Gaseous cyclobutene isomerises to butadiene in a first order process which has a 'k' value of 3.3 x

Q: Why do we use $[A]_t = 60$ instead of $40$?

In the integrated rate law, $[A]_t$ represents the concentration of the reactant that is *remaining* at time $t$, not the amount that has reacted. If $40\%$ has reacted, then $100\% - 40\% = 60\%$ is left.

Q: How do I calculate $\ln(100/60)$ without a calculator?

In JEE exams, standard log values like $\log_{10}(2)$ and $\log_{10}(3)$ are usually provided. You can convert natural log to base 10 using $\ln(x) = 2.303 \log_{10}(x)$. So, $\ln(5/3) = 2.303 \times (\log_{10}(5) - \log_{10}(3))$. Remember $\log_{10}(5) = 1 - \log_{10}(2)$.

Q: Why is the unit of $k$ given as $\text{s}^{-1}$?

The unit of the rate constant $k$ depends on the order of the reaction. For a first-order reaction, the rate is proportional to concentration to the power of 1. Therefore, $k$ must have units of $\text{time}^{-1}$ to balance the equation.

Q: What if the question asked for $40\%$ remaining instead of $40\%$ completion?

If the question stated that $40\%$ of the reactant was remaining, then you would directly use $[A]_t = 40$ in your formula.

Q: Why did we round off to $26$?

Our calculated value was $25.79$ minutes. The question explicitly asked to round off to the nearest integer. Since the decimal part $.79$ is greater than $0.5$, we round up to $26$.

Question

Gaseous cyclobutene isomerises to butadiene in a first order process which has a '$k$' value of $3.3 	imes 10^{-4} 	ext{ s}^{-1}$ at $153^\circ	ext{C}$. The time in minutes it takes for the isomerisation to proceed $40\%$ to completion at this temperature is ....... .
(Rounded off to the nearest integer)

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are observing a closed vessel at $153^\circ	ext{C}$ where gaseous cyclobutene is undergoing a structural transformation. It is isomerising into 1,3-butadiene. The problem explicitly states that this is a **first-order process**.

This is a crucial piece of information because it immediately tells us which mathematical tools we need to deploy. We are also given the rate constant for this reaction, $k = 3.3 	imes 10^{-4} 	ext{ s}^{-1}$. Our mission is to find out exactly how many minutes it will take for $40\%$ of the cyclobutene to react and turn into butadiene.

### The Master Equation

For any first-order reaction, the relationship between time, rate constant, and concentration is governed by the integrated rate law:

$$kt = \ln\left(\frac{[A]_0}{[A]_t}ight)$$

Here, $[A]_0$ represents the initial concentration of our reactant (cyclobutene), and $[A]_t$ represents the concentration remaining at time $t$.

To make our calculations straightforward, let's assume the initial concentration $[A]_0$ is $100\%$. The problem states that the reaction proceeds $40\%$ to completion. This means $40\%$ of the cyclobutene has been consumed.

Therefore, the amount remaining, $[A]_t$, is simply $100 - 40 = 60\%$.

### Final Calculation

Now, we substitute these values into our integrated rate law:

$$3.3 	imes 10^{-4} 	imes t = \ln\left(\frac{100}{60}ight)$$

This simplifies to:

$$3.3 	imes 10^{-4} 	imes t = \ln\left(\frac{5}{3}ight)$$

We know that $\ln(5/3) \approx 0.5108$. Now, we isolate $t$:

$$t = \frac{0.5108}{3.3 	imes 10^{-4}} \approx 1547.95 	ext{ s}$$

**Watch out for the trap!** The time we just calculated is in seconds because our rate constant $k$ was given in $	ext{s}^{-1}$. However, the question specifically asks for the time in **minutes**.

To convert seconds to minutes, we divide by $60$:

$$t = \frac{1547.95}{60} \approx 25.79 	ext{ min}$$

Finally, the question asks us to round off to the nearest integer. Since $25.79$ is closer to $26$, our final answer is **$26$ minutes**.

Enter Numerical Value:

Visualized Solution

\text{First Order Isomerisation}

\text{Integrated Rate Law}

\text{Concentration Values}

\text{Substituting Values}

\text{Solving for Time (s)}

\text{Converting to Minutes}

\text{Conclusion}

The Sigma Insight: Rate of Chemical Reaction

Analyzing the Setup

The Master Equation

Final Calculation

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