_{a_2} / RT}$.
Substituting the values: $10^8 e^{-600 / RT} = 10^{10} e^{-1800 / RT}$.
Rearranging the terms: $\frac{e^{-600 / RT}}{e^{-1800 / RT}} = \frac{10^{10}}{10^8}$.
$e^{(1800 - 600) / RT} = 10^2$.
$e^{1200 / RT} = 100$.
Taking the natural logarithm on both sides: $\frac{1200}{RT} = \ln(100) = 2 \ln(10)$.
Using $\ln(10) \approx 2.303$ and $R = 2 \ cal/K \cdot mol$:
$\frac{1200}{2 \times T} = 2 \times 2.303$.
$\frac{600}{T} = 4.606$.
$T = \frac{600}{4.606} \ K$.

(A) The overall rate constant $k$ is given by $k = (\frac{k_1 k_2}{k_3})^{2/3}$.
Using the Arrhenius equation $k = A e^{-E_a/RT}$,we substitute the expressions for each rate constant:
$e^{-E_a/RT} = [\frac{e^{-E_{a1}/RT} \times e^{-E_{a2}/RT}}{e^{-E_{a3}/RT}}]^{2/3}$
Taking the natural logarithm on both sides:
$-\frac{E_a}{RT} = \frac{2}{3} [-\frac{E_{a1}}{RT} - \frac{E_{a2}}{RT} + \frac{E_{a3}}{RT}]$
$E_a = \frac{2}{3} [E_{a1} + E_{a2} - E_{a3}]$
Substituting the given values:
$E_a = \frac{2}{3} [180 + 80 - 50] = \frac{2}{3} [210] = 140 \ kJ/mol$.

(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: $\log \, k = \log \, A - \frac{E_a}{2.303 \, R} \times \frac{1}{T}$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = \log \, k$ and $x = \frac{1}{T}$,the slope $m = -\frac{E_a}{2.303 \, R}$.
Therefore,a plot of $\log \, k$ versus $\frac{1}{T}$ yields a straight line.

(A) The temperature coefficient $\mu$ is defined as the ratio of rate constants at temperatures $(T + 10) \ K$ and $T \ K$.
From the Arrhenius equation,$k = Ae^{-E_a / RT}$,it can be derived that $\log \mu \propto E_a$.
This implies that the temperature coefficient $\mu$ increases as the activation energy $E_a$ increases.
Given $E_{a_1} > E_{a_2}$,it follows that $\mu_1 > \mu_2$.

(C) The Arrhenius equation is given by $k = A \times e^{-E_a / (RT)}$.
Given that the activation energy $E_a = 0$.
Substituting $E_a = 0$ into the equation,we get $k = A \times e^0 = A \times 1 = A$.
This implies that the rate constant $k$ is equal to the frequency factor $A$ and is independent of temperature $T$.
Since $k = 3.2 \times 10^8 \ s^{-1}$ at $300 \ K$,the frequency factor $A$ is $3.2 \times 10^8 \ s^{-1}$.
Therefore,at $400 \ K$,the value of the frequency factor $A$ remains $3.2 \times 10^8 \ s^{-1}$.

(C) The temperature coefficient $\mu$ is given as $3$ for a $10^\circ C$ rise in temperature.
The change in temperature $\Delta T$ is $60^\circ C$.
The increase in the reaction rate is calculated using the formula: $\text{Rate increase} = \mu^{(\Delta T / 10)}$.
Substituting the values: $\text{Rate increase} = 3^{(60 / 10)} = 3^6$.
Calculating the value: $3^6 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 729$.

(A) The Arrhenius equation is given by $\ln(\frac{k_2}{k_1}) = \frac{E_a}{R} (\frac{1}{T_1} - \frac{1}{T_2})$.
Given: $T_1 = 27 + 273 = 300 \ K$,$T_2 = 327 + 273 = 600 \ K$,$E_a = 24.942 \ kJ \ mol^{-1} = 24942 \ J \ mol^{-1}$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$.
Substituting the values: $\ln(\frac{k_2}{k_1}) = \frac{24942}{8.314} (\frac{1}{300} - \frac{1}{600}) = 3000 \times (\frac{2-1}{600}) = 3000 \times \frac{1}{600} = 5$.
Therefore,$\frac{k_2}{k_1} = e^5 = 150$.
Since the rate of a first-order reaction is proportional to the rate constant $(Rate = k[A])$,the ratio of rates is equal to the ratio of rate constants.
$Rate_2 = 150 \times Rate_1 = 150 \times 2.4 \times 10^{-3} = 360 \times 10^{-3} = 0.36 \ mol \ L^{-1} \ s^{-1}$.

