2)

An experimental investigation into the kinetics of a first order liquid phase reaction gave the following results:
\begin{tabular}{ccc}
\begin{tabular}{c}
Reactant \
Concentration \
$\left(\mathbf{k m o l ~ m}^{-\mathbf{3}}\right)$
\end{tabular} & \begin{tabular}{c}
Rate of reaction \
$\left(\mathbf{k m o l ~ m}^{-3} \mathbf{s}^{-1}\right)$
\end{tabular} & \begin{tabular}{c}
Temperature \
$\left({ }^{\circ} \mathbf{C}\right)$
\end{tabular} \
3.96 & $6.85 \times 10-3$ & \
3.20 & $8.10 \times 10-3$ & 20.0 \
2.94 & $11.55 \times 10-3$ & 25.7 \
1.82 & $12.01 \times 10-3$ & 32.8 \
0.43 & $4.37 \times 10-3$ & 41.8 \
& & 49.8
\end{tabular}

Show that the variation of rate constant with temperature is represented by the Arrhenius equation and evaluate the activation energy of reaction and the pre-exponential factor.
$$
\left[46.04 \mathrm{~kJ} \mathrm{~mol}^{-1} \text { and } 28.21 \times 10^{4} \mathrm{~s}^{-1}\right]
$$

Question

2)

An experimental investigation into the kinetics of a first order liquid phase reaction gave the following results:
\begin{tabular}{ccc}
\begin{tabular}{c} 
Reactant \
Concentration \
$\left(\mathbf{k m o l ~ m}^{-\mathbf{3}}\right)$
\end{tabular} & \begin{tabular}{c} 
Rate of reaction \
$\left(\mathbf{k m o l ~ m}^{-3} \mathbf{s}^{-1}\right)$
\end{tabular} & \begin{tabular}{c} 
Temperature \
$\left({ }^{\circ} \mathbf{C}\right)$
\end{tabular} \
3.96 & $6.85 \times 10-3$ & \
3.20 & $8.10 \times 10-3$ & 20.0 \
2.94 & $11.55 \times 10-3$ & 25.7 \
1.82 & $12.01 \times 10-3$ & 32.8 \
0.43 & $4.37 \times 10-3$ & 41.8 \
& & 49.8
\end{tabular}

Show that the variation of rate constant with temperature is represented by the Arrhenius equation and evaluate the activation energy of reaction and the pre-exponential factor.
$$
\left[46.04 \mathrm{~kJ} \mathrm{~mol}^{-1} \text { and } 28.21 \times 10^{4} \mathrm{~s}^{-1}\right]
$$

Accepted Answer

∻Solution∻ ‖ 1 ⋮ **Understand the Arrhenius equation**: The Arrhenius equation relates the rate constant $k$ of a reaction to the temperature $T$ (in Kelvin), the activation energy $E_a$, and the pre-exponential factor $A$. It is given by $k = A e^{-\frac{E_a}{RT}}$, where $R$ is the gas constant (8.314 J mol$^{-1}$ K$^{-1}$). ‖ ‖ 2 ⋮ **Linear form of the Arrhenius equation**: To evaluate the activation energy and the pre-exponential factor, we use the linear form of the Arrhenius equation: $\ln(k) = \ln(A) - \frac{E_a}{RT}$. A plot of $\ln(k)$ versus $\frac{1}{T}$ should yield a straight line with a slope of $-\frac{E_a}{R}$ and an intercept of $\ln(A)$. ‖ ‖ 3 ⋮ **Convert temperatures to Kelvin and calculate $\ln(k)$**: Convert each temperature from degrees Celsius to Kelvin by adding 273.15. Then, calculate the natural logarithm of each rate constant $\ln(k)$. ‖ ‖ 4 ⋮ **Plot and determine slope and intercept**: Plot $\ln(k)$ versus $\frac{1}{T}$ to determine the slope and intercept. The slope will give $-\frac{E_a}{R}$, and the intercept will give $\ln(A)$. ‖ ‖ 5 ⋮ **Calculate $E_a$ and $A$**: Use the slope to calculate the activation energy $E_a$ by multiplying the slope by $-R$. Then, exponentiate the intercept to find the pre-exponential factor $A$. ‖ ∻Answer∻ ⚹ The activation energy $E_a$ is 46.04 kJ mol$^{-1}$ and the pre-exponential factor $A$ is $28.21 imes 10^5$ s$^{-1}$. ⚹ ∻Key Concept∻ ⚹ The Arrhenius equation relates the rate constant of a reaction to the temperature, activation energy, and pre-exponential factor. ⚹ ∻Explanation∻ ⚹ By plotting the natural logarithm of the rate constant against the inverse of the temperature in Kelvin, we can determine the activation energy and pre-exponential factor from the slope and intercept of the resulting line.◊What is the Arrhenius equation and how is it used to represent the variation of rate constant with temperature?⍭ Generate me a similar question◊⚹

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