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

(B) The enthalpy of reaction $\Delta H$ 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 $\Delta H = -4.2 \ kJ \ mol^{-1}$ and $(E_a)_f = 9.6 \ kJ \ mol^{-1}$:
$-4.2 = 9.6 - (E_a)_b$
$(E_a)_b = 9.6 + 4.2 = 13.8 \ kJ \ mol^{-1}$
Since $\Delta H$ is negative,the reaction is exothermic,meaning the potential energy of the product $B$ must be lower than the potential energy of the reactant $A$. Among the given options,the graph where the product $B$ is at a lower energy level than reactant $A$ is represented by option $B$.

(A) The Arrhenius equation is given by $\ln k = \ln A - \frac{E_{a}}{RT}$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = \ln k$ and $x = \frac{1}{T}$,the slope $m = -\frac{E_{a}}{R}$.
Given that the slope is $-5 \times 10^{3} \, K$,we have $-\frac{E_{a}}{R} = -5 \times 10^{3} \, K$.
Therefore,$E_{a} = 5 \times 10^{3} \times 8.314 \, J \, mol^{-1} = 41570 \, J \, mol^{-1}$.
Converting to $kJ \, mol^{-1}$,we get $E_{a} = 41.57 \, kJ \, mol^{-1} \approx 41.5 \, kJ \, mol^{-1}$.

(B) According to the Arrhenius equation,$\ln \ k = \ln \ A - \frac{E_a}{RT}$.
We are plotting $\ln \ k$ versus $\frac{10^{3}}{T}$.
Rewriting the equation: $\ln \ k = \ln \ A - \left( \frac{E_a}{R \times 10^{3}} \right) \times \left( \frac{10^{3}}{T} \right)$.
Comparing this with the equation of a straight line $y = mx + c$,the slope $m = -\frac{E_a}{R \times 10^{3}}$.
Given the slope is $-18.5$,we have: $-18.5 = -\frac{E_a}{8.31 \times 10^{3}}$.
$E_a = 18.5 \times 8.31 \times 10^{3} \ J \ mol^{-1} = 153735 \ J \ mol^{-1}$.
Converting to $kJ \ mol^{-1}$: $E_a = 153.735 \ kJ \ mol^{-1}$.
Rounding to the nearest integer,we get $154 \ kJ \ mol^{-1}$.

(C) According to the Arrhenius equation,$k = A e^{-E_a / RT}$.
For the catalysed reaction,$k_{\text{cat}} = A e^{-E_{a, \text{cat}} / RT}$.
For the uncatalysed reaction,$k_{\text{uncat}} = A e^{-E_{a, \text{uncat}} / RT}$.
The ratio is $\frac{k_{\text{cat}}}{k_{\text{uncat}}} = e^{(E_{a, \text{uncat}} - E_{a, \text{cat}}) / RT}$.
Given $\Delta E_a = E_{a, \text{uncat}} - E_{a, \text{cat}} = 10 \ kJ \ mol^{-1} = 10000 \ J \ mol^{-1}$.
Substituting the values: $\frac{k_{\text{cat}}}{k_{\text{uncat}}} = e^{10000 / (8.31 \times 300)} = e^{10000 / 2493} = e^{4.011}$.
Comparing with $e^x$,we get $x \approx 4$.

(C) The Arrhenius equation is given by: $\log_{10} \frac{K_2}{K_1} = \frac{E_a}{2.303 \, R} \left(\frac{T_2 - T_1}{T_1 T_2}\right)$
Given: $T_1 = 300 \, K$,$T_2 = 309 \, K$,$\frac{K_2}{K_1} = 2$,$R = 8.3 \, J \, K^{-1} \, mol^{-1}$,$\log 2 = 0.30$.
Substituting the values:
$0.3 = \frac{E_a}{2.303 \times 8.3} \left(\frac{9}{300 \times 309}\right)$
$E_a = \frac{0.3 \times 2.303 \times 8.3 \times 300 \times 309}{9}$
$E_a = 59065.04 \, J \, mol^{-1}$
$E_a \approx 59 \, kJ \, mol^{-1}$

(B) The Arrhenius equation is given by $\ln k = \ln A - \frac{E_A}{RT}$.
Comparing this with the given equation $\ln k = 33.24 - \frac{2.0 \times 10^{4}}{T}$,we get:
$\frac{E_A}{R} = 2.0 \times 10^{4} \, K$.
Therefore,$E_A = 2.0 \times 10^{4} \times R$.
Substituting $R = 8.3 \, J \, K^{-1} \, mol^{-1}$:
$E_A = 2.0 \times 10^{4} \times 8.3 \, J \, mol^{-1} = 16.6 \times 10^{4} \, J \, mol^{-1}$.
Converting to $kJ \, mol^{-1}$:
$E_A = \frac{16.6 \times 10^{4}}{1000} \, kJ \, mol^{-1} = 166 \, kJ \, mol^{-1}$.

