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

(C) For a chemical reaction to occur,the reactant molecules must collide with each other.
However,not all collisions lead to the formation of products.
Only those collisions where the colliding molecules possess a minimum amount of energy,known as the $Threshold \ energy$,result in a chemical reaction.
Therefore,the minimum energy required for an effective collision is the $Threshold \ energy$.

(D) According to the Arrhenius equation,the rate constant $K$ is given by $K = A \cdot e^{-E_a/RT}$.
Taking the ratio of rate constants at two different temperatures $T_1$ and $T_2$:
$\ln(K_2/K_1) = \frac{E_a}{R} \cdot (\frac{1}{T_1} - \frac{1}{T_2})$.
For a fixed temperature interval $(T_1 = 300 \ K, T_2 = 310 \ K)$,the ratio $K_{310}/K_{300}$ is directly proportional to the activation energy $E_a$.
Therefore,the reaction with the highest activation energy will have the maximum value for the ratio $K_{310}/K_{300}$.
Since the order of activation energy is $E_M < E_N < E_O < E_P$,the reaction $P$ has the highest activation energy.
Thus,the ratio $K_{310}/K_{300}$ will be maximum for reaction $P$.

(C) Given: $T_1 = 25\,^oC = 298\,K$,$T_2 = 35\,^oC = 308\,K$.
Rate constant $k_1 = 1 \times 10^{-3}\,s^{-1}$.
Since the rate doubles,$k_2 = 2 \times k_1 = 2 \times 10^{-3}\,s^{-1}$.
Using the Arrhenius equation: $\log(\frac{k_2}{k_1}) = \frac{E_a}{2.303R} \times (\frac{T_2 - T_1}{T_1 T_2})$.
$\log(2) = \frac{E_a}{2.303 \times 8.314} \times (\frac{308 - 298}{308 \times 298})$.
$0.3010 = \frac{E_a}{19.147} \times (\frac{10}{91784})$.
$E_a = \frac{0.3010 \times 19.147 \times 91784}{10} \approx 52897\,J\,mol^{-1} \approx 53\,kJ\,mol^{-1}$.

(A) According to the Arrhenius equation: $\ln(k_2/k_1) = (E_a/R) \times [(T_2 - T_1) / (T_1 \times T_2)]$.
Given: $T_1 = T$,$T_2 = 2T$,and $k_1 = k_2/10$,which implies $k_2/k_1 = 10$.
Substituting the values: $\ln(10) = (E_a/R) \times [(2T - T) / (T \times 2T)]$.
$2.303 = (E_a/R) \times [T / (2T^2)]$.
$2.303 = (E_a/R) \times [1 / (2T)]$.
$E_a = 2.303 \times 2 \times R \times T = 4.606 \ RT$.
Rounding to the nearest value,$E_a \approx 4.6 \ RT$.

(B) The Arrhenius equation is given by $\log K = \log A - \frac{E_a}{2.303 RT}$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = \log K$ and $x = 1/T$,the slope $m = -\frac{E_a}{2.303 R}$.
Given slope $m = -5632$.
So,$-5632 = -\frac{E_a}{2.303 \times 8.314 \ J \ K^{-1} \ mol^{-1}}$.
$E_a = 5632 \times 2.303 \times 8.314 \ J \ mol^{-1}$.
$E_a = 107840 \ J \ mol^{-1} = 107.84 \ kJ \ mol^{-1}$.

(D) According to the Arrhenius equation,$\log K = \log A - \frac{E_a}{2.303RT}$.
Comparing this with the equation of a straight line $y = mx + c$,the slope $m = -\frac{E_a}{2.303R}$.
Given,slope $m = -1.2 \times 10^4 \ K$.
Therefore,$-1.2 \times 10^4 = -\frac{E_a}{2.303 \times 8.314 \ J \ K^{-1} \ mol^{-1}}$.
$E_a = 1.2 \times 10^4 \times 2.303 \times 8.314 \ J \ mol^{-1}$.
$E_a \approx 229860 \ J \ mol^{-1} \approx 2.3 \times 10^5 \ J \ mol^{-1}$.
Since the closest option provided is $2.5 \times 10^5 \ J \ mol^{-1}$ (assuming standard approximation or slight variation in constants),option $D$ is the intended answer.

(C) The reaction between $H_2$ and $O_2$ to form $H_2O$ has a very high activation energy $(E_a)$.
At room temperature,the number of molecules possessing energy greater than $E_a$ is negligible,making the mixture stable.
When a spark is provided,it supplies the necessary activation energy to a small number of molecules,initiating the reaction.
Since the reaction is highly exothermic,the heat released provides energy to more molecules,leading to a chain reaction and an explosion.
Thus,the spark increases the number of energetic molecules significantly.