(A) Given $k_1 = k_2$:
$10^{15} e^{-25000 / 8.314\, T} = 10^{14} e^{-15000 / 8.314\, T}$
$\frac{10^{15}}{10^{14}} = e^{(25000 - 15000) / 8.314\, T}$
$10 = e^{10000 / 8.314\, T}$
Taking natural logarithm on both sides:
$ln(10) = \frac{10000}{8.314\, T}$
Given $ln(10) = 2.3$:
$2.3 = \frac{10000}{8.314\, T}$
$T = \frac{10000}{8.314 \times 2.3} \approx 522\, K$
Thus,the temperature is $522\, K$.

(B) In the Arrhenius equation modified by collision theory,$r = P Z_{AB} e^{-E_a/RT}$,the term $P$ is known as the steric factor or probability factor.
It accounts for the requirement that molecules must be properly oriented during a collision to result in a chemical reaction.

(C) Using the Arrhenius equation: $\log \frac{K_{2}}{K_{1}} = \frac{E_{a}}{2.303 R} \left[ \frac{T_{2} - T_{1}}{T_{1} T_{2}} \right]$
Given: $E_{a} = 65,000 \, J/mol$,$R = 8.314 \, J/K \cdot mol$,$T_{1} = 273 \, K$,$T_{2} = 298 \, K$.
Substituting the values:
$\log \frac{K_{2}}{K_{1}} = \frac{65000}{2.303 \times 8.314} \left[ \frac{298 - 273}{273 \times 298} \right]$
$\log \frac{K_{2}}{K_{1}} = \frac{65000}{19.147} \times \frac{25}{81354}$
$\log \frac{K_{2}}{K_{1}} = 3394.78 \times 0.000307 \approx 1.043$
$\frac{K_{2}}{K_{1}} = 10^{1.043} \approx 11$
Thus,the reaction proceeds approximately $11$ times faster.

(A, B) The Arrhenius equation is given by $k = A e^{-E_a/RT}$.
For the rate constant $k$ to be equal to the Arrhenius constant $A$,the exponential term $e^{-E_a/RT}$ must be equal to $1$.
This occurs when the exponent is $0$,i.e.,$-E_a/RT = 0$.
This condition is satisfied if $E_a = 0$ (activation energy is zero).
Alternatively,as $T \to \infty$,the term $E_a/RT \to 0$,so $e^0 = 1$,which also makes $k = A$.

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

(C) The Arrhenius equation is given by $K = A \cdot e^{-E_a / RT}$.
Comparing this with the given equation $K = A \cdot e^{-40000/T}$,we get $\frac{E_a}{R} = 40000$.
Given that the gas constant $R \approx 2 \, cal \cdot K^{-1} \cdot mol^{-1}$.
Therefore,$E_a = 40000 \times 2 = 80000 \, cal$.

(D) The rate of reaction increases by a factor of $2^{\frac{\Delta T}{10}}$ for every $10 ^\circ C$ rise in temperature.
Here,$\Delta T = 100 ^\circ C - 10 ^\circ C = 90 ^\circ C$.
Therefore,the rate of reaction will become $(2)^{\frac{90}{10}}$ times.
$= 2^9 = 512$ times.

(A) Using the Arrhenius equation: $\log \frac{K_{2}}{K_{1}} = \frac{E_a}{2.303 \ R} \left( \frac{T_{2} - T_{1}}{T_{1} T_{2}} \right)$
Given: $K_{2} = 2K_{1}$,$T_{1} = 300 \ K$,$T_{2} = 310 \ K$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$.
$\log 2 = \frac{E_a}{2.303 \times 8.314} \left( \frac{310 - 300}{300 \times 310} \right)$
$0.3010 = \frac{E_a}{19.147} \left( \frac{10}{93000} \right)$
$E_a = \frac{0.3010 \times 19.147 \times 93000}{10} \ J/mol$
$E_a \approx 53598 \ J/mol = 53.6 \ kJ/mol$.

