A reaction has a half-life of 1 min. The time required for 99.9% completion of the reaction is .....

Q: Why do we assume it's a first-order reaction?

The half-life is given as a constant value ($1 \text{ min}$), which is a unique characteristic of first-order kinetics. For zero or second-order reactions, the half-life depends on the initial concentration.

Q: Why is $C_t$ taken as $0.1$ and not $99.9$?

In the integrated rate law, $C_t$ represents the *remaining* concentration of the reactant at time $t$, not the amount that has already reacted. If $99.9\%$ has reacted, then $100\% - 99.9\% = 0.1\%$ remains.

Q: Can we use $\log_{10}$ instead of $\ln$?

Yes, the formula becomes $t = \frac{2.303}{k} \log_{10}\left(\frac{C_0}{C_t}\right)$. However, since the problem explicitly provides the values for $\ln 2$ and $\ln 10$, it is faster and more direct to stick with natural logarithms.

Q: Is there a shortcut for this specific calculation?

Absolutely! For any first-order reaction, the time required for $99.9\%$ completion is always approximately $10$ times its half-life ($t_{99.9\%} \approx 10 \times t_{1/2}$).

Q: What if the reaction was zero-order?

For a zero-order reaction, the half-life is directly proportional to the initial concentration ($t_{1/2} = \frac{C_0}{2k}$). It wouldn't be a single constant time unless specified for a particular starting concentration.

Question

A reaction has a half-life of 1 min. The time required for 99.9% completion of the reaction is ......... min (Round off to the nearest integer).
[Use : $\ln 2 = 0.69$, $\ln 10 = 2.3$]

Sigma Percentile · Accepted Answer

### The Signature of First-Order Kinetics

Let's visualize the decay of our reactant. We are given that the half-life is exactly $1 	ext{ min}$. This means every minute, the concentration drops to half of its previous value. This constant half-life is a classic signature of a **first-order reaction**.

### The Heartbeat of the Reaction: Rate Constant

Now, the rate constant, $k$, is the heartbeat of this reaction. For a first-order process, $k$ is simply the natural log of two divided by the half-life.

Since the half-life is $1 	ext{ min}$, our rate constant is just:

$$k = \frac{\ln 2}{t_{1/2}} = \frac{\ln 2}{1} = \ln 2 	ext{ min}^{-1}$$

### The Master Equation: Integrated Rate Law

We need to find the time for $99.9\%$ completion. Let's bring in our master tool: the integrated rate law.

If we start with an initial concentration of $C_0 = 100$, then after $99.9\%$ reacts, the remaining concentration $C_t$ is what we care about. Don't make a silly mistake here by putting $99.9$ as the final concentration! The remaining amount is:

$$C_t = 100 - 99.9 = 0.1$$

### The Final Calculation

Let's substitute our values into the integrated rate law equation:

$$t = \frac{1}{k} \ln \left( \frac{C_0}{C_t} ight)$$

We plug in $\ln 2$ for $k$, $100$ for the initial concentration, and $0.1$ for the final concentration. $100$ divided by $0.1$ gives us $1000$ inside the logarithm.

$$t = \frac{1}{\ln 2} \ln \left( \frac{100}{0.1} ight) = \frac{\ln (1000)}{\ln 2}$$

Now for the math. $1000$ is just $10^3$. Using the power rule of logarithms, we can bring that $3$ to the front. So, we have $3$ times the natural log of $10$, all divided by the natural log of $2$.

$$t = \frac{3 \ln 10}{\ln 2}$$

Finally, we use the given values. $\ln 10$ is $2.3$, and $\ln 2$ is $0.69$.

$$t = \frac{3 	imes 2.3}{0.69} = \frac{6.9}{0.69} = 10 	ext{ min}$$

So, it takes exactly $10 	ext{ minutes}$ for the reaction to reach $99.9\%$ completion.

### A Golden Shortcut

Here is a pro-tip for your exams. For any first-order reaction, the time for $99.9\%$ completion is always approximately **10 times its half-life** ($t_{99.9\%} \approx 10 	imes t_{1/2}$). Remembering this shortcut can save you precious seconds in competitive exams!

Animated Solution for Chemistry - Chemical Kinetics: A reaction has a half-life of 1 min. The time required for 99.9% completion of the reaction is ......... min (Round off to the nearest integer). [Use : $ln 2 = 0.69$ , $ln 10 = 2.3$ ]

Enter Numerical Value:

Visualized Solution

$First Order Kinetics$

$Rate Constant k$

$Integrated Rate Law$

$Substitution$

$Calculation$

$Final Answer$

$The Way Forward$

The Sigma Insight: Order and Molecularity

The Signature of First-Order Kinetics

The Heartbeat of the Reaction

The Master Equation

The Final Calculation

A Golden Shortcut

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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; } First Order Kinetics

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

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

.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; } Calculation

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

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