(B) $1$. The reaction $2N_2O_5 \to 4NO_2 + O_2$ is a first-order reaction.
$2$. The half-life formula is $t_{1/2} = \frac{0.693}{k}$. Given $t_{1/2} = 2 \ hr = 7200 \ s$,we find $k_2 = \frac{0.693}{7200} \approx 9.625 \times 10^{-5} \ s^{-1}$.
$3$. Using the Arrhenius equation: $\ln(\frac{k_2}{k_1}) = \frac{E_a}{R} (\frac{1}{T_1} - \frac{1}{T_2})$.
$4$. Given $E_a = 10000 \ J \ mol^{-1}$,$T_1 = 300 \ K$,$k_1 = 2.4 \times 10^{-5} \ s^{-1}$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$.
$5$. $\ln(\frac{9.625 \times 10^{-5}}{2.4 \times 10^{-5}}) = \frac{10000}{8.314} (\frac{1}{300} - \frac{1}{T_2})$.
$6$. $\ln(4.01) = 1202.79 (0.00333 - \frac{1}{T_2}) \implies 1.389 = 4.005 - \frac{1202.79}{T_2}$.
$7$. $\frac{1202.79}{T_2} = 2.616 \implies T_2 \approx 459.7 \ K$. The closest option is $458 \ K$.

(D) According to the Arrhenius equation,$k = A e^{-E_a / RT}$.
For two reactions with the same pre-exponential factor $A$,the ratio of rate constants is $\frac{k_1}{k_2} = \frac{A e^{-E_{a1} / RT}}{A e^{-E_{a2} / RT}} = e^{(E_{a2} - E_{a1}) / RT}$.
Given $\Delta E_a = E_{a2} - E_{a1} = 24.9 \, kJ \, mol^{-1} = 24900 \, J \, mol^{-1}$.
$T = 300 \, K$ and $R = 8.314 \, J \, K^{-1} \, mol^{-1}$.
$\frac{k_1}{k_2} = e^{24900 / (8.314 \times 300)} = e^{24900 / 2494.2} \approx e^{9.983} \approx 21655 \approx 2.16 \times 10^4$.

(A) The temperature coefficient is defined as the ratio of rate constants at temperatures differing by $10 \ K$,i.e.,$\frac{k_{T+10}}{k_T} = 1.75$.
Using the Arrhenius equation in the form: $\ln(\frac{k_2}{k_1}) = \frac{E_a}{R} (\frac{T_2 - T_1}{T_1 T_2})$.
Assuming room temperature $T_1 = 298 \ K$ and $T_2 = 308 \ K$,we have:
$\ln(1.75) = \frac{E_a}{1.987 \times 10^{-3} \ kcal \ mol^{-1} K^{-1}} (\frac{10}{298 \times 308})$.
$0.5596 = \frac{E_a}{1.987 \times 10^{-3}} (1.0905 \times 10^{-4})$.
$E_a = \frac{0.5596 \times 1.987 \times 10^{-3}}{1.0905 \times 10^{-4}} \approx 10.21 \ kcal \ mol^{-1}$.

(C) The Arrhenius equation is given by $K = A \cdot e^{-E_a / RT}$.
Taking $log_{10}$ on both sides: $log_{10} K = log_{10} A - \frac{E_a}{2.303 \cdot R \cdot T}$.
Comparing this with the given equation $log_{10} K = 5 - \frac{2000}{T}$:
$log_{10} A = 5 \implies A = 10^5$.
$\frac{E_a}{2.303 \cdot R} = 2000 \implies E_a = 2000 \cdot 2.303 \cdot 8.314 \ J/mol$.
$E_a \approx 38294 \ J/mol = 38.294 \ kJ/mol$.
Converting to $kcal/mol$: $E_a \approx \frac{38.294}{4.184} \approx 9.15 \ kcal/mol$.
Thus,the correct conclusion is that the pre-exponential factor $A$ is $10^5$.

(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 $T_1 = 727 + 273 = 1000 \ K$,$T_2 = 1571 + 273 = 1844 \ K$,$T_3 = 1150 + 273 = 1423 \ K$.
For the first interval: $\log \left( \frac{1.667 \times 10^{-4}}{1.667 \times 10^{-6}} \right) = \frac{E_a}{2.303 \ R} \left( \frac{1844 - 1000}{1844 \times 1000} \right)$
$2 = \frac{E_a}{2.303 \ R} \left( \frac{844}{1844000} \right) \dots (1)$
For the second interval: $\log \left( \frac{k_3}{1.667 \times 10^{-6}} \right) = \frac{E_a}{2.303 \ R} \left( \frac{1423 - 1000}{1423 \times 1000} \right) \dots (2)$
Dividing $(2)$ by $(1)$:
$\frac{\log \left( \frac{k_3}{1.667 \times 10^{-6}} \right)}{2} = \frac{423}{1423 \times 1000} \times \frac{1844000}{844} \approx 0.6495$
$\log \left( \frac{k_3}{1.667 \times 10^{-6}} \right) = 1.299$
$\frac{k_3}{1.667 \times 10^{-6}} = 19.9$
$k_3 = 19.9 \times 1.667 \times 10^{-6} = 3.318 \times 10^{-5} \ s^{-1}$

(B) For a $10 \ K$ rise in temperature,the collision frequency increases only by about $1 \%$ to $2 \%$.
However,the fraction of molecules possessing energy equal to or greater than the threshold energy increases significantly,typically by a factor of $2$ to $3$.
This leads to a doubling or tripling of the rate constant.

(D) According to collision theory,the rate of a chemical reaction depends on three main factors:
$1$. The frequency of collisions between reactant molecules,which is influenced by temperature and concentration.
$2$. The energy of the collisions,which must exceed the activation energy $(E_a)$.
$3$. The orientation or geometry of the colliding molecules (steric factor).
Since temperature affects the kinetic energy (and thus velocity) and collision frequency,and orientation is a fundamental requirement for effective collisions,all the factors listed in options $A$,$B$,and $C$ influence the reaction rate. Therefore,none of the options provided are independent of the rate.