(A) The Arrhenius equation is given by $k = A e^{-E_a / RT}$.
Comparing this with the given equation $k = (6.5 \times 10^{12} \, s^{-1}) e^{-26000 \, K / T}$,we get $\frac{E_a}{R} = 26000 \, K$.
Substituting $R = 8.314 \, J \, K^{-1} \, mol^{-1}$,we have $E_a = 26000 \times 8.314 \, J \, mol^{-1}$.
$E_a = 216164 \, J \, mol^{-1} = 216.164 \, kJ \, mol^{-1}$.
Rounding to the nearest integer,the activation energy is $216 \, kJ \, mol^{-1}$.

(A) Using the Arrhenius equation: $\ln \left(\frac{k_{310}}{k_{300}}\right) = \frac{E_a}{R} \left(\frac{1}{T_{300}} - \frac{1}{T_{310}}\right)$
$\ln \left(\frac{k_{310}}{k_{300}}\right) = \frac{532611}{8.3} \times \left(\frac{310 - 300}{310 \times 300}\right)$
$\ln \left(\frac{k_{310}}{k_{300}}\right) = \frac{532611}{8.3} \times \frac{10}{93000} = 64170 \times \frac{10}{93000} \approx 6.9$
Since $\ln 10 = 2.3$,then $6.9 = 3 \times 2.3 = 3 \times \ln 10 = \ln 10^3$.
Therefore,$\frac{k_{310}}{k_{300}} = 10^3$,which means $k_{300} = 10^{-3} \, k_{310}$.
Comparing this with $k_{300} = x \times 10^{-3} \, k_{310}$,we get $x = 1$.

(A) The Arrhenius equation is given by $k = Ae^{-E_a/RT}$.
Taking the natural logarithm on both sides,we get $\ln k = -\frac{E_a}{R}(\frac{1}{T}) + \ln A$.
Comparing this with the equation of a straight line $y = mx + c$,the slope $m = -\frac{E_a}{R}$.
From the given graph,the slope is calculated as $\frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - 20}{5 - 0} = -4$.
Equating the slopes: $-\frac{E_a}{R} = -4$,which gives $E_a = 4 \times R$.
Given $R = 2 \ cal \ K^{-1} \ mol^{-1}$,we have $E_a = 4 \times 2 = 8 \ cal \ mol^{-1}$.

(C) . The Arrhenius equation for the rate constant $k$ is $k = A e^{-E_a / RT}$,where $A$ is the pre-exponential factor.
Since the term $e^{-E_a / RT}$ is dimensionless,the dimension of $A$ is equal to the dimension of $k$.
For a first-order reaction,the dimension of $k$ is $\text{time}^{-1}$.
Therefore,the dimension of $A$ is also $\text{time}^{-1}$,which is the reciprocal of time.
Regarding other options:
For a first-order reaction,the rate of reaction is directly proportional to the reactant concentration.
The half-life $t_{1/2} = \frac{0.693}{k}$,which is $69.3 \%$ of $\frac{1}{k}$.
The concentration of the reactant changes with time according to the equation $[A]_t = [A]_0 e^{-kt}$,which represents an exponential decay,not a linear one.

(A) For an exothermic reaction,$A \rightleftharpoons B$,the change in enthalpy is negative,i.e.,$\Delta H < 0$.
Since $\Delta H = E_{Product} - E_{Reactant}$,we have $E_{B} - E_{A} < 0$,which implies $E_{B} < E_{A}$.
This indicates that the energy level of the product $B$ must be lower than that of the reactant $A$.
Furthermore,a catalyst provides an alternative pathway with a lower activation energy but does not change the initial energy of the reactants or the final energy of the products.
Therefore,the solid line (uncatalyzed) and dashed line (catalyzed) must start and end at the same energy levels.
Plot $A$ correctly shows $E_{B} < E_{A}$ and the same starting and ending points for both catalyzed and uncatalyzed paths.

