Arrhenius Equation — Revision Notes

⚡ 30-Second Revision

Arrhenius Equation: — $k = A e^{-E_a/RT}$
Logarithmic Form: — $\ln k = \ln A - \frac{E_a}{RT}$
Two-Point Form: — $\ln \frac{k_2}{k_1} = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)$
Arrhenius Plot: — $\ln k$ vs $1/T$ is a straight line.
Slope of Arrhenius Plot: — $-\frac{E_a}{R}$
Y-intercept of Arrhenius Plot: — $\ln A$
Units: — $T$ in Kelvin (K), $E_a$ in $\text{J mol}^{-1}$ (if $R = 8.314\,\text{J mol}^{-1}\text{K}^{-1}$).
Catalyst Effect: — Lowers $E_a$, increases $k$.

2-Minute Revision

The Arrhenius equation, $k = A e^{-E_a/RT}$, is central to understanding how reaction rates change with temperature. The rate constant ($k$) increases exponentially with absolute temperature ($T$) because a higher temperature means a greater fraction of molecules possess the necessary activation energy ($E_a$) to react.

$E_a$ is the minimum energy barrier for a reaction, while $A$ (pre-exponential factor) accounts for collision frequency and proper orientation. For calculations, always convert temperature to Kelvin and ensure $E_a$ and the gas constant ($R = 8.

314\,\text{J mol}^{-1}\text{K}^{-1}$) are in consistent units (Joules). The logarithmic form,$\ln k = \ln A - \frac{E_a}{RT}$, shows that a plot of$\ln k$versus$1/T$yields a straight line with a negative slope equal to$-E_a/R$.

This 'Arrhenius plot' is crucial for experimentally determining $E_a$. The two-point form, $\ln \frac{k_2}{k_1} = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)$, is highly useful for solving numerical problems involving rate constants at two different temperatures.

Remember that catalysts accelerate reactions by lowering $E_a$.

5-Minute Revision

The Arrhenius equation, $k = A e^{-E_a/RT}$, is the bedrock for understanding the temperature dependence of reaction rates. It quantitatively links the rate constant ($k$) to absolute temperature ($T$), activation energy ($E_a$), and the pre-exponential factor ($A$).

Key Components & Their Roles:

Rate Constant ($k$): — A measure of reaction speed. Increases with temperature.
Pre-exponential Factor ($A$): — Represents the frequency of effective collisions (considering both collision rate and correct molecular orientation). It has the same units as $k$.
Activation Energy ($E_a$): — The minimum energy required for reactants to form products. A higher $E_a$ means a slower reaction. Catalysts lower $E_a$.
Gas Constant ($R$): — $8.314\,\text{J mol}^{-1}\text{K}^{-1}$ (or $1.987\,\text{cal mol}^{-1}\text{K}^{-1}$). Ensure units match $E_a$.
Absolute Temperature ($T$): — Always in Kelvin. Higher $T$ increases the fraction of molecules with energy $\ge E_a$, thus increasing $k$.

Forms of the Equation:

1
Basic: — $k = A e^{-E_a/RT}$
2
Logarithmic (Linear): — $\ln k = \ln A - \frac{E_a}{RT}$. This is a linear equation ($y = mx + c$) where $y = \ln k$, $x = 1/T$, slope $m = -E_a/R$, and intercept $c = \ln A$.
3
Two-Point Form: — $\ln \frac{k_2}{k_1} = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)$. This is vital for calculating $E_a$ from two rate constants at two temperatures, or finding $k_2$ given $k_1$, $T_1$, $T_2$, and $E_a$.

Arrhenius Plot: Plotting $\ln k$ (y-axis) against $1/T$ (x-axis) yields a straight line with a negative slope. From this plot, $E_a$ can be calculated from the slope (Slope $\times -R = E_a$), and $A$ from the y-intercept ($e^{\text{intercept}} = A$). A steeper negative slope indicates a higher $E_a$.

