Question

For an elementary chemical reaction, the Arrhenius plot is given below.

If the energy of activation is $6.64 \text{ kJ mol}^{-1}$ and $R = 8.3 \text{ J K}^{-1}\text{ mol}^{-1}$, the temperature at which the rate constant becomes $e^2 \text{ min}^{-1}$, is

• A. $250\text{ K}$
• B. $125\text{ K}$
• C. $150\text{ K}$
• D. $200\text{ K}$

Hint:

Always double-check your units in chemical kinetics calculations! Activation energy ($E_a$) is usually given in kJ/mol, while the gas constant $R$ is given in $\text{J/(K}\cdot\text{mol)}$. Multiplying the kilojoules by $10^3$ is a vital step to avoid being off by a factor of 1000!

Answer 1

The Correct Option is D

Concept: The dependence of the rate constant ($k$) of a chemical reaction on the absolute temperature ($T$) is mathematically described by the logarithmic form of the

Arrhenius Equation: \[ k = A \cdot e^{-\frac{E_a}{RT}} \quad \Rightarrow \quad \ln k = \ln A - \frac{E_a}{R}\left(\frac{1}{T}\right) \] Where:

• $k$ = Rate constant of the chemical reaction.

• $A$ = Arrhenius pre-exponential frequency factor.

• $E_a$ = Activation energy required for the reaction.

• $R$ = Universal Gas Constant.

• $T$ = Absolute temperature in Kelvin (K).
When plotting $\ln k$ on the vertical $y$-axis against $\frac{1}{T}$ on the horizontal $x$-axis, the equation represents a straight line matching the standard linear form $y = mx + c$, where the slope $m = -\frac{E_a}{R}$ and the $y$-intercept $c = \ln A$.

Step 1: Extracting the value of the y-intercept ($\ln A$) from the given graph.
Looking closely at the provided Arrhenius graph, the straight line intersects the vertical $\ln k$ axis exactly at the numerical value of $6$. By definition, the $y$-intercept occurs where the horizontal coordinate $\frac{1}{T} = 0$. Therefore: \[ \text{Intercept} = \ln A = 6 \]

Step 2: Converting given units to standard SI units.
The problem provides the following thermodynamic constants:

• Activation Energy, $E_a = 6.64 \text{ kJ mol}^{-1} = 6.64 \times 10^3 \text{ J mol}^{-1} = 6640 \text{ J mol}^{-1}$

• Universal Gas Constant, $R = 8.3 \text{ J K}^{-1}\text{ mol}^{-1}$
It is essential to convert $E_a$ from kilojoules to joules so that the units cancel out perfectly with the units of the gas constant $R$.

Step 3: Setting up the specific Arrhenius equation for the target rate constant.
We are asked to find the specific temperature $T$ where the rate constant $k$ reaches a value of $e^2 \text{ min}^{-1}$. Let us take the natural logarithm of this target rate constant: \[ k = e^2 \quad \Rightarrow \quad \ln k = \ln(e^2) = 2 \] Now, substitute the values $\ln k = 2$ and $\ln A = 6$ into the logarithmic Arrhenius equation: \[ \ln k = \ln A - \frac{E_a}{RT} \] \[ 2 = 6 - \frac{E_a}{RT} \]

Step 4: Isolating and computing the absolute temperature $T$.
Rearranging the algebraic terms to isolate the temperature component on one side: \[ \frac{E_a}{RT} = 6 - 2 \] \[ \frac{E_a}{RT} = 4 \] Cross-multiplying to solve explicitly for $T$: \[ T = \frac{E_a}{4R} \] Substituting the numerical values of $E_a$ and $R$ that we prepared in Step 2: \[ T = \frac{6640}{4 \times 8.3} \] Let us calculate the product in the denominator first: \[ 4 \times 8.3 = 33.2 \] Now, divide the numerator by this product: \[ T = \frac{6640}{33.2} \] To make the division cleaner, multiply both the numerator and the denominator by 10 to eliminate the decimal: \[ T = \frac{66400}{332} \] Observing that $332 \times 2 = 664$: \[ T = 200\text{ K} \] The calculated absolute temperature is exactly $200\text{ K}$, which perfectly corresponds to Option (4).