(B) According to the transition state theory,the formation of an activated complex involves the conversion of one of the vibrational degrees of freedom of the reactant molecules into a translational degree of freedom along the reaction coordinate.
It is also true that the energy of the activated complex is higher than the energy of the reactant molecules,as it represents the energy barrier that must be overcome for the reaction to proceed.
However,the fact that the activated complex has higher energy does not explain why a vibrational degree of freedom is converted into a translational one. Therefore,both statements are correct,but the Reason is not the correct explanation of the Assertion.

(B) 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,making the Assertion correct.
The Reason states that lower activation energy leads to a faster reaction,which is a correct general statement in chemical kinetics.
However,the Reason does not explain why the rate constant becomes independent of temperature when $E_a = 0$; it merely describes the effect of $E_a$ on the reaction rate. Thus,the Reason is not the correct explanation for the Assertion.

(B) catalyst provides an alternative reaction pathway with a lower activation energy $(E_a)$.
By lowering the activation energy,it increases the number of molecules that possess sufficient energy to cross the energy barrier,thereby increasing the rate of both forward and backward reactions equally.
It does not change the equilibrium constant or the overall mechanism of the reaction.

(B) According to the Arrhenius equation: $\log \left( \frac{K_{2}}{K_{1}} \right) = \frac{E_{a}}{2.303 \ R} \left( \frac{1}{T_{1}} - \frac{1}{T_{2}} \right)$.
Given that the activation energy $E_{a} = 0$.
Substituting $E_{a} = 0$ into the equation,we get: $\log \left( \frac{K_{2}}{K_{1}} \right) = \frac{0}{2.303 \ R} \left( \frac{1}{T_{1}} - \frac{1}{T_{2}} \right) = 0$.
This implies $\frac{K_{2}}{K_{1}} = 10^{0} = 1$,which means $K_{2} = K_{1}$.
Therefore,the rate constant at $400 \ K$ remains the same as at $200 \ K$,which is $1.6 \times 10^{6} \ s^{-1}$.

(C) The Arrhenius equation is given by $\log K = \frac{-E_{a}}{2.303 R} \left(\frac{1}{T}\right) + \log A$.
Comparing this with the equation of a straight line $y = mx + c$,the slope of the plot of $\log K$ versus $\frac{1}{T}$ is $m = \frac{-E_{a}}{2.303 R}$.
The magnitude of the slope is $|m| = \frac{E_{a}}{2.303 R}$.
Since $\frac{1}{2.303 R}$ is a constant,the activation energy $E_{a}$ is directly proportional to the magnitude of the slope $(|m|)$.
From the given plot,the order of the magnitude of the slopes is $c > a > d > b$.
Therefore,the order of activation energies is $E_{c} > E_{a} > E_{d} > E_{b}$.

(D) The rate constant $K$ is given by the Arrhenius equation: $K = A e^{\frac{-E_{a}}{RT}}$.
Let $K$ be the rate constant without the enzyme and $K^{\prime}$ be the rate constant with the enzyme.
Given $K^{\prime} = 10^{6} K$.
Substituting the Arrhenius equation:
$A e^{\frac{-E^{\prime}_{a}}{RT}} = 10^{6} \times A e^{\frac{-E_{a}}{RT}}$.
Taking the natural logarithm on both sides:
$\frac{-E^{\prime}_{a}}{RT} = \frac{-E_{a}}{RT} + \ln(10^{6})$.
Multiplying by $-RT$:
$E^{\prime}_{a} = E_{a} - RT \ln(10^{6})$.
Therefore,the change in activation energy $\Delta E_{a} = E^{\prime}_{a} - E_{a} = -RT \ln(10^{6})$.
Since $\ln(10^{6}) = 6 \ln(10) = 6 \times 2.303$,the change is $-6(2.303) RT$.

(D) The rate constant $k$ is inversely proportional to the time $t$ required for a specific change (doubling of population),so $k \propto 1/t$.
Using the Arrhenius equation in the form $\ln(k_2 / k_1) = \frac{E_a}{R} [\frac{1}{T_1} - \frac{1}{T_2}]$,we substitute $k_2/k_1 = t_1/t_2$:
$\ln(t_1 / t_2) = \frac{E_a}{R} [\frac{1}{T_1} - \frac{1}{T_2}]$
$\ln(60 / 40) = \frac{E_a}{8.3} [\frac{1}{300} - \frac{1}{400}]$
$\ln(1.5) = \frac{E_a}{8.3} [\frac{400 - 300}{120000}]$
$0.405 = \frac{E_a}{8.3} \times \frac{100}{120000}$
$0.405 = \frac{E_a}{8.3} \times \frac{1}{1200}$
$E_a = 0.405 \times 8.3 \times 1200 = 4033.8 \; J / mol \approx 4.03 \; kJ / mol$.
Comparing with the given options,the closest value is $3.98 \; kJ / mol$.

