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

(C) 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 $(E_a)$.
By lowering the activation energy, a greater fraction of reactant molecules possess sufficient energy to cross the energy barrier, thereby increasing the reaction rate.
It does not change the enthalpy $(\Delta H)$ of the reaction or the equilibrium constant.

(C) Given: $k_1 = 1.5 \times 10^7 \text{ s}^{-1}$, $T_1 = 300 \text{ K}$, $k_2 = 3.0 \times 10^7 \text{ s}^{-1}$, $T_2 = 330 \text{ K}$.
Using the Arrhenius equation: $\log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left( \frac{T_2 - T_1}{T_1 T_2} \right)$.
$\log \left( \frac{3.0 \times 10^7}{1.5 \times 10^7} \right) = \frac{E_a}{19.15} \left( \frac{330 - 300}{300 \times 330} \right)$.
$\log(2) = \frac{E_a}{19.15} \left( \frac{30}{99000} \right)$.
$0.3010 = \frac{E_a}{19.15} \times \frac{1}{3300}$.
$E_a = 0.3010 \times 19.15 \times 3300 \text{ J mol}^{-1}$.
$E_a = 19023.225 \text{ J mol}^{-1} \approx 19.02 \text{ kJ mol}^{-1}$.

(C) $1$. The rate of a chemical reaction is directly proportional to the rate constant $k$.
$2$. According to the Arrhenius equation $k = Ae^{-E_a/RT}$, the rate constant $k$ depends on the activation energy $E_a$ and temperature $T$.
$3$. As the activation energy $E_a$ decreases, the term $e^{-E_a/RT}$ increases, which leads to an increase in the value of $k$.
$4$. Therefore, a decrease in $E_a$ results in a faster reaction rate.
$5$. Thus, option $C$ is correct.

(D) $1$. For a chemical reaction to occur, reactant molecules must collide with a certain minimum amount of kinetic energy.
$2$. This minimum energy threshold is known as the activation energy $(E_a)$.
$3$. Molecules with kinetic energy less than $E_a$ do not react upon collision, while those with energy equal to or greater than $E_a$ can successfully form products.

(D) $1$. $A$ catalyst is a substance that increases the rate of a chemical reaction without being consumed in the process.
$2$. It functions by providing an alternative reaction pathway with a lower activation energy $(E_a)$.
$3$. By lowering the activation energy, a greater fraction of reactant molecules possess sufficient energy to cross the energy barrier, thereby increasing the reaction rate.
$4$. Therefore, the correct role is to decrease the activation energy of a reaction.

(A) The overall rate constant is given by $K = \frac{k_1 k_2}{k_3}$.
Taking the natural logarithm on both sides: $\ln K = \ln k_1 + \ln k_2 - \ln k_3$.
Differentiating with respect to temperature $T$: $\frac{d(\ln K)}{dT} = \frac{d(\ln k_1)}{dT} + \frac{d(\ln k_2)}{dT} - \frac{d(\ln k_3)}{dT}$.
Using the Arrhenius equation $\frac{d(\ln k)}{dT} = \frac{E_a}{RT^2}$, we get: $\frac{E_a}{RT^2} = \frac{E_{a1}}{RT^2} + \frac{E_{a2}}{RT^2} - \frac{E_{a3}}{RT^2}$.
Thus, $E_a = E_{a1} + E_{a2} - E_{a3}$.
Substituting the given values: $E_a = 40 + 50 - 60 = 30 \text{ kJ mol}^{-1}$.

(B) The relationship between enthalpy change $\Delta H$, activation energy of the forward reaction $(E_a)_f$, and activation energy of the backward reaction $(E_a)_b$ is given by: $\Delta H = (E_a)_f - (E_a)_b$.
Given: $(E_a)_f = 150 \text{ kJ mol}^{-1}$ and $\Delta H = -135 \text{ kJ mol}^{-1}$.
Substituting the values: $-135 = 150 - (E_a)_b$.
Rearranging for $(E_a)_b$: $(E_a)_b = 150 + 135 = 285 \text{ kJ mol}^{-1}$.