Two reactions R1 and R2 have identical preexponential factors. Activation energy of R1 exceeds that

Q: Why did we divide $k_2$ by $k_1$ instead of $k_1$ by $k_2$?

We divided $k_2$ by $k_1$ simply because the question specifically asked for the value of $\ln(k_2/k_1)$. If you did it the other way around, you would get $-4$, which is not what was asked.

Q: Why did the pre-exponential factor $A$ disappear?

The problem explicitly stated that both reactions have 'identical preexponential factors'. Because they are the same value, when we divide the two rate equations, $A/A$ cancels out to $1$.

Q: Why did we multiply $10$ by $1000$?

This is a classic unit trap! The activation energy difference was given in kilojoules (kJ), but the gas constant $R$ is in joules (J). To make the units cancel out properly in the calculation, we must convert $10 \text{ kJ}$ to $10,000 \text{ J}$.

Q: What is the physical meaning of the pre-exponential factor $A$?

The pre-exponential factor, also known as the frequency factor, represents the total number of collisions (per second) between reactant molecules that have the correct spatial orientation to react, regardless of their energy.

Q: Does temperature affect the activation energy?

For the scope of standard chemical kinetics, activation energy ($E_a$) is considered a constant that is independent of temperature. Temperature affects the *fraction* of molecules that possess this energy, not the energy barrier itself.

Question

Two reactions $R_1$ and $R_2$ have identical preexponential factors. Activation energy of $R_1$ exceeds that of $R_2$ by $10	ext{ kJ mol}^{-1}$. If $k_1$ and $k_2$ are rate constants for reactions $R_1$ and $R_2$, respectively at $300	ext{ K}$, then $\ln\left(\frac{k_2}{k_1}ight)$ is equal to ($R = 8.314	ext{ J mol}^{-1}	ext{K}^{-1}$)

Options:

(A) 8

(B) 12

(C) 6

(D) 4

Sigma Percentile · Accepted Answer

### Visualizing the Energy Landscape

Imagine you are looking at the energy profiles of two different chemical reactions, $R_1$ and $R_2$. The problem tells us that the activation energy of $R_1$ is higher than that of $R_2$ by exactly $10 	ext{ kJ mol}^{-1}$.

If we were to draw this, we would see two hills. The peak for $R_1$ is taller than the peak for $R_2$. The vertical distance between these two peaks represents the difference in their activation energies, which we can write as:

$$\Delta E_a = E_{a_1} - E_{a_2} = 10 	ext{ kJ mol}^{-1}$$

### The Master Equation: Arrhenius

To connect these activation energies to their respective rate constants ($k_1$ and $k_2$), we need our trusty tool: the **Arrhenius equation**. This equation beautifully links the speed of a reaction to its energy barrier and temperature.

$$k = A e^{-\frac{E_a}{RT}}$$

Here, $A$ is the pre-exponential factor (representing collision frequency), $E_a$ is the activation energy, $R$ is the universal gas constant, and $T$ is the absolute temperature.

The problem gives us a massive hint: both reactions have **identical pre-exponential factors**. This means $A$ is the same for both. Let's write the equation for each reaction:

$$k_1 = A e^{-\frac{E_{a_1}}{RT}}$$

$$k_2 = A e^{-\frac{E_{a_2}}{RT}}$$

### The Power of Ratios

We are asked to find the value of $\ln\left(\frac{k_2}{k_1}ight)$. The most logical next step is to divide the equation for $k_2$ by the equation for $k_1$. Watch what happens to the pre-exponential factor $A$:

$$\frac{k_2}{k_1} = \frac{A e^{-\frac{E_{a_2}}{RT}}}{A e^{-\frac{E_{a_1}}{RT}}}$$

Because $A$ is identical, it cancels out completely! Using the laws of exponents, we can combine the terms:

$$\frac{k_2}{k_1} = e^{\frac{E_{a_1} - E_{a_2}}{RT}}$$

To bring that complex exponent down to earth, we take the natural logarithm ($\ln$) on both sides:

$$\ln\left(\frac{k_2}{k_1}ight) = \frac{E_{a_1} - E_{a_2}}{RT}$$

### The Unit Trap and Final Calculation

Now, we just need to substitute our known values. But wait! This is where many students make a fatal error. The difference in activation energy is given as $10 	ext{ kJ mol}^{-1}$, but the gas constant $R$ is given as $8.314 	ext{ J mol}^{-1}	ext{K}^{-1}$.

**You must ensure your units match!** We need to convert kilojoules to joules by multiplying by $1000$:

$$E_{a_1} - E_{a_2} = 10 	ext{ kJ mol}^{-1} = 10,000 	ext{ J mol}^{-1}$$

Now, let's plug everything into our equation:

$$\ln\left(\frac{k_2}{k_1}ight) = \frac{10000}{8.314 	imes 300}$$

Let's do the math. The denominator is $8.314 	imes 300 \approx 2494.2$. Dividing $10,000$ by $2494.2$ gives us a very clean number:

$$\ln\left(\frac{k_2}{k_1}ight) \approx 4.009 \approx 4$$

And there we have it! The natural logarithm of the ratio of their rate constants is exactly $4$.

Select Answer:

Visualized Solution

$E_{a_{1}} - E_{a_{2}} = 10 kJ mol^{- 1}$

$k = A e^{- \frac{E _{a}}{R T}}$

$k_{1} and k_{2}$

$\frac{k _{2}}{k _{1}}$

$ln (\frac{k _{2}}{k _{1}})$

$Substitution$

$Final Answer$

$What if A_{1} \neq = A_{2} ?$

The Sigma Insight: Theories of Chemical Reaction

Visualizing the Energy Landscape

The Master Equation

The Power of Ratios

The Unit Trap and Final Calculation

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Select Answer:

Visualized Solution

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k1​ and k2​

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substitution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer

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