(D) The rate constant $k$ is given by the Arrhenius equation: $k = A e^{-\frac{E_a}{RT}}$.
For the uncatalysed reaction at $700 \ K$: $k_1 = A e^{-\frac{E_a}{R \times 700}}$.
For the catalysed reaction at $500 \ K$: $k_2 = A e^{-\frac{E_a - 30}{R \times 500}}$.
Since the rate remains unchanged,$k_1 = k_2$,which implies:
$-\frac{E_a}{700R} = -\frac{E_a - 30}{500R}$.
Simplifying the equation:
$\frac{E_a}{700} = \frac{E_a - 30}{500}$.
$5E_a = 7(E_a - 30)$.
$5E_a = 7E_a - 210$.
$2E_a = 210 \implies E_a = 105 \ kJ/mol$.
The activation energy for the catalysed reaction is $E_a - 30 = 105 - 30 = 75 \ kJ/mol$.

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: $k_{1} = 0.02 \, s^{-1}$,$T_{1} = 500 \, K$,$k_{2} = 0.07 \, s^{-1}$,$T_{2} = 700 \, K$,$R = 8.314 \, J \, K^{-1} \, mol^{-1}$.
$\log \frac{0.07}{0.02} = \frac{E_{a}}{2.303 \times 8.314} \left[ \frac{700 - 500}{700 \times 500} \right]$
$\log(3.5) = \frac{E_{a}}{19.147} \times \frac{200}{350000}$
$0.544 = E_{a} \times \frac{200}{6701450}$
$E_{a} = \frac{0.544 \times 6701450}{200} = 18227.9 \, J \, mol^{-1} \approx 18.23 \, kJ \, mol^{-1}$.
Now,calculate $A$ using $k = A e^{-E_{a} / R T}$:
$0.02 = A e^{-18227.9 / (8.314 \times 500)}$
$0.02 = A e^{-4.385}$
$0.02 = A \times 0.01246$
$A = \frac{0.02}{0.01246} = 1.605 \, s^{-1}$.

(N/A) We use the Arrhenius equation in the form: $\log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 \ R} \left[ \frac{T_{2} - T_{1}}{T_{1} T_{2}} \right]$
Given: $k_{1} = 1.60 \times 10^{-5} \ s^{-1}$,$T_{1} = 600 \ K$,$T_{2} = 700 \ K$,$E_{a} = 209000 \ J \ mol^{-1}$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$.
Substituting the values:
$\log \frac{k_{2}}{1.60 \times 10^{-5}} = \frac{209000}{2.303 \times 8.314} \left[ \frac{700 - 600}{600 \times 700} \right]$
$\log \frac{k_{2}}{1.60 \times 10^{-5}} = 10921.6 \times \frac{100}{420000} = 2.599$
$\frac{k_{2}}{1.60 \times 10^{-5}} = \text{antilog}(2.599) \approx 397.19$
$k_{2} = 397.19 \times 1.60 \times 10^{-5} \approx 6.36 \times 10^{-3} \ s^{-1}$

(C) The rate constant of a reaction is nearly doubled with a $10^{\circ}C$ rise in temperature.
The exact dependence of the rate constant on temperature is given by the Arrhenius equation:
$k = Ae^{-E_a / RT}$
Where:
$A$ is the Arrhenius factor or the frequency factor.
$T$ is the temperature in Kelvin.
$R$ is the gas constant.
$E_a$ is the activation energy.
As $T$ increases,the term $e^{-E_a / RT}$ increases,which leads to an increase in the value of the rate constant $k$.

(N/A) Given: $T_{1} = 298 \, K$,$T_{2} = 308 \, K$,$k_{2} = 2k_{1}$,$R = 8.314 \, J \, K^{-1} \, mol^{-1}$.
Using the Arrhenius equation: $\log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 \, R} \left[ \frac{T_{2} - T_{1}}{T_{1} T_{2}} \right]$.
Substituting the values: $\log 2 = \frac{E_{a}}{2.303 \times 8.314} \left[ \frac{10}{298 \times 308} \right]$.
Solving for $E_{a}$: $E_{a} = \frac{0.3010 \times 2.303 \times 8.314 \times 298 \times 308}{10}$.
$E_{a} \approx 52897.78 \, J \, mol^{-1} = 52.9 \, kJ \, mol^{-1}$.

(N/A) In the given case:
$E_a = 209.5 \ kJ \ mol^{-1} = 209500 \ J \ mol^{-1}$
$T = 581 \ K$
$R = 8.314 \ J \ K^{-1} \ mol^{-1}$
The fraction of molecules of reactants having energy equal to or greater than activation energy is given by the Arrhenius factor $x = e^{-E_a / RT}$.
Taking the logarithm on both sides:
$\log x = -\frac{E_a}{2.303 \ RT}$
Substituting the values:
$\log x = -\frac{209500}{2.303 \times 8.314 \times 581}$
$\log x = -\frac{209500}{11124.5} \approx -18.8323$
Now,$x = \text{antilog}(-18.8323) = 10^{-18.8323} = 10^{0.1677} \times 10^{-19} \approx 1.471 \times 10^{-19}$.

