A first order reaction has a specific reaction rate of $10^{-2} \, sec^{-1}$. How much time will it take for $20\, g$ of the reactant to reduce to $5\, g$ ?

Question

In 2-chloro-3,4-dimethylhexane, how many chiral carbon atoms are present?

Answer 1

To determine how much time it will take for a first-order reaction to decrease the concentration of a reactant, we can use the first-order reaction formula for the half-life calculation. However, we don't need the half-life here directly, but our problem can be solved using the integrated rate law for first-order reactions. The formula is as follows:

k = \frac{\ln(\frac{[A]_0}{[A]})}{t}

Where:

k is the rate constant.
[A]_0 is the initial concentration of the reactant.
[A] is the concentration of the reactant at time t.

Given:

k = 10^{-2} \, \text{sec}^{-1}
[A]_0 = 20 \, g
[A] = 5 \, g

We need to find t.

Substitute the values into the equation:

10^{-2} = \frac{\ln(\frac{20}{5})}{t}

First, calculate the natural logarithm part:

\ln(\frac{20}{5}) = \ln(4)

The value of \ln(4) \approx 1.386.

Now substitute back into the rate equation:

10^{-2} = \frac{1.386}{t}

Rearrange to solve for t:

t = \frac{1.386}{10^{-2}}

Calculate t:

t = 138.6 \, \text{sec}

Therefore, the time taken for the reactant to reduce from 20 g to 5 g is 138.6 seconds.

Answer: 138.6 sec

Answer 2