Question

$t_{1/2}$ of first order reaction is $10$ min. Starting with $10$ $mol{L}^{-1}$, rate after $20$ min is :

• A. $0.0693$ $mol{L}^{-1}{min}^{-1}$
• B. $0.0693\times 2.5$ $mol{L}^{-1}{min}^{-1}$
• C. $0.0693\times 5$ $mol{L}^{-1}{min}^{-1}$
• D. $0.0693\times 10$ $mol{L}^{-1}{min}^{-1}$

Answer 1

The correct option is B $0.0693\times 2.5$ $mol{L}^{-1}{min}^{-1}$

For a first order reaction, we have

Initial concentration $=10mol$ $L^{-1}$

$\therefore$ Concentration after $20$ min (two half lives) $=2.5mol$ $L^{-1}$

Now, $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{10}$ or $0.0693{min}^{-1}$

$\therefore$ Rate$=k\times \left[reactant\right]$ $=0.0693\times 2.5mol$ $L^{-1}{min}^{-1}$

Option (B) is correct.