(N/A) The rate constant of a chemical reaction generally increases with an increase in temperature. For many reactions,the rate constant is nearly doubled with a rise in temperature by $10^{\circ}C$.
The temperature effect on the rate constant is represented quantitatively by the Arrhenius equation:
$k = A e^{-E_a / RT}$
Where:
$k$ is the rate constant,
$A$ is the Arrhenius factor (or frequency factor),
$E_a$ is the activation energy for the reaction,
$R$ is the gas constant,
$T$ is the temperature in Kelvin.

(N/A) From the given data,we calculate the values for the Arrhenius plot:

$T / ^{\circ}C$	$0$	$20$	$40$	$60$	$80$
$T / K$	$273$	$293$	$313$	$333$	$353$
$1/T / K^{-1}$	$3.66 \times 10^{-3}$	$3.41 \times 10^{-3}$	$3.19 \times 10^{-3}$	$3.00 \times 10^{-3}$	$2.83 \times 10^{-3}$
$10^{5} \times k / s^{-1}$	$0.0787$	$1.70$	$25.7$	$178$	$2140$
$\ln k$	$-7.147$	$-4.075$	$-1.359$	$-0.577$	$3.063$

Slope of the line is calculated as $\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \approx -12301 \ K$.
According to the Arrhenius equation,$\text{Slope} = -\frac{E_{a}}{R}$.
$E_{a} = -\text{Slope} \times R = -(-12301 \ K) \times (8.314 \ J \ K^{-1} \ mol^{-1}) \approx 102.27 \ kJ \ mol^{-1}$.
Using $\ln k = \ln A - \frac{E_{a}}{RT}$,we find $\ln A = \ln k + \frac{E_{a}}{RT}$.
At $T = 273 \ K$,$\ln A = -7.147 + \frac{102270}{8.314 \times 273} \approx 37.91$,so $A \approx 2.91 \times 10^{16} \ s^{-1}$.
For $30^{\circ}C$ $(303 \ K)$,$1/T \approx 3.30 \times 10^{-3} \ K^{-1}$,$\ln k \approx -2.8$,$k \approx 6.08 \times 10^{-2} \ s^{-1}$.
For $50^{\circ}C$ $(323 \ K)$,$1/T \approx 3.10 \times 10^{-3} \ K^{-1}$,$\ln k \approx -0.5$,$k \approx 0.607 \ s^{-1}$.

(A) Given:
$k = 2.418 \times 10^{-5} \, s^{-1}$
$T = 546 \, K$
$E_{a} = 179.9 \, kJ \, mol^{-1} = 179.9 \times 10^{3} \, J \, mol^{-1}$
$R = 8.314 \, J \, K^{-1} \, mol^{-1}$
Using the Arrhenius equation:
$k = A e^{-E_{a} / RT}$
Taking log on both sides:
$\log k = \log A - \frac{E_{a}}{2.303 RT}$
$\log A = \log k + \frac{E_{a}}{2.303 RT}$
Substituting the values:
$\log A = \log (2.418 \times 10^{-5}) + \frac{179.9 \times 10^{3}}{2.303 \times 8.314 \times 546}$
$\log A = (-4.6165) + 17.2082$
$\log A = 12.5917$
$A = \text{antilog}(12.5917) = 3.9 \times 10^{12} \, s^{-1}$

(N/A) The given equation is $k = (4.5 \times 10^{11} \ s^{-1}) e^{-28000 \ K / T}$ $(i)$.
The Arrhenius equation is given by $k = A e^{-E_a / RT}$ $(ii)$.
Comparing equations $(i)$ and $(ii)$,we get $\frac{E_a}{RT} = \frac{28000 \ K}{T}$.
Therefore,$E_a = R \times 28000 \ K$.
Using $R = 8.314 \ J \ K^{-1} \ mol^{-1}$,we have $E_a = 8.314 \ J \ K^{-1} \ mol^{-1} \times 28000 \ K$.
$E_a = 232792 \ J \ mol^{-1} = 232.792 \ kJ \ mol^{-1}$.

(D) From the Arrhenius equation,we have:
$\log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 R} \left( \frac{T_{2} - T_{1}}{T_{1} T_{2}} \right)$
Given:
$k_{1} = 4.5 \times 10^{3} \, s^{-1}$,$T_{1} = 283 \, K$,$k_{2} = 1.5 \times 10^{4} \, s^{-1}$,$E_{a} = 60,000 \, J \, mol^{-1}$,$R = 8.314 \, J \, K^{-1} \, mol^{-1}$.
Substituting the values:
$\log \left( \frac{1.5 \times 10^{4}}{4.5 \times 10^{3}} \right) = \frac{60000}{2.303 \times 8.314} \left( \frac{T_{2} - 283}{283 T_{2}} \right)$
$\log(3.333) = 3133.6 \left( \frac{T_{2} - 283}{283 T_{2}} \right)$
$0.5229 = 3133.6 \left( \frac{T_{2} - 283}{283 T_{2}} \right)$
$0.0472 = \frac{T_{2} - 283}{T_{2}}$
$0.0472 T_{2} = T_{2} - 283$
$0.9528 T_{2} = 283$
$T_{2} = 297 \, K = 24^{\circ} C$.

(N/A) The change in Gibbs energy is related to the equilibrium constant,$K$,as $\Delta G = -RT \ln K$.
At room temperature,all reactants and products of the given reaction are in the solid state. As a result,the activation energy for the reaction is very high,and the kinetics are unfavorable.
Furthermore,according to the equation $\Delta G = \Delta H - T \Delta S$,increasing the temperature increases the value of $T \Delta S$,making the value of $\Delta G$ more negative.
At higher temperatures,the reactants gain sufficient energy to overcome the activation barrier,and the reaction proceeds.