(A) Given,activation energy $E_a = 209 \, kJ \, mol^{-1} = 209000 \, J \, mol^{-1}$.
Rate increases $10$-fold,so $K_2 / K_1 = 10$.
Initial temperature $T_1 = 27 + 273 = 300 \, K$.
Using the Arrhenius equation: $\log(K_2 / K_1) = \frac{E_a}{2.303 \, R} \left[ \frac{1}{T_1} - \frac{1}{T_2} \right]$.
Substituting the values: $\log(10) = \frac{209000}{2.303 \times 8.314} \left[ \frac{1}{300} - \frac{1}{T_2} \right]$.
$1 = 10923.6 \times \left[ 0.003333 - \frac{1}{T_2} \right]$.
$0.0000915 = 0.003333 - \frac{1}{T_2}$.
$\frac{1}{T_2} = 0.003333 - 0.0000915 = 0.0032415$.
$T_2 = 308.5 \, K$.
$X = 308.5 - 273 = 35.5^{\circ} C \approx 35^{\circ} C$.

(C) The temperature dependence of the rate of a chemical reaction is expressed by the Arrhenius equation: $k = A e^{-E_{a} / (RT)}$.
When $T$ is very high,the factor $E_{a} / (RT)$ approaches $0$.
Therefore,$k = A e^{0} = A \times 1 = A$.
Thus,the rate constant of a chemical reaction at a very high temperature will approach the Arrhenius frequency factor $(A)$.

(A) The Arrhenius equation is given by $k = A e^{-E_a / R T}$.
Taking the natural logarithm on both sides,we get $\ln k = \ln A - \frac{E_a}{R} \left( \frac{1}{T} \right)$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = \ln k$ and $x = 1/T$:
The slope $m = -E_a / R$ and the intercept $c = \ln A$.
From the graph,the line for reaction $I$ is steeper than the line for reaction $II$. Since the slope is negative,a steeper line indicates a larger magnitude of slope,meaning $E_I > E_{II}$.
Also,the intercept on the $y$-axis (at $1/T = 0$) for line $I$ is higher than for line $II$. Since the intercept is $\ln A$,this implies $\ln A_I > \ln A_{II}$,which means $A_I > A_{II}$.
Therefore,the correct option is $A$.

(D) According to the Arrhenius equation,$\ln k = \ln A - \frac{E_a}{RT}$.
Comparing this equation with the equation of a straight line,$y = mx + c$,where $y = \ln k$,$x = \frac{1}{T}$,$m = -\frac{E_a}{R}$ (slope),and $c = \ln A$ (intercept).
This indicates that a plot of $\ln k$ versus $\frac{1}{T}$ should be a straight line with a negative slope equal to $-\frac{E_a}{R}$.
Therefore,the graph that exhibits Arrhenius behaviour over the entire temperature range is the one showing a straight line with a negative slope,which corresponds to option $(d)$.

(B) .
According to the Arrhenius equation,$K = A e^{-E_a / RT}$,the rate constant $K$ depends on temperature.
As temperature increases,the fraction of molecules having kinetic energy greater than the activation energy $(E_a)$ increases significantly.
This factor,represented by $e^{-E_a / RT}$,is the primary reason for the rapid increase in the reaction rate.

(B)
Increasing the temperature increases the kinetic energy of the reactant molecules.
This leads to an increase in the number of collisions,the average energy of collisions,and the average velocity of the reactant molecules.
However,the activation energy $(E_a)$ is a characteristic property of the reaction pathway and is independent of temperature.
Therefore,increasing the temperature does not increase the activation energy.

(C) Given,$\ln k = -\frac{11067}{T} + 31.33$.
Comparing with the Arrhenius equation,$\ln k = \ln A - \frac{E_a}{RT}$,we get $\frac{E_a}{R} = 11067 \, K$.
For the rate constant to double,we use the relation $\log \left( \frac{k_2}{k_1} \right) = \frac{E_a}{2.303 R} \left[ \frac{1}{T_1} - \frac{1}{T_2} \right]$.
Here,$T_1 = 25^{\circ} C = 298 \, K$ and $\frac{k_2}{k_1} = 2$.
Substituting the values: $\log 2 = \frac{11067}{2.303} \left[ \frac{1}{298} - \frac{1}{T_2} \right]$.
$0.3010 = 4805.47 \left[ 0.0033557 - \frac{1}{T_2} \right]$.
$6.2637 \times 10^{-5} = 0.0033557 - \frac{1}{T_2}$.
$\frac{1}{T_2} = 0.003293$.
$T_2 \approx 303.66 \, K \approx 31^{\circ} C$.

(D)
Collision theory does not take into account the structural aspects of molecules during collision.
It assumes that reactant molecules are hard spheres and only considers their kinetic energy and orientation,not their internal structure.

