The rate of a reaction decreased by 3.555 times when the temperature was changed from 40°C to 30°C.

Q: Why do we convert temperature to Kelvin?

The gas constant $R$ is in units of $\text{J K}^{-1} \text{mol}^{-1}$. To maintain dimensional consistency, temperature must be in Kelvin.

Q: What if I took $T_1 = 40^\circ\text{C}$ and $T_2 = 30^\circ\text{C}$?

Then the rate ratio $k_2/k_1$ would be $1/3.555$. The math still works out exactly the same, but dealing with fractions inside the logarithm can be slightly more prone to calculation errors.

Q: Why does the rate decrease when temperature decreases?

Lower temperature means fewer molecules have kinetic energy greater than the activation energy, leading to fewer effective collisions per second.

Q: How did we get the value of $\ln(3.555)$?

It is provided in the question! Always check the given data at the end of the problem statement in JEE questions.

Q: Why is the final answer 100 and not 99.98?

The question asks for the nearest integer or a rounded-off value typically in such numerical formats. $99.98$ is extremely close to $100$, making $100$ the intended answer.

Question

The rate of a reaction decreased by $3.555$ times when the temperature was changed from $40^{\circ}	ext{C}$ to $30^{\circ}	ext{C}$. The activation energy (in $	ext{kJ mol}^{-1}$) of the reaction is ......... .
(Take; $R = 8.314 	ext{ J mol}^{-1} 	ext{K}^{-1}$, $\ln 3.555 = 1.268$)

Sigma Percentile · Accepted Answer

### The Power of Temperature in Chemical Kinetics

Imagine a chemical reaction taking place. Every reaction needs a minimum energy to start, which we call **activation energy** ($E_a$). It acts as an energy barrier that reactant molecules must overcome to transform into products.

Now, how does the rate of reaction change with temperature? For this, we rely on the elegant **Arrhenius equation**. This equation beautifully relates the rate constants at two different temperatures, showing us exactly how sensitive a reaction is to thermal changes.

### Setting Up the Arrhenius Equation

The question states that when the temperature was changed from $40^{\circ}	ext{C}$ to $30^{\circ}	ext{C}$, the rate decreased by $3.555$ times. This means if we consider $T_1$ as the lower temperature ($30^{\circ}	ext{C}$) and $T_2$ as the higher temperature ($40^{\circ}	ext{C}$), the ratio of their rate constants $\frac{k_2}{k_1}$ will be exactly $3.555$.

Don't make a silly mistake here! Temperature must always be taken in Kelvin to maintain dimensional consistency with the universal gas constant $R$.

$$T_1 = 30 + 273 = 303 	ext{ K}$$
$$T_2 = 40 + 273 = 313 	ext{ K}$$

### The Master Calculation

Let's substitute the values into the Arrhenius equation:

$$\ln\left(\frac{k_2}{k_1}ight) = \frac{E_a}{R}\left[\frac{1}{T_1} - \frac{1}{T_2}ight]$$

$$\ln(3.555) = \frac{E_a}{8.314}\left[\frac{1}{303} - \frac{1}{313}ight]$$

Now let's solve the bracketed part. Taking the LCM, the numerator will be $313 - 303$, which is $10$.

On the left side, the value of the natural log of $3.555$ is generously given as $1.268$ in the question. Putting this in, we rearrange the equation to isolate $E_a$:

$$1.268 = \frac{E_a}{8.314} 	imes \frac{10}{303 	imes 313}$$

$$E_a = \frac{1.268 	imes 8.314 	imes 303 	imes 313}{10}$$

### The Final Result

On calculating, the activation energy comes out to be $99980.7 	ext{ J mol}^{-1}$. Converting this to kilo Joules by dividing by $1000$, it becomes $99.98 	ext{ kJ mol}^{-1}$, which is approximately **$100 	ext{ kJ mol}^{-1}$**.

Did you get the feel of it? Imagine if we add a catalyst to this reaction; the activation energy barrier would decrease, and the reaction would become even faster at the exact same temperature!

Enter Numerical Value:

Visualized Solution

$E_{a} and Energy Profile$

$Arrhenius Equation$

$Given Data$

$Temperature in Kelvin$

$Substitution$

$Simplifying the Bracket$

$Solving for E_{a}$

$Final Calculation$

$Effect of Catalyst$

The Sigma Insight: Theories of Chemical Reaction

The Power of Temperature in Chemical Kinetics

Setting Up the Arrhenius Equation

The Master Calculation

The Final Result

Similar Questions

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For following reactions : $A 700 K Product$ ; $A 500 K Catalyst Product$ It was found that the $E_{a}$ is decreased by $30 kJ/mol$ in the presence of catalyst. If the rate remains unchanged, the activation energy for catalysed reaction is (Assume pre exponential factor is same):

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

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Enter Numerical Value:

Visualized Solution

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

.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; } Given Data

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

.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; } Simplifying the Bracket

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

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

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