**chapter 6. CHEMICAL KINETICS class 12 chemistry textbook solution**

### 4. Solve

vi. Show that time required for 99.9% completion of a first order reaction is three times the time required for 90% completion.

**For a first-order reaction:**

t = 2.303/k * log_{10}[A]_{0}/[A]_{t}

**i. Time is taken for 99.9% completion:**

Let the time taken for 99.9% completion of the reaction be t_{99.9%}.

Let initial concentration, [A]_{0} = a

The final concentration, [A]_{t} = a - 99.9% of a

= a - (99.9/100 * a) = 0.001 * a

t_{99.9%} = 2.303/k * log_{10}[A]_{0}/[A]_{t}

= 2.303/k * log_{10} a/(0.001 * a)

= 2.303/k * log_{10} 1000 ... (1)

**ii. Time is taken for 90% completion:**

Let the time taken for 90% completion of the reaction be t_{90%}.

Let initial concentration, [A]_{0} = a

Then, final concentration, [A]_{t} = a - 90% of a

= a - (90/100 * a) = 0.1 * a

t_{90%} = 2.303/k * log_{10}[A]_{0}/[A]_{t} = 2.303/k * log_{10} a/(0.1 * a)

= 2.303/k * log_{10} 10 ... (2)

Dividing (1) by (2), we get

(t_{99.9%}/t_{90%}) = (2.303/k * log_{10} 1000)/(2.303/k * log_{10} 10) = (log_{10} 1000)/(log_{10} 10) = 3/1

Therefore, (t_{99.9%}/t_{90%}) = 3

So, t_{99.9%} = 3 * t_{90%}

Therefore, for a first-order reaction, the time required for 99.9% completion is 3 times that required for 90% completion.