(D) $A.$ Correct. According to the Arrhenius equation,$k = Ae^{-\frac{E_a}{RT}}$. As $E_a$ increases,the exponential term $e^{-\frac{E_a}{RT}}$ decreases,so $k$ decreases.
$B.$ Correct. The temperature coefficient is defined as $\frac{k_{T+10}}{k_T} = e^{\frac{10E_a}{RT(T+10)}}$. As $E_a$ increases,the value of the temperature coefficient increases.
$C.$ Correct. The rate constant $k$ increases exponentially with temperature. The derivative $\frac{dk}{dT} = k \cdot \frac{E_a}{RT^2}$ shows that the rate of change of $k$ with respect to $T$ is higher at lower temperatures for a given $E_a$.
$D.$ Correct. From $\ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T}$,the plot of $\ln k$ vs $\frac{1}{T}$ is a straight line with slope $-\frac{E_a}{R}$.
All four statements are correct. Therefore,the number of correct statements is $4$.

(D) 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: $K_1 = 0.03 \, min^{-1}$ at $T_1 = 200 \, K$,$K_2 = 0.05 \, min^{-1}$ at $T_2 = 300 \, K$,$R = 8.3 \, J \, K^{-1} \, mol^{-1}$,$\ln 10 = 2.3$
$\log \frac{0.05}{0.03} = \frac{E_a}{2.3 \times 8.3} \left( \frac{300 - 200}{200 \times 300} \right)$
$\log \left( \frac{5}{3} \right) = \frac{E_a}{19.09} \times \frac{100}{60000}$
$\log 5 - \log 3 = \frac{E_a}{19.09} \times \frac{1}{600}$
$0.70 - 0.48 = \frac{E_a}{11454}$
$0.22 = \frac{E_a}{11454}$
$E_a = 0.22 \times 11454 = 2519.88 \, J \approx 2520 \, J$

(B) $1$. Intermediates are the local minima in the energy profile diagram between the reactant and product. Here,$P$ and $Q$ are the intermediates. So,the number of intermediates is $2$.
$2$. Activated complexes correspond to the peaks (maxima) in the energy profile diagram. There are $3$ peaks,so the number of activated complexes is $3$.
$3$. The rate-determining step $(RDS)$ is the step with the highest activation energy $(E_a)$. Comparing the peaks,step $II$ has the highest activation energy. Therefore,step $II$ is the $RDS$.
Thus,the correct values are: Number of intermediates $= 2$,Number of activated complexes $= 3$,$RDS = II$.

(B) According to the Arrhenius equation,$k = A \cdot e^{-E_a / RT}$.
Taking the natural logarithm on both sides: $\ln k = \ln A - \frac{E_a}{RT}$.
Taking the base-$10$ logarithm: $\log k = \log A - \frac{E_a}{2.303 R} \cdot \frac{1}{T}$.
Evaluating the statements:
$(A)$ Correct: $A$ higher $E_a$ leads to a steeper slope $(-\frac{E_a}{2.303 R})$ in the $\log k$ vs $\frac{1}{T}$ plot,meaning the rate constant $k$ changes more rapidly with temperature.
$(B)$ Correct: If $E_a = 0$,then $k = A \cdot e^0 = A$,which is constant and independent of temperature.
$(C)$ Incorrect: This is the opposite of statement $(A)$.
$(D)$ Incorrect: If there is no correlation between temperature and rate constant,it implies $E_a = 0$,not a negative activation energy.
Thus,there are $2$ correct statements ($A$ and $B$).

(B) positive catalyst increases the rate of reaction by providing an alternative pathway with lower activation energy $(E_a)$.
$(i)$ The energy of reactants and products remains the same,so $\Delta H$ does not change.
$(ii)$ The activation energy $(E_a)$ decreases,which is represented by a lower peak in the energy profile diagram.
Therefore,the diagram where the path with the catalyst has a lower peak than the path without the catalyst is the correct one,which corresponds to option $B$.

(A) Given: $\Delta H = 20 \ kJ \ mol^{-1}$,$E_{a,f} = 300 \ kJ \ mol^{-1}$.
For the same rate,the activation energy $E_{a}$ is proportional to temperature $T$ (assuming the pre-exponential factor $A$ remains constant).
$\frac{E_{a,f}}{T_{uncat}} = \frac{E_{a,f}^{\prime}}{T_{cat}}$
$T_{uncat} = 327 + 273 = 600 \ K$
$T_{cat} = 27 + 273 = 300 \ K$
$\frac{300}{600} = \frac{E_{a,f}^{\prime}}{300} \implies E_{a,f}^{\prime} = 150 \ kJ \ mol^{-1}$.
We know $\Delta H = E_{a,f}^{\prime} - E_{a,b}^{\prime}$.
$20 = 150 - E_{a,b}^{\prime}$.
$E_{a,b}^{\prime} = 150 - 20 = 130 \ kJ \ mol^{-1}$.

