QJEEM24Cha1d7: JEE Main 2024 Chemistry - Class 12 JEEM24Cha1d7 Solution: Time required for completion of a first order reaction is…

To determine the time required for a certain level of completion ( $x\%$ ) of a first-order reaction, we can use the formula that relates the time $t$ , the rate constant $k$ , and the concentration of the reactant. The formula for a first-order reaction, when expressed in terms of the initial concentration $[A]_0$ and the concentration at time $t$ , $[A]_t$ , is given by the integrated rate law for first-order reactions:

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

For a given percentage of completion, we use the fact that $[A]_t = [A]_0(1-\frac{x}{100})$ , where $x$ is the percentage completion. Thus, the equation becomes:

$\ln\left(\frac{[A]_0}{[A]_0(1-\frac{x}{100})}\right) = kt$

Simplifying, we get:

$\ln\left(\frac{1}{1-\frac{x}{100}}\right) = kt$

Let's calculate the time required for $99.9\%$ and $90\%$ :

• For $99.9\%$ completion ( $x = 99.9\%$ ):

$\ln\left(\frac{1}{1-\frac{99.9}{100}}\right) = kt_{99.9}$

$\ln\left(\frac{1}{0.001}\right) = kt_{99.9}$

$\ln(1000) = kt_{99.9}$

• For $90\%$ completion ( $x = 90\%$ ):

$\ln\left(\frac{1}{1-\frac{90}{100}}\right) = kt_{90}$

$\ln\left(\frac{1}{0.1}\right) = kt_{90}$

$\ln(10) = kt_{90}$

Since $k$ is a constant for a given reaction at a constant temperature, we can compare the times $t_{99.9}$ and $t_{90}$ directly by comparing the logarithmic values:

$\frac{t_{99.9}}{t_{90}} = \frac{\ln(1000)}{\ln(10)}$

Using the property of logarithms, we know that $\ln(1000) = 3\ln(10)$ because $1000 = 10^3$ , so:

$\frac{t_{99.9}}{t_{90}} = \frac{3\ln(10)}{\ln(10)} = 3$

Therefore, the time required for $99.9\%$ completion of a first-order reaction is $3$ times the time required for completion of $90\%$ of the reaction. The nearest integer to this value is $3$ .