(A) The rate of reaction changes with temperature according to the formula: $\frac{R_{2}}{R_{1}} = (\mu)^{\frac{\Delta T}{10}}$.
Here,$\Delta T = 55\,^{\circ}C - 25\,^{\circ}C = 30\,^{\circ}C$ and $\mu = 3$.
Substituting the values: $\frac{R_{2}}{R_{1}} = (3)^{\frac{30}{10}} = (3)^{3} = 27$.

(C) The Arrhenius equation is given by $k = A e^{-\frac{E_a}{RT}}$.
Taking the natural logarithm on both sides,we get $\ln k = \ln A - \frac{E_a}{RT}$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = \ln k$ and $x = \frac{1}{T}$,the slope $m = -\frac{E_a}{R}$.
Option $C$ is correct because the slope of the plot of $\ln k$ versus $\frac{1}{T}$ is $-\frac{E_a}{R}$.
Note: If the plot is $\log_{10} k$ versus $\frac{1}{T}$,the slope is $-\frac{E_a}{2.303R}$.

(A) The Arrhenius equation in logarithmic form is $\log_{10} K = \log_{10} A - \frac{E_a}{2.303RT}$.
Comparing this with the given equation $\log_{10} K = -\frac{2000}{T} + 6.0$,we get $\frac{E_a}{2.303R} = 2000$.
Given $R = 1.987 \times 10^{-3} \ kcal \ K^{-1} \ mol^{-1} \approx 2 \times 10^{-3} \ kcal \ K^{-1} \ mol^{-1}$.
$E_a = 2000 \times 2.303 \times 2 \times 10^{-3} \ kcal \ mol^{-1}$.
$E_a = 2000 \times 2.303 \times 0.002 = 9.212 \ kcal \ mol^{-1}$.
Thus,the activation energy is $9.21 \ kcal \ mol^{-1}$.

(C) For the forward reaction $A + B \to C + D$,the activation energy is $x$ and the enthalpy change is $-y$ (since it is an exothermic reaction).
For the reverse reaction $C + D \to A + B$:
$1$. The enthalpy change is equal in magnitude but opposite in sign to the forward reaction,which is $y$.
$2$. The activation energy for the reverse reaction is the energy difference between the transition state and the reactants $(C + D)$,which is $x + y$.

(B) The fraction of molecules $(f)$ having energy equal to or greater than the activation energy $(E_a)$ is given by the Arrhenius equation: $f = e^{-E_a/RT}$.
Given:
$E_a = 200 \ kJ \ mol^{-1} = 200,000 \ J \ mol^{-1}$
$R = 8.314 \ J \ K^{-1} \ mol^{-1}$
$T = 600 \ K$
Substituting the values:
$f = e^{-(200,000) / (8.314 \times 600)}$
$f = e^{-(200,000) / (4988.4)}$
$f \approx e^{-40.09} \approx e^{-40}$.

(A) $1$. The rate of a reaction is inversely proportional to the activation energy $(E_a)$. $A$ lower activation energy means a faster reaction rate.
$2$. From the provided energy profile diagrams,it is clear that the activation energy for reaction $A$ $(E_{a1})$ is lower than the activation energy for reaction $B$ $(E_{a2})$,i.e.,$E_{a1} < E_{a2}$. Therefore,reaction $A$ is faster than reaction $B$.
$3$. Exergonic reactions are those where the free energy of the products is lower than that of the reactants (negative $\Delta G$). Both reactions are exergonic as the energy of the products is lower than the reactants.
$4$. Comparing the overall energy change $(\Delta E)$,the drop in energy for reaction $B$ is greater than for reaction $A$,making reaction $B$ more exergonic than reaction $A$.
$5$. Thus,reaction $A$ is faster,but reaction $B$ is more exergonic. This implies that reaction $A$ is faster and less exergonic than reaction $B$.

(A) The rate-determining step $(R.D.S.)$ of a multi-step reaction is the step with the highest activation energy $(E_a)$.
In the given energy profile diagram,the peak labeled $1$ represents the highest energy barrier relative to the preceding intermediate or reactant.
Therefore,the step corresponding to the highest peak,which is $A \to B$,is the rate-determining step.