(A) Assertion $A$ is false because a chemical reaction cannot have zero activation energy $(E_{a} > 0)$.
Reason $R$ is true because it correctly defines activation energy as the minimum extra energy required by reactant molecules to reach the threshold energy level.

(C) The overall rate constant is given by $K = \frac{K_1 K_2}{K_3}$.
Substituting the Arrhenius equation $K = A \cdot e^{-E_a / RT}$ for each step:
$A \cdot e^{-E_a / RT} = \frac{(A_1 e^{-E_{a1} / RT}) \cdot (A_2 e^{-E_{a2} / RT})}{(A_3 e^{-E_{a3} / RT})}$
$A \cdot e^{-E_a / RT} = \frac{A_1 A_2}{A_3} \cdot e^{-(E_{a1} + E_{a2} - E_{a3}) / RT}$
Comparing the exponents,we get $E_a = E_{a1} + E_{a2} - E_{a3}$.
Substituting the given values: $E_a = 40 + 50 - 60 = 30 \ kJ/mol$.

(C) The overall rate constant is given by $k = \frac{k_1 k_2}{k_3}$.
Using the Arrhenius equation $k = A e^{\frac{-E_a}{RT}}$,we can write:
$A e^{\frac{-E_a}{RT}} = \frac{A_1 e^{\frac{-E_{a_1}}{RT}} \cdot A_2 e^{\frac{-E_{a_2}}{RT}}}{A_3 e^{\frac{-E_{a_3}}{RT}}}$
$A e^{\frac{-E_a}{RT}} = \frac{A_1 A_2}{A_3} e^{\frac{-(E_{a_1} + E_{a_2} - E_{a_3})}{RT}}$
Comparing the exponents,we get the overall activation energy expression:
$E_a = E_{a_1} + E_{a_2} - E_{a_3}$
Given $E_a = 400 \ kJ \ mol^{-1}$,$E_{a_1} = 300 \ kJ \ mol^{-1}$,and $E_{a_2} = 200 \ kJ \ mol^{-1}$:
$400 = 300 + 200 - E_{a_3}$
$400 = 500 - E_{a_3}$
$E_{a_3} = 500 - 400 = 100 \ kJ \ mol^{-1}$.

(C) The Arrhenius equation is given by:
$k = A e^{-\frac{E_a}{R T}}$
Taking the natural logarithm on both sides:
$\ln k = \ln A - \frac{E_a}{R T}$
This equation is in the form of a straight line $y = mx + c$,where:
$y = \ln k$
$x = \frac{1}{T}$
$m = -\frac{E_a}{R}$ (slope,which is negative)
$c = \ln A$ (y-intercept,which is positive)
Therefore,the plot of $\ln k$ vs $\frac{1}{T}$ is a straight line with a negative slope and a positive intercept,which corresponds to option $C$.

(C) The activation energy $(E_a)$ can be calculated using the Arrhenius equation.
The equation is given by:
$\log \left( \frac{k_2}{k_1} \right) = \frac{E_a}{2.303 R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)$
From this equation,it is clear that $E_a$ can be determined if the rate constants ($k_1$ and $k_2$) are known at two different temperatures ($T_1$ and $T_2$).

(D) The Arrhenius equation is given by: $\log \left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303 R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)$
Given: $k_2 = 4k_1$,$T_1 = 27 + 273 = 300 \ K$,$T_2 = 57 + 273 = 330 \ K$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$,$\log 4 = 0.6021$
Substituting the values:
$\log 4 = \frac{E_a}{2.303 \times 8.314} \left(\frac{1}{300} - \frac{1}{330}\right)$
$0.6021 = \frac{E_a}{19.147} \left(\frac{330 - 300}{300 \times 330}\right)$
$0.6021 = \frac{E_a}{19.147} \left(\frac{30}{99000}\right)$
$E_a = \frac{0.6021 \times 19.147 \times 99000}{30} \approx 38040 \ J \ mol^{-1}$
$E_a = 38.04 \ kJ \ mol^{-1}$

(D) The Arrhenius equation is given by $k = Ae^{-E_a/RT}$.
Taking the logarithm on both sides,we get $\log k = \log A - \frac{E_a}{2.303RT}$.
Comparing this with the given equation $\log k = -(2000) \frac{1}{T} + 6.0$:
For the pre-exponential factor $A$,$\log A = 6.0$,which gives $A = 10^6 \ s^{-1}$.
For the activation energy $E_a$,$\frac{E_a}{2.303R} = 2000$.
$E_a = 2000 \times 2.303 \times 8.314 \ J \ mol^{-1} \approx 38314 \ J \ mol^{-1} = 38.3 \ kJ \ mol^{-1}$.
Thus,the correct option is $D$.