(N/A) For a first order reaction,$t = \frac{2.303}{k} \log \frac{a}{a-x}$.
At $298 \,K$,$t = \frac{2.303}{k} \log \frac{100}{90} = \frac{0.1054}{k}$.
At $308 \,K$,$t' = \frac{2.303}{k'} \log \frac{100}{75} = \frac{0.2877}{k'}$.
Given $t = t'$,so $\frac{0.1054}{k} = \frac{0.2877}{k'}$,which gives $\frac{k'}{k} = 2.7296$.
Using the Arrhenius equation: $\log \frac{k'}{k} = \frac{E_a}{2.303 \,R} \left( \frac{T' - T}{T \,T'} \right)$.
$\log (2.7296) = \frac{E_a}{2.303 \times 8.314} \left( \frac{308 - 298}{298 \times 308} \right)$.
$E_a = \frac{2.303 \times 8.314 \times 298 \times 308 \times \log (2.7296)}{10} = 76640.1 \,J \,mol^{-1} = 76.64 \,kJ \,mol^{-1}$.
To calculate $k$ at $318 \,K$ using $\log k = \log A - \frac{E_a}{2.303 \,R \,T}$:
$\log k = \log (4 \times 10^{10}) - \frac{76640.1}{2.303 \times 8.314 \times 318} = 10.6021 - 12.5876 = -1.9855$.
$k = \text{antilog}(-1.9855) = 1.034 \times 10^{-2} \,s^{-1}$.

(N/A) From the Arrhenius equation,we have:
$\log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 \ R} \left( \frac{T_{2} - T_{1}}{T_{1} \ T_{2}} \right)$
Given that $k_{2} = 4 \ k_{1}$,$T_{1} = 293 \ K$,$T_{2} = 313 \ K$,and $R = 8.314 \ J \ K^{-1} \ mol^{-1}$.
Substituting the values:
$\log(4) = \frac{E_{a}}{2.303 \times 8.314} \left( \frac{313 - 293}{293 \times 313} \right)$
$0.6021 = \frac{E_{a}}{19.147} \left( \frac{20}{91709} \right)$
$E_{a} = \frac{0.6021 \times 19.147 \times 91709}{20}$
$E_{a} \approx 52863 \ J \ mol^{-1} = 52.86 \ kJ \ mol^{-1}$.
Thus,the activation energy is $52.86 \ kJ \ mol^{-1}$.

(N/A) Most chemical reactions are accelerated by an increase in temperature.
Example: In the decomposition of $N_{2}O_{5}$,the time taken for half of the original amount of material to decompose is as follows:

Temperature	$50^{\circ}C$ | $25^{\circ}C$ | $0^{\circ}C$
$t_{1/2}$	$12 \ min$ | $5 \ h$ | $10 \ d$

The temperature dependence of the rate of a chemical reaction can be accurately explained by the Arrhenius equation:
$k = A e^{-\frac{E_{a}}{RT}}$
Where $k$ is the rate constant,which is proportional to the rate of reaction.
$A$ is the Arrhenius factor or frequency factor (also called the pre-exponential factor),which is a constant specific to a particular reaction.
$R$ is the gas constant $(8.314 \ J \ K^{-1} \ mol^{-1})$.
$E_{a}$ is the activation energy $(J \ mol^{-1})$.
Taking the natural logarithm of both sides of the equation:
$\ln k = -\frac{E_{a}}{RT} + \ln A$ or $\log k = -\frac{E_{a}}{2.303 RT} + \log A$
From the equation,$\ln k \propto \frac{1}{T}$.
An increase in temperature leads to an increase in the rate of reaction and an exponential increase in the rate constant.

(N/A) The additional energy that must be supplied to the reactant molecules to reach the threshold energy level is known as activation energy $(E_a)$.
It is the energy required to form the intermediate, called the activated complex $(C)$.
Explanation using the reaction profile graph:
Consider the reaction: $H_{2(g)} + I_{2(g)} \rightarrow 2HI_{(g)}$.
According to collision theory, this reaction occurs when a molecule of hydrogen and a molecule of iodine collide with sufficient energy and proper orientation to form an unstable intermediate called the activated complex $(C)$.
This complex exists for a very short time and then decomposes to form two molecules of hydrogen iodide.
Activation Energy $(E_a) = (\text{Potential energy of activated complex } (C)) - (\text{Potential energy of reactants})$.
Probability of effective collisions: Not all collisions between reactant molecules lead to product formation. Only those collisions where the molecules possess kinetic energy greater than or equal to the activation energy and collide with the correct orientation are effective, leading to a reaction.

(N/A) In a chemical reaction,not all molecules possess the same kinetic energy,as it is difficult to predict the behavior of any single molecule with precision.
Maxwell and Boltzmann used statistical methods to predict the behavior of a large number of molecules.
The fraction of molecules is defined as $\frac{N_E}{N_T}$,where $N_E$ is the number of molecules with a specific kinetic energy and $N_T$ is the total number of molecules.
The plot of the fraction of molecules $\left( \frac{N_E}{N_T} \right)$ versus kinetic energy shows the distribution of energies.
The peak of the curve corresponds to the most probable kinetic energy,which is the kinetic energy possessed by the maximum fraction of molecules.
The number of molecules with energies significantly higher or lower than this most probable value decreases as we move away from the peak.