(C) According to the Arrhenius equation,$K = Ae^{-E_a/RT}$.
Therefore,a greater activation energy $(E_a)$ results in a smaller rate constant $(K)$.
The rate of reaction is inversely proportional to the height of the energy barrier.
Since the first step is the slowest (rate-determining step),it must have the highest activation energy barrier.
Therefore,the first peak in the energy profile diagram must be the highest,as shown in option $(C)$.

(C) The rate constant $k$ is related to the activation energy $E_a$ by the Arrhenius equation,$k = A e^{-E_a/RT}$.
To increase the rate constant $k$,the activation energy $E_a$ must be decreased.
$E_a = E_{\text{transition state}} - E_{\text{reactant}}$.
Decreasing the stability of the reactant increases its energy,which decreases $E_a$.
Increasing the stability of the transition state decreases its energy,which also decreases $E_a$.
Therefore,the rate constant increases by decreasing the stability of the reactant or by increasing the stability of the transition state.

(B) Using the Arrhenius equation: $\ln(\frac{k_2}{k_1}) = \frac{E_a}{R} [\frac{T_2 - T_1}{T_1 T_2}]$.
For reaction $A$: $\ln(2) = \frac{E_{a,A}}{R} [\frac{310 - 300}{300 \times 310}] = \frac{E_{a,A}}{R} [\frac{10}{93000}]$.
For reaction $B$: $\ln(2) = \frac{E_{a,B}}{R} [\frac{\Delta T}{300(300 + \Delta T)}]$.
Given $E_{a,B} = 2 E_{a,A}$,we substitute $\ln(2)$ from reaction $A$ into reaction $B$:
$\frac{E_{a,A}}{R} [\frac{10}{93000}] = \frac{2 E_{a,A}}{R} [\frac{\Delta T}{300(300 + \Delta T)}]$.
$\frac{10}{93000} = \frac{2 \Delta T}{300(300 + \Delta T)}$.
$\frac{1}{9300} = \frac{\Delta T}{150(300 + \Delta T)}$.
$150(300 + \Delta T) = 9300 \Delta T$.
$45000 + 150 \Delta T = 9300 \Delta T$.
$9150 \Delta T = 45000$.
$\Delta T = \frac{45000}{9150} \approx 4.92 \, K$.

(A) Using the Arrhenius equation: $\ln \frac{k_2}{k_1} = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)$
Given: $\frac{k_2}{k_1} = 4$,$T_1 = 300 \, K$,$T_2 = 310 \, K$,$R = 8.314 \, J \, mol^{-1} \, K^{-1}$,$\ln 2 = 0.693$.
$\ln 4 = \frac{E_a}{8.314} \left( \frac{310 - 300}{310 \times 300} \right)$
$2 \ln 2 = \frac{E_a}{8.314} \left( \frac{10}{93000} \right)$
$2 \times 0.693 = \frac{E_a}{8.314} \times \frac{1}{9300}$
$E_a = 1.386 \times 8.314 \times 9300 \, J \, mol^{-1} \approx 107200 \, J \, mol^{-1} = 107.2 \, kJ \, mol^{-1}$.

(A) Given,$\frac{E_f}{E_b} = \frac{2}{3}$.
We know that for a reaction,$\Delta H = E_f - E_b$.
Given $\Delta H = -40 \ kJ/mol$,so $-40 = E_f - E_b$,which implies $E_b = E_f + 40$.
Substituting this into the ratio: $\frac{E_f}{E_f + 40} = \frac{2}{3}$.
Cross-multiplying gives $3E_f = 2(E_f + 40)$.
$3E_f = 2E_f + 80$.
$E_f = 80 \ kJ/mol$.
Then,$E_b = 80 + 40 = 120 \ kJ/mol$.

