For the first order reaction, A - 2B, 1 mole of reactant A gives 0.2 moles of B after 100 minutes.

Q: Why do we use the moles of A remaining instead of the moles of B formed in the rate law?

The integrated rate law for a first-order reaction specifically relates the concentration of the *reactant* over time. The rate of reaction depends on how much reactant is left to undergo the transformation.

Q: How did we know the reaction was first-order?

The problem explicitly states 'For the first order reaction...' in the very first sentence! Always read the question carefully to identify the kinetic order, as it dictates which equations to use.

Q: Why did we convert the natural logarithm ($\ln$) to base-10 logarithm ($\log_{10}$)?

The problem provided the value for $\ln 10 = 2.3$, which acts as the conversion factor between $\ln$ and $\log_{10}$ (since $\ln x = \ln 10 \times \log_{10} x$). This allows us to use standard base-10 log values like $\log_{10} 3 \approx 0.4771$ to solve the expression without a calculator.

Q: What if I didn't know the exact value of $\log_{10} 3$?

In JEE, standard log values like $\log_{10} 2 \approx 0.3010$ and $\log_{10} 3 \approx 0.4771$ are often expected to be memorized. However, sometimes they are provided in the instructions. Always keep these basic values handy!

Q: Why is the half-life independent of the initial concentration here?

That's a hallmark of first-order kinetics! The time it takes for half of the reactant to decay is constant ($t_{1/2} = \frac{\ln 2}{k}$), regardless of whether you start with 1 mole or 1000 moles.

Question

For the first order reaction, $A \longrightarrow 2B$, 1 mole of reactant $A$ gives 0.2 moles of $B$ after 100 minutes. The half-life of the reaction is ............... min. (Round off to the nearest integer). [Use : $\ln 2 = 0.69$, $\ln 10 = 2.3$; properties of logarithms : $\ln x^y = y \ln x$; $\ln \left(\frac{x}{y}\right) = \ln x - \ln y$]

Sigma Percentile · Accepted Answer

Title: Unlocking the Secrets of First-Order Kinetics: A Journey Through Time and Concentration

### Analyzing the Setup
Imagine you are observing a chemical transformation in real-time. We have a reactant, $A$, which is steadily decomposing to form two moles of product $B$. The reaction is governed by first-order kinetics, meaning the rate at which $A$ disappears depends solely on how much of $A$ is currently present.

We start our stopwatch at $t = 0$ with exactly $1 	ext{ mol}$ of $A$ and no $B$. Fast forward to $t = 100 	ext{ min}$, and we find that $0.2 	ext{ mol}$ of $B$ has been produced. But how much of $A$ is left?

According to the stoichiometry of the reaction $A \longrightarrow 2B$, for every $x$ moles of $A$ that react, $2x$ moles of $B$ are formed. Since $2x = 0.2 	ext{ mol}$, we can easily deduce that $x = 0.1 	ext{ mol}$. Therefore, the amount of $A$ remaining after 100 minutes is simply $1 - 0.1 = 0.9 	ext{ mol}$. This remaining concentration is the key to unlocking the rate constant.

### The Master Equation
For any first-order reaction, the relationship between time, rate constant, and concentration is beautifully captured by the integrated rate law:
$$k = \frac{1}{t} \ln\left(\frac{[A]_0}{[A]_t}ight)$$

Let's substitute our known values into this master equation. We know the time $t = 100 	ext{ min}$, the initial concentration $[A]_0 = 1 	ext{ mol}$, and the final concentration $[A]_t = 0.9 	ext{ mol}$. 
$$k = \frac{1}{100} \ln\left(\frac{1}{0.9}ight) = \frac{1}{100} \ln\left(\frac{10}{9}ight)$$

This looks simple enough, but evaluating $\ln\left(\frac{10}{9}ight)$ requires a bit of mathematical gymnastics using the properties of logarithms.

### The Logarithmic Labyrinth
We are given the value of $\ln 10 = 2.3$. To evaluate $\ln\left(\frac{10}{9}ight)$, we can expand it using the quotient rule for logarithms:
$$\ln\left(\frac{10}{9}ight) = \ln 10 - \ln 9 = \ln 10 - 2 \ln 3$$

But wait, we don't have the value of $\ln 3$! However, we can use the standard base-10 logarithm $\log_{10} 3 \approx 0.47712$. Since $\ln x = \ln 10 	imes \log_{10} x$, we can rewrite our expression as:
$$\ln\left(\frac{10}{9}ight) = 2.3 	imes \log_{10}\left(\frac{10}{9}ight) = 2.3 	imes (1 - 2 \log_{10} 3)$$

Substituting the value of $\log_{10} 3$:
$$\ln\left(\frac{10}{9}ight) = 2.3 	imes (1 - 2(0.47712)) = 2.3 	imes (1 - 0.95424) = 2.3 	imes 0.04576 = 0.105248$$

Now, we can find our rate constant $k$:
$$k = \frac{0.105248}{100} = 0.00105248 	ext{ min}^{-1}$$

### Final Calculation
The ultimate goal is to find the half-life of the reaction, $t_{1/2}$. For a first-order reaction, the half-life is a constant value given by:
$$t_{1/2} = \frac{\ln 2}{k}$$

The problem provides $\ln 2 = 0.69$. Let's plug in our values:
$$t_{1/2} = \frac{0.69}{0.00105248} \approx 655.59 	ext{ min}$$

Rounding this off to the nearest integer, we arrive at our final answer: **656 minutes**.

This problem is a fantastic blend of chemical stoichiometry and mathematical precision. It teaches us that in physical chemistry, keeping track of your decimal places and mastering logarithmic properties is just as important as understanding the chemical concepts themselves!

Enter Numerical Value:

Visualized Solution

\text{Reaction Stoichiometry}

\text{Finding Remaining Reactant}

\text{First-Order Kinetics}

\text{Logarithmic Expansion}

\text{Calculating Rate Constant}

\text{Half-Life Calculation}

The Sigma Insight: Order and Molecularity

Analyzing the Setup Imagine you are observing a chemical transformation in real-time

The Master Equation

The Logarithmic Labyrinth We are given the value of $ln 10 = 2.3$

Final Calculation The ultimate goal is to find the half-life of the reaction, $t_{1/2}$

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