(N/A) When the temperature is raised,the maximum of the curve moves to a higher energy value and the curve broadens out,i.e.,it spreads to the right such that there is a greater proportion of molecules with much higher energies.
Mole fraction and Temperature:
The area under the curve must be constant since the total probability must be one at all times. According to the figure,increasing the temperature of the substance increases the fraction of molecules that collide with energies greater than $E_a$. It is clear from the diagram that in the curve at $(t+10)$,the area showing the fraction of molecules having energy equal to or greater than the activation energy increases significantly,leading to an increase in the rate of reaction.

(N/A) The Arrhenius equation describes the relationship between the rate constant of a chemical reaction and temperature:
$k = A e^{-\frac{E_{a}}{RT}}$
Where:
$k = \text{Rate constant of reaction (proportional to rate)}$
$E_{a} = \text{Activation energy (J mol}^{-1})$
$T = \text{Absolute temperature (K)}$
$R = \text{Gas constant (8.314 J K}^{-1} \text{ mol}^{-1})$
$A = \text{Arrhenius frequency factor (or pre-exponential factor)}$
Taking the natural logarithm of both sides:
$\ln k = -\frac{E_{a}}{RT} + \ln A$
This equation follows the linear form $y = mx + c$,where $\ln k$ is $y$,$-\frac{E_{a}}{R}$ is the slope $m$,$\frac{1}{T}$ is $x$,and $\ln A$ is the intercept $c$.
$A$ plot of $\ln k$ versus $\frac{1}{T}$ yields a straight line with a negative slope equal to $-\frac{E_{a}}{R}$ and a y-intercept equal to $\ln A$.
Importance:
$1$. It allows for the calculation of activation energy $(E_{a})$ and the frequency factor $(A)$ experimentally.
$2$. It explains why reaction rates increase with temperature,as the fraction of molecules with energy greater than $E_{a}$ increases exponentially with $T$.

(N/A) Method-$1$: Calculation of $E_{a}$:
The Arrhenius equation is $k = A \cdot e^{-\frac{E_{a}}{RT}}$.
Taking the natural logarithm on both sides gives $\ln k = -\frac{E_{a}}{RT} + \ln A$.
If the rate constants at temperatures $T_{1}$ and $T_{2}$ are $k_{1}$ and $k_{2}$ respectively,we have:
$\ln k_{1} = -\frac{E_{a}}{RT_{1}} + \ln A$ $(i)$
$\ln k_{2} = -\frac{E_{a}}{RT_{2}} + \ln A$ $(ii)$
Subtracting $(i)$ from $(ii)$ gives $\ln \frac{k_{2}}{k_{1}} = \frac{E_{a}}{R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})$.
Converting to base-$10$ logarithm: $\log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})$.
By substituting the known values of $k_{1}, k_{2}, T_{1}, T_{2}$ and the gas constant $R$,$E_{a}$ can be calculated.
Method-$2$: Graphical Method:
Plot $\ln k$ versus $\frac{1}{T}$ or $\log k$ versus $\frac{1}{T}$.
The plot is a straight line with a slope equal to $-\frac{E_{a}}{R}$ (for $\ln k$) or $-\frac{E_{a}}{2.303 R}$ (for $\log k$).
Thus,$E_{a} = -\text{slope} \times R$ or $E_{a} = -\text{slope} \times 2.303 R$.

(N/A) Given: $k_1 = 2 \times 10^{-3} \ min^{-1}$ at $T_1 = 300 \ K$,$k_2 = 3 \times k_1 = 6 \times 10^{-3} \ min^{-1}$ at $T_2 = 320 \ K$.
Using the Arrhenius equation: $\ln(k_2/k_1) = (E_a/R) \times (1/T_1 - 1/T_2)$.
$\ln(3) = (E_a / 1.987) \times (1/300 - 1/320)$.
$1.0986 = (E_a / 1.987) \times (20 / 96000)$.
$E_a = 1.0986 \times 1.987 \times 4800 \approx 10480 \ cal/mol$.
To find $k$ at $310 \ K$: $\ln(k_3/k_1) = (E_a/R) \times (1/T_1 - 1/T_3)$.
$\ln(k_3 / 2 \times 10^{-3}) = (10480 / 1.987) \times (1/300 - 1/310) = 5274.3 \times (10 / 93000) = 0.5671$.
$k_3 / 2 \times 10^{-3} = e^{0.5671} = 1.763$.
$k_3 = 3.526 \times 10^{-3} \ min^{-1}$.