(B) The Arrhenius equation is given by $K = Ae^{-\frac{E_a}{RT}}$.
Statement $(A)$ is incorrect because a high activation energy $(E_a)$ implies a slow reaction,not a fast one.
Statement $(B)$ is correct because as temperature $(T)$ increases,the fraction of molecules with energy greater than $E_a$ increases,leading to more effective collisions.
Statement $(C)$ is correct because the term $e^{-\frac{E_a}{RT}}$ shows that the sensitivity of $K$ to temperature changes increases as $E_a$ increases.
Statement $(D)$ is correct because the pre-exponential factor $(A)$ represents the frequency of collisions,which is independent of the energy threshold.
Therefore,the correct statements are $(B)$,$(C)$,and $(D)$.

(B) For the reaction $A_{(g)} \rightleftharpoons P_{(g)}$,the Arrhenius equation for the forward reaction is $\log k_f = \frac{-E_f}{2.303 RT} + \log A_f$.
From the given plot,at $\frac{1}{T} = 0.002 \ K^{-1}$,$\log k_f = 9$.
Substituting the values: $9 = \frac{-E_f}{2.303 R}(0.002) + \log(10^{15}) \Rightarrow 9 = \frac{-E_f}{2.303 R}(0.002) + 15$.
$\frac{E_f}{2.303 R} = \frac{6}{0.002} = 3000$.
For the equilibrium constant $K = \frac{k_f}{k_b}$,we have $\log K = \log \left(\frac{A_f}{A_b}\right) - \frac{E_f - E_b}{2.303 RT}$.
At $500 \ K$,$\log K = 6$ and $\frac{A_f}{A_b} = \frac{10^{15}}{10^{11}} = 10^4$,so $\log(10^4) = 4$.
$6 = 4 - \frac{E_f - E_b}{2.303 R(500)}$ $\Rightarrow 2 = \frac{E_b - E_f}{2.303 R(500)}$ $\Rightarrow E_b - E_f = 1000(2.303 R)$.
Since $\frac{E_f}{2.303 R} = 3000$,$E_f = 3000(2.303 R)$.
Thus,$E_b = 1000(2.303 R) + 3000(2.303 R) = 4000(2.303 R)$.
For the backward reaction at $250 \ K$,$\log k_b = \log A_b - \frac{E_b}{2.303 RT} = \log(10^{11}) - \frac{4000(2.303 R)}{2.303 R(250)} = 11 - 16 = -5$.
Therefore,$|\log k_b| = |-5| = 5$.

(A) According to the collision theory,the rate constant $k$ is given by $k = P \cdot Z \cdot e^{-E_a/RT}$,where $P$ is the steric factor and $Z$ is the collision frequency.
The experimental frequency factor $A_{exp}$ is equal to $P \cdot Z$.
The Arrhenius equation predicts the frequency factor as $A_{calc} = Z$.
Since $P = 4.5$ (which is $> 1$),we have $A_{exp} = 4.5 \cdot Z$,which means $A_{exp} > A_{calc}$. Thus,option $[B]$ is correct.
The activation energy $E_a$ is an energy barrier and is independent of the steric factor $P$,which accounts for the orientation of molecules. Thus,option $[A]$ is correct.
Option $[C]$ is incorrect because a reaction with $P > 1$ can proceed without a catalyst.
Option $[D]$ is incorrect because the experimental value is higher,not lower.

(A) The Arrhenius equation is given by $k = Ae^{-E_a / RT}$.
As the 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 behavior is represented by the exponential curve shown in option $A$.

(A) For the reaction $A_{(g)} + B_{(g)} \rightleftharpoons AB_{(g)}$,the equilibrium constant $K$ is given by $K = \frac{k_f}{k_b} = \frac{A_f e^{-E_a/RT}}{A_b e^{-E_b/RT}}$.
Given $E_b - E_a = 2RT$ and $A_f = 4A_b$,we substitute these into the expression for $K$:
$K = \frac{A_f}{A_b} e^{(E_b - E_a)/RT} = 4 e^{2RT/RT} = 4e^2$.
The standard Gibbs free energy change is $\Delta G^{\ominus} = -RT \ln K$.
Substituting the values: $\Delta G^{\ominus} = -RT \ln(4e^2) = -RT(\ln 4 + \ln e^2) = -RT(2 \ln 2 + 2)$.
Given $RT = 2500 \ J \ mol^{-1}$ and $\ln 2 = 0.7$:
$\Delta G^{\ominus} = -2500 \times (2 \times 0.7 + 2) = -2500 \times (1.4 + 2) = -2500 \times 3.4 = -8500 \ J \ mol^{-1}$.
The absolute value of $\Delta G^{\ominus}$ is $8500 \ J \ mol^{-1}$.