(B) According to the Arrhenius equation:
$\log \frac{k_2}{k_1} = \frac{E_a}{2.303 \ R} \left( \frac{T_2 - T_1}{T_1 T_2} \right)$
Given: $k_1 = 1.3 \times 10^{-4} \ M^{-1} \ s^{-1}$,$T_1 = 100 + 273 = 373 \ K$
$k_2 = 1.3 \times 10^{-3} \ M^{-1} \ s^{-1}$,$T_2 = 150 + 273 = 423 \ K$
$\log \left( \frac{1.3 \times 10^{-3}}{1.3 \times 10^{-4}} \right) = \frac{E_a}{2.303 \times 8.314} \left( \frac{423 - 373}{373 \times 423} \right)$
$\log(10) = \frac{E_a}{19.147} \left( \frac{50}{157779} \right)$
$1 = \frac{E_a \times 50}{3020755.7}$
$E_a = \frac{3020755.7}{50} \approx 60415 \ J \ mol^{-1} \approx 60.4 \ kJ \ mol^{-1}$
Thus,the closest value is $60 \ kJ \ mol^{-1}$.

(B) For a reaction $X \to Y$,the enthalpy change $\Delta H$ 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:
$E_{a(f)} = 150\,kJ\,mol^{-1}$
$\Delta H = -135\,kJ\,mol^{-1}$
Substituting the values into the equation:
$-135 = 150 - E_{a(b)}$
Rearranging to solve for $E_{a(b)}$:
$E_{a(b)} = 150 + 135 = 285\,kJ\,mol^{-1}$

(C) The activation energy $(E_a)$ can be calculated using the Arrhenius equation: $\log(\frac{K_2}{K_1}) = \frac{E_a}{2.303 \ R} \times \frac{T_2 - T_1}{T_1 \times T_2}$.
Given: $\frac{K_2}{K_1} = 2$,$T_1 = 298 \ K$,$T_2 = 308 \ K$,and $R = 8.314 \ J \ K^{-1} \ mol^{-1}$.
Substituting the values: $\log(2) = \frac{E_a}{2.303 \times 8.314} \times \frac{308 - 298}{308 \times 298}$.
$0.3010 = \frac{E_a}{19.147} \times \frac{10}{91784}$.
$E_a = \frac{0.3010 \times 19.147 \times 91784}{10} \approx 52897 \ J \ mol^{-1} = 52.9 \ kJ \ mol^{-1}$.

(A) The Arrhenius equation is given by $k = A e^{-E_a/RT}$.
Plot $I$: As $E_a$ increases,the term $e^{-E_a/RT}$ decreases,so $k$ decreases. Thus,the plot of $k$ versus $E_a$ is an exponential decay curve,which is correct.
Plot $II$: As temperature $T$ increases,the term $e^{-E_a/RT}$ increases,so $k$ increases exponentially with $T$. The provided plot $II$ shows a curve that does not represent the standard exponential increase of $k$ with $T$ correctly (it looks more like a linear or different growth). Therefore,$II$ is incorrect.
Hence,$I$ is right but $II$ is wrong.

(A) The Arrhenius equation is given by $\ln k = -\frac{E_a}{RT} + \ln A$.
Comparing this with the equation of a straight line $y = mx + c$,where $x = 1/(RT)$,the slope $m$ is equal to $-E_a$.
Given that the gradient is $-y$,we have $-E_a = -y$.
Therefore,the activation energy $E_a$ is equal to $y \ unit$.

(C) From the Arrhenius equation,$\ln k = \ln A - \frac{E_a}{R} \left( \frac{1}{T} \right)$.
Comparing this with the equation of a straight line $y = mx + c$,the slope $m = -\frac{E_a}{R}$.
Given slope $= -4606 \ K$,so $\frac{E_a}{R} = 4606 \ K$.
Using the integrated Arrhenius equation: $\ln \left( \frac{k_2}{k_1} \right) = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)$.
Here,$T_1 = 400 \ K$,$k_1 = 10^{-5} \ s^{-1}$,$T_2 = 500 \ K$,and $k_2 = ?$.
$\ln \left( \frac{k_2}{10^{-5}} \right) = 4606 \times \left( \frac{1}{400} - \frac{1}{500} \right) = 4606 \times \left( \frac{500 - 400}{200000} \right) = 4606 \times \frac{100}{200000} = 4606 \times \frac{1}{2000} = 2.303$.
Since $\ln 10 \approx 2.303$,we have $\ln \left( \frac{k_2}{10^{-5}} \right) = \ln 10$.
Therefore,$\frac{k_2}{10^{-5}} = 10$,which gives $k_2 = 10^{-4} \ s^{-1}$.