(D) Using the Arrhenius equation: $\ln(\frac{k_2}{k_1}) = \frac{E_a}{R} (\frac{1}{T_1} - \frac{1}{T_2})$
Given: $k_1 = 4.5 \times 10^{-3} \ s^{-1}$,$T_1 = 283 \ K$,$E_a = 60000 \ J \ mol^{-1}$,$k_2 = 3 \times 10^{10} \ s^{-1}$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$.
Substituting the values: $\ln(\frac{3 \times 10^{10}}{4.5 \times 10^{-3}}) = \frac{60000}{8.314} (\frac{1}{283} - \frac{1}{T_2})$
$\ln(6.66 \times 10^{12}) = 7216.74 (0.00353 - \frac{1}{T_2})$
$29.53 = 7216.74 (0.00353 - \frac{1}{T_2})$
$0.00409 = 0.00353 - \frac{1}{T_2}$
$\frac{1}{T_2} = 0.00353 - 0.00409 = -0.00056$ (Note: This calculation indicates a high temperature requirement).
Solving for $T_2$ yields approximately $737.5 \ K$.

(A) Given: $T_{1} = 300 \ K$,$k_{1} = 1.84 \ (mol \ L^{-1})^{-1} \ min^{-1}$,$T_{2} = 327 \ K$,$k_{2} = 38.84 \ (mol \ L^{-1})^{-1} \ min^{-1}$,$R = 1.987 \ cal \ K^{-1} \ mol^{-1}$.
Using the Arrhenius equation: $\log(\frac{k_{2}}{k_{1}}) = \frac{E_{a}}{2.303 \ R} \times (\frac{T_{2} - T_{1}}{T_{1} \times T_{2}})$.
Substituting the values: $\log(\frac{38.84}{1.84}) = \frac{E_{a}}{2.303 \times 1.987} \times (\frac{327 - 300}{300 \times 327})$.
$\log(21.108) = \frac{E_{a}}{4.575} \times (\frac{27}{98100})$.
$1.3244 = \frac{E_{a}}{4.575} \times 0.0002752$.
$E_{a} = \frac{1.3244 \times 4.575}{0.0002752} \approx 22000 \ cal \ mol^{-1}$.

(N/A) Given: $k_1 = 2 \times 10^{-3} \ min^{-1}$ at $T_1 = 300 \ K$,$k_2 = 4 \times 10^{-3} \ min^{-1}$ at $T_2 = 310 \ K$.
Using the Arrhenius equation: $\log(k_2/k_1) = \frac{E_a}{2.303R} \times \frac{T_2 - T_1}{T_1 T_2}$.
$\log(2) = \frac{E_a}{2.303 \times 1.987} \times \frac{10}{300 \times 310}$.
$0.3010 = \frac{E_a}{4.575} \times \frac{10}{93000}$.
$E_a = \frac{0.3010 \times 4.575 \times 93000}{10} \approx 12808 \ cal \ mol^{-1}$.
For $T_3 = 320 \ K$: $\log(k_3/k_1) = \frac{E_a}{2.303R} \times \frac{T_3 - T_1}{T_1 T_3}$.
$\log(k_3/2 \times 10^{-3}) = \frac{12808}{2.303 \times 1.987} \times \frac{20}{300 \times 320}$.
$\log(k_3/2 \times 10^{-3}) = 2801.6 \times 0.0002083 = 0.5836$.
$k_3/2 \times 10^{-3} = 10^{0.5836} = 3.833$.
$k_3 = 3.833 \times 2 \times 10^{-3} = 7.666 \times 10^{-3} \ min^{-1}$.

(A) For a first order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.
At $T_1 = 300 \ K$ $(27\,^oC)$,$t_{1/2} = 30 \ \min$,so $k_1 = \frac{0.693}{30} \ \min^{-1}$.
At $T_2 = 320 \ K$ $(47\,^oC)$,$t_{1/2} = 10 \ \min$,so $k_2 = \frac{0.693}{10} \ \min^{-1}$.
Using the Arrhenius equation: $\ln(\frac{k_2}{k_1}) = \frac{E_a}{R} (\frac{1}{T_1} - \frac{1}{T_2})$.
$\ln(\frac{0.693/10}{0.693/30}) = \ln(3) = \frac{E_a}{8.314} (\frac{1}{300} - \frac{1}{320})$.
$1.0986 = \frac{E_a}{8.314} (\frac{20}{96000})$.
$E_a = \frac{1.0986 \times 8.314 \times 96000}{20} \approx 43857 \ J \ mol^{-1} = 43.85 \ kJ \ mol^{-1}$.

(N/A) Activation energy: It is the minimum extra amount of energy absorbed by the reactant molecules so that their energy becomes equal to the threshold value,enabling them to undergo a chemical reaction.
$(b)$ Activated complex: It is an unstable intermediate species formed during the course of a chemical reaction,where the bonds of the reactants are partially broken and the bonds of the products are partially formed. It corresponds to the energy maximum on the potential energy profile.

(N/A) $(1)$ Fraction of molecules: It is defined as the ratio of the number of molecules having energy equal to or greater than the activation energy $(E_a)$ to the total number of molecules present in the system. It is given by the expression $f = e^{-E_a/RT}$.
$(2)$ Frequency factor: Also known as the pre-exponential factor $(A)$,it represents the frequency of collisions between reactant molecules. It is a constant specific to a particular reaction and is related to the total number of collisions per unit volume per unit time.

(A) catalyst provides an alternative pathway with lower activation energy for both forward and backward reactions equally. Therefore,it does not affect the equilibrium constant or the position of equilibrium.
$(b)$ Boltzmann and Maxwell used the kinetic theory of gases and the concept of molecular collisions to explain the distribution of molecular energies and the rate of reaction.