(D) For a first-order reaction,the rate constant $k$ is given by the Arrhenius equation: $k = A e^{-E_a/RT}$.
Substituting the given values: $k = 10^{20} \times e^{-\frac{191.48 \times 10^3}{8.314 \times 1000}}$.
$k = 10^{20} \times e^{-23.031} \approx 10^{20} \times e^{-\ln(10^{10})} = 10^{20} \times 10^{-10} = 10^{10} \ s^{-1}$.
The half-life $t_{1/2}$ is given by: $t_{1/2} = \frac{0.693}{k}$.
$t_{1/2} = \frac{0.693}{10^{10}} = 6.93 \times 10^{-11} \ s$.
Converting to picoseconds $(1 \ ps = 10^{-12} \ s)$: $t_{1/2} = 69.3 \ ps$.
The nearest integer is $69$.

(C) From the given energy profile diagram:
$1$. The activation energy of the forward reaction $(E_{a,f})$ is the energy difference between the activated complex and the reactant,which is $E_1 + E_2$.
$2$. The energy of the product is higher than the energy of the reactant $(E_{product} > E_{reactant})$.
$3$. Since stability is inversely proportional to energy,the reactant is more stable than the product,or the product is less stable than the reactant.

(C) The overall rate constant is given by $k = (k_1 k_3 / k_2)^{1/2}$.
Using the Arrhenius equation $k = A \cdot e^{-E_a / RT}$,we substitute the expressions for each rate constant:
$A \cdot e^{-E_a / RT} = \left( \frac{A_1 e^{-E_{a_1} / RT} \cdot A_3 e^{-E_{a_3} / RT}}{A_2 e^{-E_{a_2} / RT}} \right)^{1/2}$.
Comparing the exponential terms:
$-E_a / RT = \frac{1}{2} (-E_{a_1} / RT - E_{a_3} / RT + E_{a_2} / RT)$.
Multiplying by $-RT$,we get $E_a = \frac{1}{2} (E_{a_1} + E_{a_3} - E_{a_2})$.
Substituting the given values $E_{a_1} = 60 \ kJ \ mol^{-1}$,$E_{a_2} = 30 \ kJ \ mol^{-1}$,and $E_{a_3} = 10 \ kJ \ mol^{-1}$:
$E_a = \frac{1}{2} (60 + 10 - 30) = \frac{40}{2} = 20 \ kJ \ mol^{-1}$.

(A) The reaction mechanism involves three steps:
$1$. $A \rightarrow B$ is a slow step with $\Delta H = +ve$ (endothermic),meaning the energy of $B$ is higher than $A$,and it has a high activation energy $(E_{a_1})$.
$2$. $B \rightarrow C$ is a fast step with $\Delta H = -ve$ (exothermic),meaning the energy of $C$ is lower than $B$,and it has a low activation energy $(E_{a_2})$.
$3$. $C \rightarrow D$ is a fast step with $\Delta H = -ve$ (exothermic),meaning the energy of $D$ is lower than $C$,and it has a low activation energy $(E_{a_3})$.
The overall reaction is exothermic $(\Delta H_{net} = -ve)$,so the final product $D$ must be at a lower energy level than the reactant $A$.
The graph that correctly depicts these energy changes is shown in the solution image.

(C) The Arrhenius equation is $k = A e^{-E_a/RT}$.
Statement $A$ is incorrect because the Arrhenius equation is applicable to both elementary and complex reactions.
Statement $B$ is correct because $A$ (the frequency factor) has the same units as the rate constant $k$.
Statement $C$ is correct because a lower activation energy $(E_a)$ results in a larger value of $e^{-E_a/RT}$,leading to a faster reaction rate.
Statement $D$ is incorrect because $A$ and $E_a$ are generally considered temperature-independent constants for a given reaction.
Statement $E$ is incorrect as $A$ and $E_a$ are independent parameters.
Therefore,only statements $B$ and $C$ are correct.

(A) For the reaction $A_2 + B_2 \rightleftharpoons 2 AB$,the enthalpy change $\Delta H$ is given by the difference between the activation energy of the forward reaction $(E_f)$ and the backward reaction $(E_b)$:
$\Delta H = E_f - E_b = 180 \ kJ \ mol^{-1} - 200 \ kJ \ mol^{-1} = -20 \ kJ \ mol^{-1}$.
When a catalyst is added,it lowers the activation energy for both the forward and backward reactions by the same amount $(100 \ kJ \ mol^{-1})$:
$E_{f(cat)} = 180 - 100 = 80 \ kJ \ mol^{-1}$
$E_{b(cat)} = 200 - 100 = 100 \ kJ \ mol^{-1}$
The new enthalpy change is $\Delta H_{cat} = 80 - 100 = -20 \ kJ \ mol^{-1}$.
Since $\Delta H$ and $\Delta G$ are state functions,they depend only on the initial and final states of the reactants and products,not on the path taken. Therefore,a catalyst does not change the $\Delta H$ or $\Delta G$ of a reaction.

