Question

$T_{50\%}$ 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 $0.0693\times 2.5$ $mol$ $L^{-1}{min}^{-1}$

$t_{1/2}=10mink=\frac{0.693}{t_{1/2}}=\frac{0.693}{10}mi{n}^{-1}$

Now,

$A=A_o{e}^{-kt}A=10.e^{\frac{0.693}{10}.20}$

$=10e^{-2\times 0.693}⟶$ Concentration at $20min.$

So,

$v=kA=\frac{0.693}{10}.{10e}^{-2\times 0.693}=0.0693\times {10e}^{-2\times 0.693}=0.0693\times 2.5mol{L}^{-1}mi{n}^{-1}$