Example: If a reaction's rate constant doubles for every $10^\circ\text{C}$ rise in temperature from $298\,\text{K}$ to $308\,\text{K}$, and $k_1$ at $298\,\text{K}$ is $1.0 \times 10^{-3},\text{s}^{-1}$, then $k_2$ at $308\,\text{K}$ is $2.

0 \times 10^{-3},\text{s}^{-1}$. Using the two-point form:$\ln \frac{2.0 \times 10^{-3}}{1.0 \times 10^{-3}} = \frac{E_a}{8.314} \left( \frac{1}{298} - \frac{1}{308} \right)$$\ln 2 = \frac{E_a}{8.314} \left( \frac{308 - 298}{298 \times 308} \right)$$0.

693 = \frac{E_a}{8.314} \left( \frac{10}{91784} \right)$$E_a = \frac{0.693 \times 8.314 \times 91784}{10} \approx 52930\,\text{J mol}^{-1} \approx 52.9\,\text{kJ mol}^{-1}$.

Common Pitfalls: Incorrect unit conversions (especially $T$ to Kelvin, $E_a$ to Joules), arithmetic errors with logarithms, and misinterpreting the Arrhenius plot's slope or intercept.

Prelims Revision Notes

For NEET, the Arrhenius equation is a high-yield topic in chemical kinetics. Focus on these key points for quick recall:

1
Equation Forms:

* Basic: $k = A e^{-E_a/RT}$ * Logarithmic: $\ln k = \ln A - \frac{E_a}{RT}$ (linear form) * Two-point: $\ln \frac{k_2}{k_1} = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)$

1
Variables and Units:

* $k$: Rate constant (units vary with order, e.g., $\text{s}^{-1}$ for first order). * $A$: Pre-exponential factor (same units as $k$). Represents collision frequency and orientation. * $E_a$: Activation energy.

Always positive. Units: $\text{J mol}^{-1}$ or $\text{kJ mol}^{-1}$. * $R$: Gas constant. Use $8.314\,\text{J mol}^{-1}\text{K}^{-1}$ (or $1.987\,\text{cal mol}^{-1}\text{K}^{-1}$). Match units with $E_a$.

* $T$: Absolute temperature. ALWAYS in Kelvin (K). Convert $^\circ\text{C}$ to K by adding 273.15 (or 273).

1
Temperature Dependence:

* Increasing $T$ increases $k$ exponentially, thus increasing reaction rate. * The exponential term $e^{-E_a/RT}$ represents the fraction of molecules with energy $\ge E_a$.

1
**Activation Energy ($E_a$):**

* Energy barrier for reaction. Higher $E_a \implies$ slower reaction. * Catalysts lower $E_a$, thereby increasing $k$ and reaction rate. * $E_a$ is generally constant for a given reaction, independent of $T$.

1
Arrhenius Plot:

* Plot $\ln k$ (y-axis) vs $1/T$ (x-axis). * Results in a straight line with a negative slope. * **Slope $= -E_a/R$.** So, $E_a = -\text{Slope} \times R$. * **Y-intercept $= \ln A$.** So, $A = e^{\text{intercept}}$.

1
Problem-Solving Tips:

* Always convert $T$ to Kelvin first. * Ensure $E_a$ and $R$ units are consistent. Be proficient with natural logarithms and exponentials. For 'rate doubles for every $10^\circ\text{C}$ rise' type problems, use the two-point form. * Catalyst problems: Remember catalysts lower $E_a$ and increase $k$. The ratio of rate constants can be used to find the new $E_a$.

By focusing on these points, you can quickly recall the necessary information and formulas to solve Arrhenius equation problems in NEET.

Vyyuha Quick Recall

To remember the Arrhenius equation $k = A e^{-E_a/RT}$, think: King Arrhenius Explains Energy Required for Transformation.

King: $k$ (rate constant)
Arrhenius: $A$ (pre-exponential factor)
Explains: $e$ (base of natural logarithm)
Energy: $-E_a$ (negative activation energy)
Required: $/R$ (divided by gas constant)
Transformation: $T$ (absolute temperature)