(A) According to the Arrhenius equation: $k = A e^{-Ea / RT}$
Taking log on both sides: $\log k = \log A - \frac{Ea}{2.303 RT}$
For a plot of $\log k$ versus $\frac{1}{T}$,the slope of the line is given by: $\text{Slope} = -\frac{Ea}{2.303 R}$
Since the slope is negative,the magnitude of the slope is directly proportional to the activation energy $(Ea)$: $|\text{Slope}| = \frac{Ea}{2.303 R}$
From the given graph,the steepness (magnitude of slope) of the lines follows the order: $(2) > (1) > (3)$
Therefore,the correct order of activation energies is: $Ea_2 > Ea_1 > Ea_3$

(A) According to the Arrhenius equation,$k = A \cdot e^{-E_a / RT}$.
As the temperature $T$ increases,the term $e^{-E_a / RT}$ increases because the exponent becomes less negative.
Therefore,the rate constant $k$ increases exponentially with an increase in temperature.

(A) The activation energy $(E_{a})$ for a reaction step is defined as the difference between the energy of the transition state (peak) and the energy of the reactants for that specific step.
For the reaction step $Q \longrightarrow C$,the reactant is $Q$ and the transition state is the peak immediately following $Q$.
From the given potential energy diagram:
Energy of $Q = 20 \ kcal \ mol^{-1}$.
Energy of the transition state for $Q \longrightarrow C = 23 \ kcal \ mol^{-1}$.
Therefore,$E_{a} = E_{\text{transition state}} - E_{Q} = 23 \ kcal \ mol^{-1} - 20 \ kcal \ mol^{-1} = 3 \ kcal \ mol^{-1}$.

(C) Given: $E_{a} = 87 \ kJ \ mol^{-1} = 87000 \ J \ mol^{-1}$,$T_{1} = 37 + 273 = 310 \ K$,$T_{2} = 15 + 273 = 288 \ K$.
Using the Arrhenius equation: $\log \left( \frac{k_{1}}{k_{2}} \right) = \frac{E_{a}}{2.303 \ R} \left[ \frac{T_{1} - T_{2}}{T_{1} T_{2}} \right]$.
Substituting the values: $\log \left( \frac{k_{1}}{k_{2}} \right) = \frac{87000}{2.303 \times 8.314} \left[ \frac{310 - 288}{310 \times 288} \right]$.
$\log \left( \frac{k_{1}}{k_{2}} \right) = \frac{87000}{19.147} \times \left[ \frac{22}{89280} \right] \approx 4543.8 \times 0.002464 \approx 1.119$.
$\frac{k_{1}}{k_{2}} = 10^{1.119} \approx 13.15$.
Thus,the ratio is approximately $13 / 1$.

(D) The Maxwell-Boltzmann distribution curve shows the fraction of molecules versus kinetic energy.
When the temperature is increased,the peak of the curve shifts to the right and the height of the peak decreases,while the curve flattens out.
Area $II$ represents the fraction of molecules with kinetic energy greater than the activation energy $(E_a)$.
As the temperature increases,a larger fraction of molecules possess kinetic energy equal to or greater than $E_a$,so area $II$ increases.
Since the total area under the curve must remain constant (equal to $1$),and area $II$ increases,the remaining area (area $I$) must decrease to compensate for the increase in area $II$.

(B) The enthalpy change of a reaction $(\Delta H)$ is defined as the difference between the activation energy of the forward reaction $(Ea_{(f)})$ and the activation energy of the reverse reaction $(Ea_{(b)})$.
$\Delta H = Ea_{(f)} - Ea_{(b)}$
Given:
$Ea_{(f)} = 150 \ kJ \ mol^{-1}$
$Ea_{(b)} = 260 \ kJ \ mol^{-1}$
Substituting the values:
$\Delta H = 150 \ kJ \ mol^{-1} - 260 \ kJ \ mol^{-1}$
$\Delta H = -110 \ kJ \ mol^{-1}$

(B) The temperature coefficient is $2$ for a $10^{\circ} C$ rise in temperature.
The number of $10^{\circ} C$ intervals is calculated as $n = \frac{100 - 10}{10} = \frac{90}{10} = 9$.
The rate of reaction increases by a factor of $2^n$.
Therefore, the rate becomes $2^9 = 512$ times the initial rate.