(D) From the graph:
Enthalpy of $(A+B) = 5\,kJ\,mol^{-1}$.
Enthalpy of $D = 10\,kJ\,mol^{-1}$.
Enthalpy of $C = 0\,kJ\,mol^{-1}$.
Activation energy for $(A+B \to D) = 15 - 5 = 10\,kJ\,mol^{-1}$.
Activation energy for $(A+B \to C) = 20 - 5 = 15\,kJ\,mol^{-1}$.
Activation energy for $(C \to A+B) = 20 - 0 = 20\,kJ\,mol^{-1}$.
Activation energy for $(D \to A+B) = 15 - 10 = 5\,kJ\,mol^{-1}$.
Comparing the activation energies,the formation of $(A+B)$ from $C$ has the highest activation energy $(20\,kJ\,mol^{-1})$.
$C$ is the most stable product (lowest enthalpy).
$D$ is formed faster (lower activation energy),so it is the kinetically stable product.
Activation enthalpy to form $C$ $(15\,kJ\,mol^{-1})$ is $5\,kJ\,mol^{-1}$ more than that to form $D$ $(10\,kJ\,mol^{-1})$.
Therefore,statement $D$ is incorrect.

(B) 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:
$K_1 = 2.5 \times 10^{-4} \ dm^3 \ mol^{-1} \ s^{-1}$,$T_1 = 327 + 273 = 600 \ K$
$K_2 = 1.0 \ dm^3 \ mol^{-1} \ s^{-1}$,$T_2 = 527 + 273 = 800 \ K$
$R = 8.314 \ J \ K^{-1} \ mol^{-1}$
Substituting the values:
$\log \left( \frac{1}{2.5 \times 10^{-4}} \right) = \frac{E_a}{2.303 \times 8.314} \left( \frac{800 - 600}{600 \times 800} \right)$
$\log (4000) = \frac{E_a}{19.147} \left( \frac{200}{480000} \right)$
$3.602 = \frac{E_a}{19.147} \times 4.167 \times 10^{-4}$
$E_a = \frac{3.602 \times 19.147}{4.167 \times 10^{-4}} \approx 165500 \ J \ mol^{-1} = 165.5 \ kJ \ mol^{-1} \approx 166 \ kJ \ mol^{-1}$

(D) The Arrhenius equation is given by $K = A \ e^{-E_a / (RT)}$.
Comparing this with the given expression $K = A \ e^{-1000}$,we get $E_a / (RT) = 1000$.
Given $T = 500 \ K$ and the gas constant $R = 2 \ cal \ K^{-1} \ mol^{-1}$.
Substituting the values: $E_a = 1000 \times R \times T = 1000 \times 2 \times 500 = 1,000,000 \ cal/mol = 10^6 \ cal/mol$.

(B) Activation energy $(E_a)$ is defined as the difference between the energy of the transition state and the energy of the reactants.
For a reaction to have zero activation energy,the energy of the reactants must be equal to the energy of the transition state.
In the provided graphs,Graph $B$ shows a linear decrease in energy,which is not a standard representation of a reaction coordinate. However,in the context of such problems,if we consider the energy profile where the energy remains constant or decreases without a barrier,it represents a reaction with no activation barrier.
Given the standard options provided in such chemistry problems,Graph $B$ represents a case where the energy of the system decreases linearly,implying no energy barrier exists for the reaction to proceed.
Therefore,option $B$ is correct.

(C) Using the Arrhenius equation: $\log \frac{K_{2}}{K_{1}} = \frac{E_{a}}{2.303 R} \left( \frac{T_{2} - T_{1}}{T_{1} T_{2}} \right)$
Given: $E_{a} = 65000\, J/mol$,$T_{1} = 273\, K$,$T_{2} = 298\, K$,$R = 8.314\, J/mol\cdot K$.
Substituting the values:
$\log \frac{K_{2}}{K_{1}} = \frac{65000}{2.303 \times 8.314} \left( \frac{298 - 273}{273 \times 298} \right)$
$\log \frac{K_{2}}{K_{1}} = \frac{65000}{19.147} \times \left( \frac{25}{81354} \right)$
$\log \frac{K_{2}}{K_{1}} = 3394.78 \times 0.000307 = 1.043$
$\frac{K_{2}}{K_{1}} = 10^{1.043} \approx 11$
Thus,the reaction is $11$ times faster.