The decomposition of formic acid on gold surface follows first order kinetics. If the rate constant

Q: Why do we use 2.303 in the Arrhenius formula?

The original Arrhenius equation uses the natural logarithm ($\ln$). To convert the natural logarithm to a base-10 logarithm ($\log_{10}$), we multiply by $2.303$. This makes manual calculations much easier.

Q: Why did we convert $E_a$ from kJ/mol to J/mol?

The universal gas constant $R$ is given as $8.314\text{ J mol}^{-1}\text{ K}^{-1}$. For the units to cancel out correctly in the exponent, the activation energy must also be in Joules.

Q: What if the temperatures were given in Celsius?

You must always convert temperatures to Kelvin when using thermodynamic and kinetic equations like the Arrhenius equation. Using Celsius will yield completely incorrect results.

Q: How did $10^{-4}$ become $10 \times 10^{-5}$?

It is a simple scientific notation adjustment. $10^{-4}$ is mathematically identical to $10 \times 10^{-5}$. We did this strictly to match the format requested in the question text.

Q: Is the rate constant always higher at a higher temperature?

Yes, for elementary reactions, increasing the temperature increases the kinetic energy of the molecules, leading to more effective collisions and a higher rate constant.

Question

The decomposition of formic acid on gold surface follows first order kinetics. If the rate constant at $300	ext{ K}$ is $1.0 	imes 10^{-3}	ext{ s}^{-1}$ and the activation energy, $E_a = 11.488	ext{ kJ mol}^{-1}$, the rate constant at $200	ext{ K}$ is ...... $	imes 10^{-5}	ext{ s}^{-1}$ (Round off to the nearest integer).
[Given, $R = 8.314	ext{ J mol}^{-1}	ext{ K}^{-1}$]

Sigma Percentile · Accepted Answer

### The Temperature Dependence of Kinetics

Imagine you are observing a chemical reaction—the decomposition of formic acid on a gold surface. As you heat the system, the reaction speeds up. This isn't just a random observation; it is a fundamental principle of chemical kinetics governed by the **Arrhenius Equation**.

In this problem, we are given the rate constant at a higher temperature ($300	ext{ K}$) and asked to find the rate constant at a lower temperature ($200	ext{ K}$). We are also provided with the activation energy, which is the energy barrier the molecules must overcome to react.

### The Master Equation

To connect the rate constants at two different temperatures, we use the two-point form of the Arrhenius equation:

$$\log \left( \frac{k_2}{k_1} ight) = \frac{E_a}{2.303 R} \left( \frac{1}{T_1} - \frac{1}{T_2} ight)$$

Before we plug in the numbers, there is a critical trap we must avoid: **Unit Consistency**. The universal gas constant $R$ is given as $8.314	ext{ J mol}^{-1}	ext{ K}^{-1}$, but our activation energy $E_a$ is in $	ext{kJ mol}^{-1}$. We must convert $E_a$ to Joules by multiplying by $1000$:

$$E_a = 11.488	ext{ kJ mol}^{-1} = 11488	ext{ J mol}^{-1}$$

### The Magic of Cancellation

Now, let's substitute our values into the equation:

$$\log \left( \frac{10^{-3}}{k_1} ight) = \frac{11488}{2.303 	imes 8.314} \left( \frac{1}{200} - \frac{1}{300} ight)$$

At first glance, the numbers look terrifying. But let's break them down. The denominator of the constant term is $2.303 	imes 8.314$, which equals approximately $19.147$. If we divide $11488$ by $19.147$, we get exactly $600$!

Next, let's look at the temperature bracket:

$$\frac{1}{200} - \frac{1}{300} = \frac{3 - 2}{600} = \frac{1}{600}$$

This is where the beauty of the problem shines. The $600$ from the constant term perfectly cancels out the $600$ in the denominator of the temperature term:

$$\log \left( \frac{10^{-3}}{k_1} ight) = 600 	imes \frac{1}{600} = 1$$

### The Final Transformation

We are left with a beautifully simple logarithmic equation. To solve for $k_1$, we take the antilog (base 10) of both sides:

$$\frac{10^{-3}}{k_1} = 10^1$$

Rearranging the equation to isolate $k_1$:

$$k_1 = \frac{10^{-3}}{10} = 10^{-4}	ext{ s}^{-1}$$

The question asks for the answer in the format of $x 	imes 10^{-5}	ext{ s}^{-1}$. To match this, we simply multiply and divide by $10$:

$$k_1 = 10 	imes 10^{-5}	ext{ s}^{-1}$$

Thus, the value of $x$ is **$10$**. This problem is a perfect example of how intimidating numbers in physical chemistry often collapse into elegant, simple integers if you trust the math and keep your units consistent.

Enter Numerical Value:

Visualized Solution

$T_{1}, T_{2} and k_{2} Setup$

$Arrhenius Equation$

$Substituting Values$

$Evaluating the Constant Term$

$Evaluating the Temperature Term$

$Combining the Terms$

$Solving for k_{1}$

$Conclusion$

The Sigma Insight: Theories of Chemical Reaction

The Temperature Dependence of Kinetics

The Master Equation

The Magic of Cancellation

The Final Transformation

Similar Questions

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The rate constant ( $k$ ) of a reaction is measured at different temperature ( $T$ ), and the data are plotted in the given figure. The activation energy of the reaction in $kJ mol^{- 1}$ is ( $R$ is gas constant)

A reactant (A) forms two products $A k_{1} B$ , Activation energy $E_{a_{1}}$ $A k_{2} C$ , Activation energy $E_{a_{2}}$ If $E_{a_{2}} = 2 E_{a_{1}}$ , then $k_{1}$ and $k_{2}$ are related as

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T1​,T2​ and k2​ Setup

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

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

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

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

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

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

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